Probability The definition – probability of an Event Applies only to the special case when 1.The sample space has a finite no.of outcomes, and 2.Each

Probability

The definition – probability of an Event

no. of outcomes in =

total no. of outcomes

n E n E EP E

n S N

Applies only to the special case when

1. The sample space has a finite no.of outcomes, and

2. Each outcome is equi-probable

If this is not true a more general definition of probability is required.

Summary of the Rules of Probability

The additive rule

P[A B] = P[A] + P[B] – P[A B]

and

if P[A B] = P[A B] = P[A] + P[B]

The Rule for complements

for any event E

1P E P E

P A BP A B

P B

Conditional probability

if 0

if 0

P A P B A P AP A B

P B P A B P B

The multiplicative rule of probability

and

P A B P A P B

if A and B are independent.

This is the definition of independent

Counting techniques

Summary of counting results

Rule 1

n(A1 A2 A3 …. ) = n(A1) + n(A2) + n(A3) + …

if the sets A1, A2, A3, … are pairwise mutually exclusive

(i.e. Ai Aj = )

Rule 2

n1 = the number of ways the first operation can be performed

n2 = the number of ways the second operation can be performed once the first operation has been completed.

N = n1 n2 = the number of ways that two operations can be performed in sequence if

Rule 3

n1 = the number of ways the first operation can be performed

ni = the number of ways the ith operation can be performed once the first (i - 1) operations have been completed. i = 2, 3, … , k

N = n1n2 … nk

= the number of ways the k operations can be performed in sequence if

Basic counting formulae

!

!n k

nP

n k

1. Orderings

! the number of ways you can order objectsn n

2. Permutations

The number of ways that you can choose k objects from n in a specific order

!

! !n k

n nC

k k n k

3. Combinations

The number of ways that you can choose k objects from n (order of selection irrelevant)

Applications to some counting problems

• The trick is to use the basic counting formulae together with the Rules

• We will illustrate this with examples• Counting problems are not easy. The more practice

better the techniques

Random Variables

Numerical Quantities whose values are determine by the outcome of a random

experiment

Random variables are either• Discrete

– Integer valued – The set of possible values for X are integers

• Continuous– The set of possible values for X are all real

numbers – Range over a continuum.

Examples• Discrete

– A die is rolled and X = number of spots showing on the upper face.

– Two dice are rolled and X = Total number of spots showing on the two upper faces.

– A coin is tossed n = 100 times and X = number of times the coin toss resulted in a head.

– We observe X, the number of hurricanes in the Carribean from April 1 to September 30 for a given year

Examples• Continuous

– A person is selected at random from a population and X = weight of that individual.

– A patient who has received who has revieved a kidney transplant is measured for his serum creatinine level, X, 7 days after transplant.

– A sample of n = 100 individuals are selected at random from a population (i.e. all samples of n = 100 have the same probability of being selected) . X = the average weight of the 100 individuals.

The Probability distribution of A random variable

A Mathematical description of the possible values of the random variable together with

the probabilities of those values

The probability distribution of a discrete random variable is describe by its :

probability function p(x).

p(x) = the probability that X takes on the value x.

This can be given in either a tabular form or in the form of an equation.

It can also be displayed in a graph.

Example 1• Discrete

– A die is rolled and X = number of spots showing on the upper face.

x 1 2 3 4 5 6

p(x) 1/6 1/6 1/6 1/6 1/6 1/6

formula

– p(x) = 1/6 if x = 1, 2, 3, 4, 5, 6

Graphs

To plot a graph of p(x), draw bars of height p(x) above each value of x.

Rolling a die

0

1 2 3 4 5 6

Example 2

– Two dice are rolled and X = Total number of spots showing on the two upper faces.

x 2 3 4 5 6 7 8 9 10 11 12

p(x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36

12,3,4,5,6

36( )13

7,8,9,10,11,1226

xx

p xx

x

Formula:

Rolling two dice

0

36 possible outcome for rolling two dice

Comments:Every probability function must satisfy:

1)(0 xp

1. The probability assigned to each value of the random variable must be between 0 and 1, inclusive:

x

xp

1)(

2. The sum of the probabilities assigned to all the values of the random variable must equal 1:

b

ax

xpbXaP )(3.

