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Chapter 6 Random Variables and the Normal Distribution
61 Discrete Random Variables
62 Binomial Probability Distribution
63 Continuous Random Variables and the Normal Probability Distribution
64 Standard Normal Distribution
65 Applications of the Normal Distribution
61 Discrete Random Variables
Objectives
By the end of this section I will be
able tohellip
1) Identify random variables
2) Explain what a discrete probability distribution is and construct probability distribution tables and graphs
3) Calculate the mean variance and standard deviation of a discrete random variable
Random Variables
A variable whose values are determined by chance
Chance in the definition of a random variable is crucial
Example 62 - Notation for random variables
Suppose our experiment is to toss a single fair
die and we are interested in the number
rolled We define our random variable X to be
the outcome of a single die roll
a Why is the variable X a random variable
b What are the possible values that the
random variable X can take
c What is the notation used for rolling a 5
d Use random variable notation to express the probability of rolling a 5
Example 62 continued
Solution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Types of Random Variables
Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process
Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions
Identify each as a discrete or continuous random variable
(a) Total amount in ounces of soft drinks you consumed in the past year
(b) The number of cans of soft drinks that you consumed in the past year
Example
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
61 Discrete Random Variables
Objectives
By the end of this section I will be
able tohellip
1) Identify random variables
2) Explain what a discrete probability distribution is and construct probability distribution tables and graphs
3) Calculate the mean variance and standard deviation of a discrete random variable
Random Variables
A variable whose values are determined by chance
Chance in the definition of a random variable is crucial
Example 62 - Notation for random variables
Suppose our experiment is to toss a single fair
die and we are interested in the number
rolled We define our random variable X to be
the outcome of a single die roll
a Why is the variable X a random variable
b What are the possible values that the
random variable X can take
c What is the notation used for rolling a 5
d Use random variable notation to express the probability of rolling a 5
Example 62 continued
Solution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Types of Random Variables
Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process
Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions
Identify each as a discrete or continuous random variable
(a) Total amount in ounces of soft drinks you consumed in the past year
(b) The number of cans of soft drinks that you consumed in the past year
Example
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Random Variables
A variable whose values are determined by chance
Chance in the definition of a random variable is crucial
Example 62 - Notation for random variables
Suppose our experiment is to toss a single fair
die and we are interested in the number
rolled We define our random variable X to be
the outcome of a single die roll
a Why is the variable X a random variable
b What are the possible values that the
random variable X can take
c What is the notation used for rolling a 5
d Use random variable notation to express the probability of rolling a 5
Example 62 continued
Solution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Types of Random Variables
Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process
Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions
Identify each as a discrete or continuous random variable
(a) Total amount in ounces of soft drinks you consumed in the past year
(b) The number of cans of soft drinks that you consumed in the past year
Example
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example 62 - Notation for random variables
Suppose our experiment is to toss a single fair
die and we are interested in the number
rolled We define our random variable X to be
the outcome of a single die roll
a Why is the variable X a random variable
b What are the possible values that the
random variable X can take
c What is the notation used for rolling a 5
d Use random variable notation to express the probability of rolling a 5
Example 62 continued
Solution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Types of Random Variables
Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process
Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions
Identify each as a discrete or continuous random variable
(a) Total amount in ounces of soft drinks you consumed in the past year
(b) The number of cans of soft drinks that you consumed in the past year
Example
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example 62 continued
Solution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Types of Random Variables
Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process
Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions
Identify each as a discrete or continuous random variable
(a) Total amount in ounces of soft drinks you consumed in the past year
(b) The number of cans of soft drinks that you consumed in the past year
Example
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Types of Random Variables
Discrete random variable - either a finite number of values or countable number of values where ldquocountablerdquo refers to the fact that there might be infinitely many values but they result from a counting process
Continuous random variable infinitely many values and those values can be associated with measurements on a continuous scale without gaps or interruptions
Identify each as a discrete or continuous random variable
(a) Total amount in ounces of soft drinks you consumed in the past year
(b) The number of cans of soft drinks that you consumed in the past year
Example
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Identify each as a discrete or continuous random variable
