IMSE 213: Probability and Statistics for Engineers

Instructor: Steven E. Guffey, PhD, CIH© 2002-2011

2

Introduction

• Will cover most common discrete probability distributions

• Simplest distribution:– Equal probability distribution– I.e., all events equally likely– If there are k values with equal probability, then:

kkxf /1);( • Examples:

– Toin toss– Random selection from a group

3

Theorem 5.1: Mean and Variance of Discrete Uniform Distribution

)(2

1

2 ifxk

ii

)(1

ifxk

ii

k

xk

ii

2

12

k

xk

ii

1

f(i) = n/N = 1/k

4

5.3 Binomial Distribution

• Bernoulli Process:– Experiment consists of n repeated trials– Each trial outcome can be classified as “success” or “failure”

(i.e., only 2 possible outcomes)– Probability of success, p, remains constant from trial to trial– The repeated trials are independent

• Number of successes, X, is called a binomial random variable– b(x; n, p)

• Its distribution is called the binomial distribution

5

Binomial Distribution

• Each success has probability, p• Each failure has probability, q

q = 1 - p• x = no. successes• n - x = no. failures• Since independent, combination probabilities

are multiplied:– Probability of a specific order = px qn-x

– Total number of sample points that have n trials and x successes and n-x failures is

!( ; , )

!( )!x n xn

b x n p p qx n x

Graph of Binomial

6

f(x)

65 87 109 121121 43 130

7

Example

),,(b

• The probability that a component will survive a given shock test is ¾. Find the probability that exactly 2 of the next 4 components will survive.

• Solution:– Assuming the tests are independent– And p = ¾ for each of the 4 tests, then:

211.0

xnx qpxnx

npnxb

)!(!

!),;(

24

4

1

2

4

3

)!24(!2

!42 4 3/4

8

Sum of Probabilites Must Equal Unity

x

x

pnxbpnxB0

1),;(),;(

• As with all other distributions, the sum of all probabilities must equal unity:

9

Table A.1: Binomial Cumulative DistributionProportion, p

Trials Success 0.1 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9

1 0 0.9000 0.8000 0.7500 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000

1 1 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

2 0 0.8100 0.6400 0.5625 0.4900 0.3600 0.2500 0.1600 0.0900 0.0400 0.0100

2 1 0.9900 0.9600 0.9375 0.9100 0.8400 0.7500 0.6400 0.5100 0.3600 0.1900

2 2 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

3 0 0.7290 0.5120 0.4219 0.3430 0.2160 0.1250 0.0640 0.0270 0.0080 0.0010

3 1 0.9720 0.8960 0.8438 0.7840 0.6480 0.5000 0.3520 0.2160 0.1040 0.0280

3 2 0.9990 0.9920 0.9844 0.9730 0.9360 0.8750 0.7840 0.6570 0.4880 0.2710

3 3 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

4 0 0.6561 0.4096 0.3164 0.2401 0.1296 0.0625 0.0256 0.0081 0.0016 0.0001

4 1 0.9477 0.8192 0.7383 0.6517 0.4752 0.3125 0.1792 0.0837 0.0272 0.0037

4 2 0.9963 0.9728 0.9492 0.9163 0.8208 0.6875 0.5248 0.3483 0.1808 0.0523

4 3 0.9999 0.9984 0.9961 0.9919 0.9744 0.9375 0.8704 0.7599 0.5904 0.3439

4 4 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000

Trials Success 0.1 0.2 0.25 0.3 0.4 0.5

15 5 0.9978 0.9389 0.8516 0.7216 0.4032 0.1509

15 6 0.9997 0.9819 0.9434 0.8689 0.6098 0.3036

15 7 1.0000 0.9958 0.9827 0.9500 0.7869 0.5000

15 8 1.0000 0.9992 0.9958 0.9848 0.9050 0.6964

15 9 1.0000 0.9999 0.9992 0.9963 0.9662 0.8491

15 10 1.0000 1.0000 0.9999 0.9993 0.9907 0.9408 10

Example Disease• The probability that a patient recovers from a disease is p= 0.4.

For n=15 with the disease, what is the probability:– A) at least 10 survive?

P(X > 10) = 1- P(X < 10)

x

b ),,(1 x 15 0.4 1- 0.9662 = 0.03380

9What is included?

11

Example Disease -continued

The prob that a patient recovers from a rare blood disease is 0.4.