)()1()( bpapap

x 0 1 2 3

p(x) 6/14 4/14 3/14 1/14

ExampleIn baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below:

P X( )the random variable equals 2 p ( ) 23

14

Note: This chart implies the only values x takes on are 0, 1, 2, and 3.

If the random variable X is observed repeatedly the probabilities, p(x), represents the proportion times the value x appears in that sequence.

2least at is variablerandom the XP 32 pp 14

4

14

1

14

3

A Bar Graph

No. of persons on base when a home run is hit

0.429

0.286

0.214

0.071

0.000

0.100

0.200

0.300

0.400

0.500

0 1 2 3

# on base

p(x)

Discrete Random Variables

Discrete Random Variable: A random variable usually assuming an integer value.

• a discrete random variable assumes values that are isolated points along the real line. That is neighbouring values are not “possible values” for a discrete random variable

Note: Usually associated with counting

• The number of times a head occurs in 10 tosses of a coin

• The number of auto accidents occurring on a weekend

• The size of a family

Continuous Random Variables

Continuous Random Variable: A quantitative random variable that can vary over a continuum

• A continuous random variable can assume any value along a line interval, including every possible value between any two points on the line

Note: Usually associated with a measurement

• Blood Pressure

• Weight gain

• Height

Probability Distributionsof Continuous Random Variables

Probability Density Function

The probability distribution of a continuous random variable is describe by probability density curve f(x).

Notes:

The Total Area under the probability density curve is 1. The Area under the probability density curve is from a to

b is P[a < X < b].

xa b

P a x b( )

Normal Probability Distributions(Bell shaped curve)

Mean and Variance (standard deviation) of a

Discrete Probability Distribution

• Describe the center and spread of a probability distribution

• The mean (denoted by greek letter (mu)), measures the centre of the distribution.

• The variance (2) and the standard deviation () measure the spread of the distribution.

is the greek letter for s.

Mean of a Discrete Random Variable

• The mean, , of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together:

Notes: The mean is a weighted average of the values of X.

x

xxp

kk xpxxpxxpx 2211

The mean is the long-run average value of the random variable.

The mean is centre of gravity of the probability distribution of the random variable

-

0.1

0.2

0.3

1 2 3 4 5 6 7 8 9 10 11

2

Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x )2, by its own probability and then adding all the products together:

Standard Deviation of a Discrete Random Variable: The positive square root of the variance:

x

xpx 22

2

2

xx

xxpxpx

22 x

xpx

Example

The number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.

x p (x ) xp(x) x 2 x 2 p(x)

0 0.429 0.000 0 0.0001 0.286 0.286 1 0.2862 0.214 0.429 4 0.8573 0.071 0.214 9 0.643

Total 1.000 0.929 1.786

)(xp )(xxp )(2 xpx

• Computing the mean:

Note:

• 0.929 is the long-run average value of the random variable

• 0.929 is the centre of gravity value of the probability distribution of the random variable

929.0x

xxp

• Computing the variance:

x

xpx 22

2

2

xx

xxpxpx

923.0929.786.1 2

• Computing the standard deviation:

2

961.0923.0

Random Variables


experiment

Random variables are either• Discrete

– Integer valued – The set of possible values for X are integers

• Continuous– The set of possible values for X are all real

numbers – Range over a continuum.

The Probability distribution of A random variable






This can be given in either a tabular form or in the form of an equation.

It can also be displayed in a graph.

x 0 1 2 3

p(x) 6/14 4/14 3/14 1/14

ExampleIn baseball the number of individuals, X, on base when a home run is hit ranges in value from 0 to 3. The probability distribution is known and is given below:

P X( )the random variable equals 2 p ( ) 23

14

Note: This chart implies the only values x takes on are 0, 1, 2, and 3.