(a) Total amount in ounces of soft drinks you consumed in the past year
(b) The number of cans of soft drinks that you consumed in the past year
Example
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(a) continuous
(b) discrete
Example
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Identify each as a discrete or continuous random variable
(a) The number of movies currently playing in US theaters
(b) The running time of a randomly selected movie
(c) The cost of making a randomly selected movie
Example
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(a) discrete
(b) continuous
(c) continuous
Example
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Discrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The probability distribution table of
the number of heads observed when
tossing a fair coin twice
Table 62 in your textbook
Example
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Let the random variable x represent the number of girls in a family of four children Construct a table describing the probability distribution
Example
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Determine the outcomes with a tree diagram
Example
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Total number of outcomes is 16
Total number of ways to have 0 girls is 1
Total number of ways to have 1 girl is 4
Total number of ways to have 2 girls is 6
06250161girls) 0(P
25000164girl) 1(P
3750166girls) 2(P
Example
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Total number of ways to have 3 girls is 4
Total number of ways to have 4 girls is 1
25000164girls) 3(P
06250161girls) 4(P
Example
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Distribution is
NOTE
x P(x)
0 00625
1 02500
2 03750
3 02500
4 00625
1)(xP
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Mean of a Discrete Random Variable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Finding the Mean of a Discrete Random Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Variability of a Discrete Random Variable Formulas for the Variance and Standard
Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
02)(xxP
)(xPx 2x )(2 xPx
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
x P(x)
0 00625 0 0 0
1 02500 025 1 02500
2 03750 075 4 15000
3 02500 075 9 22500
4 00625 025 16 10000
)(xPx 2x )(2 xPx
000010000400005)( 222 xPx
0100001
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Discrete Probability Distribution as a Graph Graphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Page 270
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Probability distribution (table)
Omit graph
x P(x)
0 025
1 035
2 025
3 015
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Page 270
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
ANSWER
(a) Probability X is fewer than 3
850
250350250
)2P()1( )0(
)2 1 0(
XXPXP
XXXP
scored goals ofnumber X
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
ANSWER
(b) The most likely number of goals is the
expected value (or mean) of X
She will most likely score one goal
x P(x)
0 025 0
1 035 035
2 025 050
3 015 045
)(xPx
131)(xxP
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
ANSWER
(c) Probability X is at least one
750
150250350
)3P()2( )1(
)3 2 1(
XXPXP
XXXP
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
62 Binomial Probability Distribution
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
62 Binomial Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Explain what constitutes a binomial experiment
2) Compute probabilities using the binomial probability formula
3) Find probabilities using the binomial tables
4) Calculate and interpret the mean variance and standard deviation of the binomial random variable
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Factorial symbol
For any integer n ge 0 the factorial symbol n is defined as follows
0 = 1
1 = 1
n = n(n - 1)(n - 2) middot middot middot 3 middot 2 middot 1
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Find each of the following
1 4
2 7
Example
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
Example
504012345677 2
2412344 1
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Factorial on Calculator
Calculator
7 MATH PRB 4
which is
Enter gives the result 5040
7
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Combinations
An arrangement of items in which
r items are chosen from n distinct items
repetition of items is not allowed (each item is distinct)
the order of the items is not important
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The number of different possible 5 card
poker hands Verify this is a combination
by checking each of the three properties
Identify r and n
Example of a Combination
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Five cards will be drawn at random from a deck of cards is depicted below
Example
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
An arrangement of items in which
5 cards are chosen from 52 distinct items
repetition of cards is not allowed (each card is distinct)
the order of the cards is not important
Example
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Combination Formula
The number of combinations of r items chosen
from n different items is denoted as nCr and
given by the formula
n r
nC
r n r
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Find the value of
Example
47 C
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
Example
35
57
123
567
)123()1234(
1234567
34
7
)47(4
747 C
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Combinations on Calculator
Calculator
7 MATH PRB 3nCr 4
To get
Then Enter gives 35
47 C
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Determine the number of different
possible 5 card poker hands
Example of a Combination
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
9605982552C
Example
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Motivational Example
Genetics bull In mice an allele A for agouti (gray-brown
grizzled fur) is dominant over the allele a which determines a non-agouti color Suppose each parent has the genotype Aa and 4 offspring are produced What is the probability that exactly 3 of these have agouti fur
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Motivational Example
