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8/3/2019 Statistics- Random Variables Working With Uncertain Numbers
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Chapter 7
Random Variables: Working with
Uncertain Numbers
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Random Variable
A specification or description of a numerical result
from a random experiment
The number is the observation of the random variable
The random variable is the meaningof the number
The observed value is 17 for the random variable
Last weeks warranty returns (number of customers)
Examples
Todays stock market close
The number of defective parts produced today Next quarters sales
Summaries: Q = mean (expected value)
W = standard deviation
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Examples ofRandom Variables
Random Standard
Variable Mean Deviation
X= $1.40 $1.40 $0
Y= $1.50 $0.50
Z= $1.90 $13.30
$1 prob 0.5
$2 prob 0.5{$0 prob 0.98$95 prob 0.02{
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Discrete Random Variable
Can list all possible outcomes
Probability Distribution
The list of values and probabilities. Use it to compute!
Mean (expected value) of random variableXQ =Sum ofValue v Prob ofValue =E(X) =XP(X)
Gives a typical or central value of the random variable
Standard deviation of random variableX
W = Sum of (ValueQ)2 v Probability ofValue
=
Tells about how far from expected this random variable will be
Q )(P)( 2 XX
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Example: Investment Payoffs
Probability distribution of investment payoffs
Q =0 v 0.98 + 95 v 0.02 = $1.90
The expected payoff is $1.90
A compromise between $0 (most of the time) and $95 (rarely)
W = (01.90)2
v 0.98 + (951.90)2
v 0.02 = $13.30 Actual payoffs approximately $13.30 above or below expected
A compromise between being $1.90below average (most of
the time) and $93.10 above average (rarely)
Payoff (Value) Probability$0 0.98
$95 0.02
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Example: Profit Scenarios
0.0
0.1
0.2
0.3
0.4
0.5
-5 0 5 10
Probability
Profit ($millions)
Lousy
OK
Good
Great
Expected profit: $3.65 millionStandard deviation:
$4.40 million
Fig 7.1.1
Scenario Profit ($millions) Probability
0.20
0.40
0.25
0.15
$10
5
1
4
Great
Good
OK
Lousy
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Binomial Distribution
A special type of discrete random variable e.g., Interview 50 random customers
How many like the new product? 0, 1, 2, , 49, or50
e.g., What percent of stocks went up yesterday?
Xis binomial if it is the number of occurrences ofsome event, out ofn trials, provided that
The probability T is the same for all trials, and The trials are independent of one another
so that each trial brings new, independent information
Binomial proportion or percent: p =X/n This is the relative frequency of the event
e.g., ifX=35 ofn =50people like product, p = 35/50 = 0.70
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Binomial Probabilities
A discrete distribution
Skewed ifT is close to 0
Symmetric ifT is 0.5 (or close)
Approximately normal
Skewed ifT is close to 1
0
0.1
0.2
0.3
0 5 10 15 20 25
0
0.1
0.2
0.3
0 5 10 15 20 25
T=0.05, n=25
T=0.5, n=25
T=0.9, n=25
0
0.2
0.4
0 5 10 15 20 25
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Binomial Mean and Std. Deviation
Shortcut to find mean and standard deviation
quickly for binomial random variables (Xorp)
No need to compute the probability distribution
work directly from n and T
Mean
Std. Dev.
Proportion or
Percent, p=X/n
Qp
= T
Wp= T(1T)/n
Number of
Occurrences,X
QX= nT
WX= nT(1T)
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Example: Sampling Defective Parts
Draw a random sample ofn=200 from the days production
Suppose that T =7% of this production is defective
Number of
Defects,X
QX= nT= 200v0.07 = 14
Expect to see 14
defects, on average
WX
= nT(1T)
= 200v0.07v0.93
= 3.61
Typically expect approx.
3.61 more or less than 14
Mean
Std. Dev.
Proportion orPercent
Defective, p=X/n
Qp = T = 0.07
Expect to see 7%
defective, on average
Wp
= T(1T)/n
= 0.07v0.93/200
= 0.018
Typically expect approx.
1.8% more or less than 7%
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Computing Binomial Probabilities
Binomial probability thatXequals a
Example: probability that exactly a= 2 of yourn= 6
major customers will call tomorrow (assuming that
T=0.25 is the probability that each one will call)
ana
ana
ana
ana
nana
n
a
naX
TTvvvvvvvv
vvvv!
TT
!
TT
!!
)1()](...321][...321[
...321
)1(
)!(!
!
)1()(P
297.0316406.00625.015
)25.01(25.02
6)2(P 262
!vv!
!!
X
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Normal Distribution
A special continuous distribution (notdiscrete)
For every mean Q and (positive) standard
deviation W there is a normal distribution The mean Q moves the curve left and right
The standard deviation W widens and narrows the curve
Q
WW
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Probability: Area Under the Curve
Probability of observing a value between a and b
is area under the curve
Note: total area = 1Probability = 0.50
Probability = 0.95
(two std. devs.)
Probability = 0.68
(one std. dev.)
a b a b
More likely
Less likely
WW
Q
WW
Q
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Standard Normal Distribution
Normal with mean Q = 0 and std. deviation W = 1
Standard normal probability table
Gives probability that a standard normal is less than a
given valueExamples
-3 -2 -1 0 1 2 3
Value =0.5
Probability
=0.3085
-3 -2 -1 0 1 2 3
Value1
0
1
2
Probability0.1587
0.5
0.8413
0.9772
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Finding Normal Probabilities
Example: Sales are forecast as $80 million (mean) with a standarddeviation of$10 million. Find the probability that sales will exceed
$86 million, assuming a normal distribution
Figure out the question
Find Prob(X>
86) where Q=
80 and W=
10 Standardize (Subtract Q, divide by W to get std. normal)
Prob
= Prob
Draw Picture
Use tables, find answer: 10.7257 = 0.274
-3 -2 -1 0 1 2 3
0.60
XQ
W
8680
10>( )
Standardnormal >0.60( )
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Normal Approx. to the Binomial
Idea: to make it easierto compute binomialprobabilities
If
n is large, and
T is not too close to 0 or1,
Then
Binomial probabilities forXare close to normal
probabilities with Q =Q
X= nT W =W
X= nT(1T)
Similarly forp=X/n with
Q =Qp= T W =W
p= T(1T)/n
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Example: an Opinion Poll
Suppose T= 55% approve presidents performance We interview n=400 chosen at random
What are the chances that fewer than 200 will say they
approve? That is, find Prob (X< 200)
When you round any number less than 199.5 to the nearest
whole number, the result will be fewer than 200
The mean is QX= nT =400v0.55 =220
The standard deviation is WX= nT(1T) =9.9499
We need to compute the probability that a normalrandom
variable with this mean and this standard deviation is less than
199.5
Prob =Prob = 0.020
XQX
W
199.5220
9.9499
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Example: Poll (continued)
Still assuming 55% approve and we interview 400at random
What are the chances that more than 58% will say they
approve?
Convert from percentage p to number of peopleX
Note that 58% of400people is 0.58v400 = 232people
Numbers that round to more than 232 are 232.5 and above
We need to compute the probability that a normalrandom
variable with mean QX=220 and standard deviation W
X=
9.9499 is more than 232.5
Prob =Prob = 0.10
About 10% of the time we will find more than 58%
XQX
W
232.5220
9.9499>( ) Standard
normal>1.26( )