If the random variable X is observed repeatedly the probabilities, p(x), represents the proportion times the value x appears in that sequence.

2least at is variablerandom the XP 32 pp 14

4

14

1

14

3

A Bar Graph

No. of persons on base when a home run is hit

0.429

0.286

0.214

0.071

0.000

0.100

0.200

0.300

0.400

0.500

0 1 2 3

# on base

p(x)




Notes:


b is P[a < X < b].

Mean, Variance and standard deviation of Random Variables

Numerical descriptors of the distribution of a Random Variable


• The mean, , of a discrete random variable x is found by multiplying each possible value of x by its own probability and then adding all the products together:


x

xxp

kk xpxxpxxpx 2211



-

0.1

0.2

0.3

1 2 3 4 5 6 7 8 9 10 11

2

Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x is found by multiplying each possible value of the squared deviation from the mean, (x )2, by its own probability and then adding all the products together:


x

xpx 22

2

2

xx

xxpxpx

22 x

xpx

Example

The number of individuals, X, on base when a home run is hit ranges in value from 0 to 3.

x p (x ) xp(x) x 2 x 2 p(x)

0 0.429 0.000 0 0.0001 0.286 0.286 1 0.2862 0.214 0.429 4 0.8573 0.071 0.214 9 0.643

Total 1.000 0.929 1.786

)(xp )(xxp )(2 xpx


Note:

• 0.929 is the long-run average value of the random variable

• 0.929 is the centre of gravity value of the probability distribution of the random variable

929.0x

xxp


x

xpx 22

2

2

xx

xxpxpx

923.0929.786.1 2


2

961.0923.0

The Binomial distribution

An important discrete distribution

Situation - in which the binomial distribution arises

• We have a random experiment that has two outcomes– Success (S) and failure (F)– p = P[S], q = 1 - p = P[F],

• The random experiment is repeated n times independently

• X = the number of times S occurs in the n repititions• Then X has a binomial distribution

Example

• A coin is tosses n = 20 times

– X = the number of heads

– Success (S) = {head}, failure (F) = {tail

– p = P[S] = 0.50, q = 1 - p = P[F]= 0.50

• An eye operation has %85 chance of success. It is performed n =100 times

– X = the number of Sucesses (S)

– p = P[S] = 0.85, q = 1 - p = P[F]= 0.15

• In a large population %30 support the death penalty. A sample n =50 indiviuals are selected at random

– X = the number who support the death penalty (S)

– p = P[S] = 0.30, q = 1 - p = P[F]= 0.70

The Binomial distribution1. We have an experiment with two outcomes –

Success(S) and Failure(F).

2. Let p denote the probability of S (Success).

3. In this case q=1-p denotes the probability of Failure(F).

4. This experiment is repeated n times independently.

5. X denote the number of successes occuring in the n repititions.

The possible values of X are

0, 1, 2, 3, 4, … , (n – 2), (n – 1), n

and p(x) for any of the above values of x is given by:

xnxxnx qpx

npp

x

nxp

1

X is said to have the Binomial distribution with parameters n and p.

Summary:


1. X is the number of successes occurring in the n repetitions of a Success-Failure Experiment.

2. The probability of success is p.

3. The probability function

xnx ppx

nxp

1

Example:

1. A coin is tossed n = 5 times. X is the number of heads occurring in the 5 tosses of the coin. In this case p = ½ and

3215

215

21

21

555

xxxxp xx

x 0 1 2 3 4 5

p(x)321

325

325

321

3210

3210

Note:

5 5!

! 5 !x x x

5 5!

10 0! 5 0 !

5 5! 5!

51 1! 5 1 ! 4!

5 5 45!10

2 2!3! 2 1

5 5 45!10

3 3!2! 2 1

5 5!5

4 4!1!

5 5!1

5 0!5!