bull A single offspring has genotypes
Sample Space
A a
A AA Aa
a aA aa
aaaAAaAA
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Motivational Example
bull Agouti genotype is dominant bull Event that offspring is agouti
bull Therefore for any one birth
aAAaAA
43genotype) agouti(P
41genotype) agoutinot (P
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Motivational Example
bull Let G represent the event of an agouti offspring and N represent the event of a non-agouti
bull Exactly three agouti offspring may occur in four different ways (in order of birth)
NGGG GNGG GGNG GGGN
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Motivational Example
bull Consecutive events (birth of a mouse) are independent and using multiplication rule
256
27
4
3
4
3
4
3
4
1
)()()()()( GPGPGPNPGGGNP
256
27
4
3
4
3
4
1
4
3
)()()()()( GPGPNPGPGGNGP
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Motivational Example
256
27
4
3
4
1
4
3
4
3
)()()()()( GPNPGPGPGNGGP
256
27
4
1
4
3
4
3
4
3
)()()()()( NPGPGPGPNGGGP
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Motivational Example
bull P(exactly 3 offspring has agouti fur)
4220256
108
4
1
4
34
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
4
3
4
3
4
3
4
3
4
1
N)birth fourth OR Nbirth thirdOR Nbirth second OR Nbirth first (
3
P
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Experiment
Agouti fur example may be considered a binomial experiment
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Experiment
Four Requirements
1) Each trial of the experiment has only two possible outcomes (success or failure)
2) Fixed number of trials
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Experiment
Agouti fur example may be considered a binomial experiment
1) Each trial of the experiment has only two possible outcomes (success=agouti fur or failure=non-agouti fur)
2) Fixed number of trials (4 births)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (frac34)
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Probability Distribution
When a binomial experiment is performed the set of all of possible numbers of successful outcomes of the experiment together with their associated probabilities makes a binomial probability distribution
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Probability Distribution Formula
For a binomial experiment the probability of observing exactly X successes in n trials where the probability of success for any one trial is p is
where
XnX
Xn ppCXP )1()(
XnX
nCXn
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Probability Distribution Formula
Let q=1-p
XnX
Xn qpCXP )(
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Rationale for the Binomial
Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
The number of outcomes with exactly x successes among n
trials
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Probability Formula
P(x) = bull px bull qn-x n
(n ndash x )x
Number of outcomes with exactly x successes among n
trials
The probability of x successes among n
trials for any one particular order
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Agouti Fur Genotype Example
4220256
108
4
1
64
274
4
1
4
34
4
1
4
3
3)34(
4)(
13
13
XP
fur agouti birth with a ofevent X
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
2ND VARS
Abinompdf(4 75 3)
Enter gives the result 0421875
n p x
Binomial Probability Distribution Formula Calculator
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Distribution Tables
n is the number of trials
X is the number of successes
p is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 284
Example
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
Example
00148
)50()50(15504
)50()50(
)5(
155
5205
520C
XP
heads ofnumber X
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Binomial Mean Variance and Standard Deviation
Mean (or expected value) μ = n p
Variance
Standard deviation
npqpq 2 then 1 Use
)1(2 pnp
npqpnp )1(
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
20 coin tosses
The expected number of heads
Variance and standard deviation
Example
10)500)(20(np
05)500)(500)(20(2 npq
2425
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 284
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (makes the basket or does not make the basket)
2) Fixed number of trials (50)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 584=0584)
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(a)
Example
05490
)4160()5840(
)25(
2525
2550C
XP
baskets ofnumber X
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(b) The expected value of X
The most likely number of baskets is 29
Example
229)5840)(50(np
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(c) In a random sample of 50 of OrsquoNealrsquos shots he is expected to make 292 of them
Example
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 285
Example
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Is this a Binomial Distribution
Four Requirements
1) Each trial of the experiment has only two possible outcomes (contracted AIDS through injected drug use or did not)
2) Fixed number of trials (120)
3) Experimental outcomes are independent of each other
4) Probability of observing a success remains the same from trial to trial (assumed to be 11=011)
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(a) X=number of white males who contracted AIDS through injected drug use
Example
08160
)890()110(
)10(
11010
10120C
XP
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(b) At most 3 men is the same as less than or equal to 3 men
Why do probabilities add
Example
)3()2()1()0()3( XPXPXPXPXP
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Use TI-83+ calculator 2ND VARS to
get Abinompdf(n p X)
Example
= binompdf(120 11 0) + binompdf(120 11 1)
+ binompdf(120 11 2) + binompdf(120 113)
)3()2()1()0( XPXPXPXP
0000554 4E 5553517238
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(c) Most likely number of white males is the expected value of X
Example
213)110)(120(np
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(d) In a random sample of 120 white males with AIDS it is expected that approximately 13 of them will have contracted AIDS by injected drug use
Example
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 286