0.0

0.1

0.2

0.3

0.4

1 2 3 4 5 6

number of heads

p(x

)

Computing the summary parameters for the distribution – , 2,

x p (x ) xp(x) x 2 x 2 p(x)

0 0.03125 0.000 0 0.0001 0.15625 0.156 1 0.1562 0.31250 0.625 4 1.2503 0.31250 0.938 9 2.8134 0.15625 0.625 16 2.5005 0.03125 0.156 25 0.781

Total 1.000 2.500 7.500

)(xp )(xxp )(2 xpx


5.2x

xxp


x

xpx 22

2

2

xx

xxpxpx

25.15.25.7 2


2

118.125.1

Example:• A surgeon performs a difficult operation n =

10 times.

• X is the number of times that the operation is a success.

• The success rate for the operation is 80%. In this case p = 0.80 and

• X has a Binomial distribution with n = 10 and p = 0.80.

xx

xxp

1020.080.0

10

x 0 1 2 3 4 5p (x ) 0.0000 0.0000 0.0001 0.0008 0.0055 0.0264

x 6 7 8 9 10p (x ) 0.0881 0.2013 0.3020 0.2684 0.1074

Computing p(x) for x = 0, 1, 2, 3, … , 10

The Graph

-

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8 9 10

Number of successes, x

p(x

)

Computing the summary parameters for the distribution – , 2,

)(xxp )(2 xpx

x p (x ) xp(x) x 2 x 2 p(x)

0 0.0000 0.000 0 0.0001 0.0000 0.000 1 0.0002 0.0001 0.000 4 0.0003 0.0008 0.002 9 0.0074 0.0055 0.022 16 0.0885 0.0264 0.132 25 0.6616 0.0881 0.528 36 3.1717 0.2013 1.409 49 9.8658 0.3020 2.416 64 19.3279 0.2684 2.416 81 21.743

10 0.1074 1.074 100 10.737Total 1.000 8.000 65.600


0.8x

xxp


x

xpx 22

2

2

xx

xxpxpx

60.10.86.65 2


2 118.125.1

Notes The value of many binomial probabilities are found in Tables posted

on the Stats 245 site.

The value that is tabulated for n = 1, 2, 3, …,20; 25 and various values of p is:

Hence

c

x

c

x

xx xpppx

ncXP

00

101

cpppp 210

1for valueTabledfor valueTabled cccp The other table, tabulates p(x). Thus when using this

table you will have to sum up the values

Example Suppose n = 8 and p = 0.70 and we want to

compute P[X = 5] = p(5)

Table value for n = 8, p = 0.70 and c =5 is 0.448 = P[X ≤ 5]

P[X = 5] = p(5) = P[X ≤ 5] - P[X ≤ 4] = 0.448 – 0.194 = .254

c 0.05 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 0.95

n = 5 0 0.663 0.430 0.168 0.058 0.017 0.004 0.001 0.000 0.000 0.000 0.000

1 0.943 0.813 0.503 0.255 0.106 0.035 0.009 0.001 0.000 0.000 0.000

2 0.994 0.962 0.797 0.552 0.315 0.145 0.050 0.011 0.001 0.000 0.000

3 1.000 0.995 0.944 0.806 0.594 0.363 0.174 0.058 0.010 0.000 0.000

4 1.000 1.000 0.990 0.942 0.826 0.637 0.406 0.194 0.056 0.005 0.000

5 1.000 1.000 0.999 0.989 0.950 0.855 0.685 0.448 0.203 0.038 0.006

6 1.000 1.000 1.000 0.999 0.991 0.965 0.894 0.745 0.497 0.187 0.057

7 1.000 1.000 1.000 1.000 0.999 0.996 0.983 0.942 0.832 0.570 0.337

8 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

We can also compute Binomial probabilities using Excel

=BINOMDIST(x, n, p, FALSE)

The function

will compute p(x).

=BINOMDIST(c, n, p, TRUE)

The function

will compute

c

x

c

x

xx xpppx

ncXP

00

101

cpppp 210

Mean, Variance and standard deviation of

Binomial Random Variables


• The mean, , of a discrete random variable x


x

xxp

kk xpxxpxxpx 2211



2

Variance and Standard DeviationVariance of a Discrete Random Variable: Variance, 2, of a discrete random variable x


x

xpx 22

2

2

xx

xxpxpx

22 x

xpx

The Binomial ditribution





xnx ppx

nxp

1

Mean,Variance & Standard Deviation of the Binomial

Ditribution• The mean, variance and standard deviation of the

binomial distribution can be found by using the following three formulas:

np 1.

pnpnpq 1 2. 2

pnpnpq 1 3.

Solutions:

1) n = 20, p = 0.75, q = 1 - 0.75 = 0.25

np ( )(0. )20 75 15

npq ( )(0. )(0. ) . .20 75 25 375 1936

Example:Find the mean and standard deviation of the binomial distribution when n = 20 and p = 0.75

p xx

xx x( ) (0. ) (0. )

2075 25 20 for 0, 1, 2, ... , 20

2)These values can also be calculated using the probability function:

Table of probabilitiesx p (x ) xp(x) x 2 x 2 p(x)

0 0.0000 0.000 0 0.0001 0.0000 0.000 1 0.0002 0.0000 0.000 4 0.0003 0.0000 0.000 9 0.0004 0.0000 0.000 16 0.0005 0.0000 0.000 25 0.0006 0.0000 0.000 36 0.0017 0.0002 0.001 49 0.0088 0.0008 0.006 64 0.0489 0.0030 0.027 81 0.24410 0.0099 0.099 100 0.99211 0.0271 0.298 121 3.27412 0.0609 0.731 144 8.76813 0.1124 1.461 169 18.99714 0.1686 2.361 196 33.04715 0.2023 3.035 225 45.52516 0.1897 3.035 256 48.55917 0.1339 2.276 289 38.69618 0.0669 1.205 324 21.69119 0.0211 0.402 361 7.63220 0.0032 0.063 400 1.268

Total 1.000 15.000 228.750


0.15x

xxp


x

xpx 22

2

2

xx

xxpxpx

75.30.1575.228 2


2 936.175.3

Histogram

-0.1

0.1

0.2

0.3

no. of successes

p(x)




Notes:


b is P[a < X < b].

xa b

P a x b( )

Normal Probability Distributions

Normal Probability Distributions

• The normal probability distribution is the most important distribution in all of statistics

• Many continuous random variables have normal or approximately normal distributions

2 3 2 3

The Normal Probability Distribution

Points of Inflection

Main characteristics of the Normal Distribution

• Bell Shaped, symmetric

• Points of inflection on the bell shaped curve are at – and + That is one standard deviation from the mean

• Area under the bell shaped curve between – and + is approximately 2/3.

• Area under the bell shaped curve between – 2 and + 2is approximately 95%.

There are many Normal distributions

depending on by and

0

0.01

0.02

0.03

0 50 100 150 200

x

f(x)

0

0.01

0.02

0.03

0 50 100 150 200

x

f(x)

0

0.01

0.02

0.03

0 50 100 150 200

x

f(x)

Normal = 100, = 40 Normal = 140, =20

Normal = 100, =20

The Standard Normal Distribution = 0, = 1

0

0.1

0.2

0.3

0.4

-3 -2 -1 0 1 2 3

• There are infinitely many normal probability distributions (differing in and )

• Area under the Normal distribution with mean and standard deviation can be converted to area under the standard normal distribution

• If X has a Normal distribution with mean and standard deviation than

has a standard normal distribution.

• z is called the standard score (z-score) of X.