Example
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(a)
Example
74811)890)(110)(120(2 npq
43374811
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
RECALL Outliers and z Scores
Data values are not unusual if
Otherwise they are moderately
unusual or an outlier (see page 131)
2score-2 z
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Sample Population
Z score Formulas
s
xxz x
z
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Z-score for 20 white males who contracted AIDS through injected drug use
It would not be unusual to find 20 white males who contracted AIDS through injected drug use in a random sample of 120 white males with AIDS
981433
21320z
Example
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
63 Continuous Random Variables and the Normal Probability Distribution
Objectives
By the end of this section I will be
able tohellip
1) Identify a continuous probability distribution
2) Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
(a) Relatively small sample (n = 100) with large class widths (05 lb)
FIGURE 615- Histograms
(b) Large sample (n = 200) with smaller class widths (02 lb)
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Figure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Continuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Requirements for Probability Distribution of a Continuous Random Variable
1) The total area under the density curve
must equal 1 (this is the Law of Total
Probability for Continuous Random
Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Probability for a Continuous Random Variable
Probability for Continuous Distributions
is represented by area under the curve
above an interval
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The Normal Probability Distribution
Most important probability distribution in the world
Population is said to be normally distributed the data values follow a normal probability distribution
population mean is μ
population standard deviation is σ
μ and σ are parameters of the normal distribution
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
FIGURE 619
The normal distribution is symmetric about its mean μ (bell-shaped)
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Properties of the Normal Density Curve (Normal Curve)
1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Properties of the Normal Density Curve (Normal Curve) continued
5) Symmetry also implies that the area under the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The Empirical Rule
For data sets having a distribution that is
approximately bell shaped the following
properties apply
About 68 of all values fall within 1
standard deviation of the mean
About 95 of all values fall within 2
standard deviations of the mean
About 997 of all values fall within 3
standard deviations of the mean
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
FIGURE 623 The Empirical Rule
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Drawing a Graph to Solve Normal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Steps in Drawing a Graph to Help You Solve Normal Probability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the
normal curve This part will depend on what values of X the problem is asking about
3) Proceed to find the desired area or probability using the empirical rule
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 295
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
Example
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 296
Example
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
Example
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 296
Example
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
Example
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
64 Standard Normal Distribution
Objectives
By the end of this section I will be
able tohellip
1) Find areas under the standard normal curve given a Z-value
2) Find the standard normal Z-value given an area
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The Standard Normal (Z) Distribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
When the normal curve is plotted we plot Z on the horizontal axis
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The Empirical Rule For Standard Normal Distribution
0 -3 -2 -1 1 2 3
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Find the area that lies under the standard normal curve between Z=-1 and Z=2
Example
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
034+034+0135=0815
Example
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Find the area that lies under the standard normal curve that lies above Z=3
Example
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
Since 100-997=03=0003
we take half of 0003 to get
00015
Example
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The Standard Normal (Z) Distribution
In fact we can use a table look-up method to compute the area under the curve for a standard normal distribution given any interval for Z
Table C in Appendix on pages T-9 and T-10
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
1 It is designed only for the standard normal distribution which has a mean of 0 and a standard deviation of 1
2 It is on two pages with one page for negative z-scores and the other page for positive z-scores
3 Each value in the body of the table is a cumulative area from the left up to a vertical boundary above a specific z-score
Table C Standard Normal Distribution
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Table continues to -00 in the appendix
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Table continues to 34 in the appendix
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
4 When working with a graph avoid confusion
between z-scores and areas
z Score
Distance along horizontal scale of the standard
normal distribution refer to the leftmost
column and top row of Table C
Area
Region under the curve refer to the values in
the body of Table C
5 The part of the z-score denoting hundredths is
found across the top
Table C Standard Normal Distribution
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Steps for finding areas under the standard normal curve
Table 67