X

z

Converting Area

under the Normal distribution with mean and standard deviation

to

Area under the standard normal distribution

Perform the z-transformation

then

Area under the Normal distribution with mean and standard deviation

X

z

P a X b

a X bP

a bP z


P a X b

Area under the Normal distribution with mean and standard deviation

a b

a bP z


a b

0

1

Using the tables for the Standard Normal distribution

Table, Posted on stats 245 web site

• The table contains the area under the standard normal curve between -∞ and a specific value of z

0 z0 z

Example

Find the area under the standard normal curve between z = -∞ and z = 1.45

9265.0)45.1( zP

• A portion of Table 3:

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06

1.4 0.9265

...

...

0 145.

9265.0

z0 145.

9265.0

z

P z( 0. ) 98 .01635

Example

Find the area to the left of -0.98; P(z < -0.98)

0000.98

Area asked for

0.980.98

Area asked for

0735.09265.00000.1)45.1( zP

Example

Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)

9265.0

145.

Area asked for

0 z

9265.0

145.

Area asked for

145.

Area asked forArea asked for

0 z0 z

(0 1.45) 0.9265 0.5000 0.4265P z

ExampleFind the area to the between z = 0 and of z = 1.45; P(0 < z < 1.45)

• Area between two points = differences in two tabled areas

145.0 z145.145.0 z0 z

Notes Use the fact that the area above zero and the area

below zero is 0.5000

the area above zero is 0.5000

When finding normal distribution probabilities, a sketch is always helpful

3962.01038.05000.0)026.1( zP

Example:Find the area between the mean (z = 0) and z = -1.26

0 z 1 2 6.

A r e a a s k e d f o r

0 z0 z 1 2 6.


1 2 6.

A r e a a s k e d f o rA r e a a s k e d f o r

Example: Find the area between z = -2.30 and z = 1.80

( 2.30 1.80) 0.9641 0.0107 0.9534P z

0 .. - 2 . 3 0

Area Required

1 . 8 00 .. 0 .. 0 .. - 2 . 3 0

Area Required

1 . 8 0- 2 . 3 0

Area Required Area Required

1 . 8 0

Example: Find the area between z = -1.40 and z = -0.50

2277.00808.03085.0)50.040.1( zP

Area asked for

-1.40 -0.500

Area asked for

-1.40 -0.50

Area asked for

-1.40 -0.5000

Computing Areas under the general Normal Distributions

(mean , standard deviation )

1. Convert the random variable, X, to its z-score.

Approach:

3. Convert area under the distribution of X to area under the standard normal distribution.

2. Convert the limits on random variable, X, to their z-scores.

X

z

b

za

PbXaP

Example 1: Suppose a man aged 40-45 is selected at random from a population.

• X is the Blood Pressure of the man.

• Assume that X has a Normal distribution with mean =180 and a standard deviation = 15.

• X is random variable.

The probability density of X is plotted in the graph below.

• Suppose that we are interested in the probability that X between 170 and 210.

Let

15

180

XXz

667.015

180170170

a

000.215

180210210

b

000.2667.210170 zPXP

Hence

000.2667.210170 zPXP

000.2667.210170 zPXP

Example 2

A bottling machine is adjusted to fill bottles with a mean of 32.0 oz of soda and standard deviation of 0.02. Assume the amount of fill is normally distributed and a bottle is selected at random:

1) Find the probability the bottle contains between 32.00 oz and 32.025 oz

2) Find the probability the bottle contains more than 31.97 oz

When x = 32.00

Solution part 1)

32.00 32.00 320.00

0.02z

When x = 32.025

32.025 32.025 321.25

0.02z

P X PX

P z

( . )0. 0.

.

0.

( . ) .

32.0 32 02532.0 32.0

02

32.0

02

32 025 32.0

02

0 125 0 3944

Graphical Illustration:

0 1 2 5. z3 2 . 0 x3 2 0 2 5.


0 1 2 5. z0 1 2 5. z3 2 . 0 x3 2 0 2 5.


3 2 . 0 x3 2 . 0 x3 2 0 2 5.


3 2 0 2 5.


P x Px

P z( . ).

( .

. . .

319732.0

0.023197 32.0

0.02150)

1 0000 00668 0 9332

Example 2, Part 2)

32.03197. x

0150. z32.03197. x32.03197. x

0150. z0150.150. z

SummaryRandom Variables


experiment

Types of Random Variables

• Discrete

Possible values integers• Continuous

Possible values vary over a continuum

The Probability distribution of a random variable






-

0.1

0.2

0.3

0.4

0 1 2 3 4 5 6 7 8 9 10

Number of successes, x

p(x

)

The Binomial distribution





xnx ppx

nxp

1




Notes:


b is P[a < X < b].

2 3 2 3

The Normal Probability Distribution

Points of Inflection

Normal approximation to the Binomial distribution

Using the Normal distribution to calculate Binomial probabilities

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

-

-0.5

Binomial distribution

Approximating

Normal distribution

Binomial distribution n = 20, p = 0.70

049.2

14

npq

np

Normal Approximation to the Binomial distribution

• X has a Binomial distribution with parameters n and p

2121 aYaPaXP

• Y has a Normal distribution

npq

np

correction continuity21

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

-

0.0500

0.1000

0.1500

0.2000

0.2500

0 2 4 6 8 10 12 14 16 18 20

Binomial distribution

-

0.0500

0.1000

0.1500

0.2000

0.2500

a

-

-0.5

Approximating

Normal distribution

P[X = a]

21a 2

1a

-

0.0500

0.1000

0.1500

0.2000

0.2500

a-

-0.5

2121 aYaP

-

0.0500

0.1000

0.1500

0.2000

0.2500

a

-

-0.5

P[X = a]

Example

• X has a Binomial distribution with parameters n = 20 and p = 0.70

13 want We XP

13 eexact valu The XP

1643.030.070.013

20 713

Using the Normal approximation to the Binomial distribution

Where Y has a Normal distribution with:

049.230.70.20

14)70.0(20

npq

np

2121 131213 YPXP

Hence

5.135.12 YP

049.2

145.13

049.2

14

049.2

145.12 YP

= 0.4052 - 0.2327 = 0.1725

24.073.0 ZP

Compare with 0.1643



2121 bYaP


npq

np


)()1()( bpapapbXaP

-

0.0500

0.1000

0.1500

0.2000

0.2500

a b

-

-0.5

21a 2

1b

bXaP

-

0.0500

0.1000

0.1500

0.2000

0.2500

a b

-

-0.5

21a 2

1b

2121 bYaP

Example


1411 want We XP 1411 eexact valu The XP

614911 30.070.014

2030.070.0

11

20

)14()13()12()11( pppp

5357.01916.01643.01144.00654.0

Using the Normal approximation to the Binomial distribution

Where Y has a Normal distribution with:

049.230.70.20

14)70.0(20

npq

np

2121 14101411 YPXP

Hence

5.145.10 YP

049.2

145.14

049.2

14

049.2

145.10 YP

= 0.5948 - 0.0436 = 0.5512

24.071.1 ZP

Compare with 0.5357

Comment:

• The accuracy of the normal appoximation to the binomial increases with increasing values of n



2121 bYaP


npq

np


)()1()( bpapapbXaP

Example• The success rate for an Eye operation is 85%• The operation is performed n = 2000 times

Find the probability that1. The number of successful operations is

between 1650 and 1750.2. The number of successful operations is at

most 1800.

Solution


17201680 want We XP

5.17205.1679 YP

where Y has a Normal distribution with:

969.1515.85.200

1700)85.0(2000

npq

np

17201680 Hence XP

969.15

17005.1720

969.15

1700

969.15

17005.1679 YP

= 0.9004 - 0.0436 = 0.8008

28.128.1 ZP

5.17205.1679 YP

Solution – part 2.

1800 want We XP

5.1800 YP

969.15

17005.1800

969.15

1700YP

= 1.000

29.6 ZP

Next topic: Sampling Theory

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Probability The definition – probability of an Event Applies only to the special case when 1.The sample space has a finite no.of outcomes, and 2.Each