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 307
Do problems 18 and 20
Example
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(18)
Example
Use Table C for Z=200
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
(20)
Example
Use Table C for Z=0500
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 307
Do problems 26 and 28
Example
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
26(c)
Use Table C to find that the area to the left of Z=212 is 09830 We want the area to the right of Z=212 which will be
1-09830=0017
Example
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
28(c)
Use Table C to find that the area to the left of Z=-069 is 02451 We want the area to the right of Z=-069 which will be
1-02451=07549
Example
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 308
Do problems 32 and 36
Example
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 308
Do problems 32 and 38
Example
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
32(c)
Use Table C to find that the area to the left of Z=128 is 08997
Use Table C to find that the area to the left of Z=196 is 09750
We want the area to the between
Z=128 and Z=196 which will be
09750-08997=00753
Example
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
ANSWER
38(c)
Use Table C to find that the area to the left of Z=-201 is 00222
Use Table C to find that the area to the left of Z=237 is 09911
We want the area to the between
Z=-201 and Z=237 which will be
09911-00222=09689
Example
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 308
Example
Do problem 42
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 308
Example
Do problem 52
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 308
Example
Do problem 58
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example
Note this answer is the Z value in the
table that corresponds to
1 - 05120 = 04880
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
65 Applications of the Normal
Distribution
Objectives
By the end of this section I will be
able tohellip
1) Compute probabilities for a given value of any normal random variable
2) Find the appropriate value of any normal random variable given an area or probability
3) Calculate binomial probabilities using the normal approximation to the binomial distribution (OMIT)
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Standardizing a Normal Random Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Finding Probabilities for Any Normal Distribution Step 1
Determine the random variable X the mean μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2 Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Finding Probabilities for Any Normal Distribution Step 3
Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X
Step4
Find the area under the standard normal curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 323
Example
Do problems 681012
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solutions
6) 01587
8) 00062
10) 08413
12) 08400
Example
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Calculator Use
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Calculator Use
Do problem 12 again with calculator
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Calculator Use
Do problems 68 again with calculator
Or use 10^99 as larger value and -10^99
as the smaller value
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Finding Normal Data Values for Specified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Finding Normal Data Values for Specified Probabilities
Step 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Finding Normal Data Values for Specified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 323
Example
Do problems 182022
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solutions
18) X=572
20) X=5355 and X=8645
22) X=442 and X=958
Example
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Calculator Use
Do problem 18 and 22 again with calculator
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Calculator Use
Do problem 18 and 22 again with calculator
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example 638 - Finding the X-values that mark the Boundaries of the middle 95 of X-values
Edmundscom reported that the average
amount that people were paying for a 2007
Toyota Camry XLE was $23400 Let X = price
and assume that price follows a normal
distribution with μ = $23400 and σ = $1000
Find the prices that separate the middle
95 of 2007 Toyota Camry XLE prices from
the bottom 25 and the top 25
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example 638 continued
Solution
Step 1
Determine X μ and σ and draw the normal curve for X
Let X = price μ = $23400 and σ = $1000
The middle 95 of prices are delimited by X1 and X2 as shown in Figure 649
FIGURE 649
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Example 638 continued
Solution
Step 3
Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440
X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360
The prices that separate the middle 95 of 2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 324
Example
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solution
Let X be the random variable that represents how many points McGrady scores in a randomly chosen game Find
Example
)30(XP
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solution
Example
04013
598701
)30(1)30( XPXP
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 324
Example
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solution
Example
14640)2010( XP
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 324
Example
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solution
Find so that
that is the 5th percentile (only 5 of his games did he have a lower score)
Example
050)( 1XXP
1X
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solution
Gives the value
Example
050)( 1XXP
points 84141X
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Page 324
Example
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Solution
The z-score for X=40 points is 150 Since this z-score is more than -2 and less than 2 this is not unusual
Example
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
Summary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