B) What is the probability that, of 15, P(3 < X < 8)(3 8)P X

Trials Success 0.1 0.2 0.25 0.3 0.4

15 2 0.8159 0.3980 0.2361 0.1268 0.0271

15 3 0.9444 0.6482 0.4613 0.2969 0.0905

15 4 0.9873 0.8358 0.6865 0.5155 0.2173

15 6 0.9997 0.9819 0.9434 0.8689 0.6098

15 8 1.0000 0.9992 0.9958 0.9848 0.9050

8 2

0 0

( ;15,0.4) ( ;15,0.4)x x

b x b x

0.9050 0.0271 0.8779

( ; , )x

b x p

3

8

15 0.4

12

Example Disease CThe probability that a patient recovers from a blood disease = 0.4. What is the probability that:

– C) exactly 5 of 15 survive?

Trials Success 0.1 0.2 0.25 0.3 0.4

15 2 0.8159 0.3980 0.2361 0.1268 0.0271

15 3 0.9444 0.6482 0.4613 0.2969 0.0905

15 4 0.9873 0.8358 0.6865 0.5155 0.2173

15 5 0.9978 0.9389 0.8516 0.7216 0.4032

= 0.4032 - 0.2173 = 0.1859

4

0

( ;15,0.4)x

b x

5

0

( ;15,0.4)x

b x

P(X = 5) = b(5;15,0.4)

13

Example Chain Retailer• A large chain retailer purchases a certain kind of electronic device

from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%.

• Problem (a) The inspector of the retailer randomly picks 20 items for a shipment. What is the probability that there will be at least one defective item among the 20?

• Solution (a):

– Denote X = number of defectives devices among the 20

– The distribution of the X follows: b(x; 20, 0.03)

– P( X > 1) = 1 – P(X < 1) = 1- b(0;20,0.03)

= 1 – (1)(0.03)x(1-0.03)n-x

= 1 – (1)(0.03)0(1-0.03)20-0

= 1 – (1)(1)(0.97)20

= 1 – 0.5438 = 0.4562

14

Example Chain Retailer - continued• A large chain retailer purchases a certain kind of electronic device from a

manufacturer. The manufacturer indicates that the defective rate of the device is 3%. From part a, P(X > 1) = 0.4562

• Problem (b) Suppose the retailer receives 10 shipments in a month and the inspector randomly tests 20 devices per shipment. What is the probability that there will be 3 shipments containing at least one defective device?

• Solution (b):

– Each shipment contains at least 1 defective or not– From part a, P(X > 1) = 0.4562– Let Y = number of shipments with at least one defective item– Assuming independence from shipment to shipment, the Y follows

binomial distribution with p = 0.4562

b( y ; 10, 0.4562)

P(Y = 3) = 10! (0.4562)3(1 - 0.4562)10-3 = 0.1602 3! 7!

15

Areas of Application

• Industrial engineers are interested in the proportion of defectives. Quality control measures and sampling schemes are based on the binomial distribution.

• Also used extensively for medical (survive, die) and military applications (hit, miss).

16

Mean and Variance

• The mean and variance of the binomial distribution [b(x;n,p)] are:

= n p 2 = n p q

17

Example Binomial Mean and Variance

• Find the mean and variance of the binomial random variable of Example 5.5

• Solution:

– Since was a binomial experiment with

n = 15 and p = 0.4, then:

n p = 15(0.4) = 6

2 =

=

= 3.60.5 = 1.897

n p q = 15(0.4)(0.6) = 3.6

18

Example Binomial Impure Wells

An impurity exists in perhaps 30% of wells. Ten wells tested at random.• a) What is the probability that exactly 3 wells have the impurity,

assuming the rate really is 30%:

( 3 | 10, 30%)P X n p

= 0.6496- 0.3828 = 0.2668

• b) What is the probability that more than 3 wells are impure?

Trials Success 0.1 0.2 0.25 0.310 2 0.9298 0.6778 0.5256 0.382810 3 0.9872 0.8791 0.7759 0.6496

P(X > 3|p=30%) = 1 – P(X < 3) = 1 – 0.6496 = 0.3504

3

0

( ;10,0.3)x

b x

2

0

( ;10,0.3)x

b x

0

6

( ;10,0.3)x

b x

5

0

( ;10,0.3)x

b x

19

Example Impure Wells - continued

How likely is that 6 of 10 are contaminated if p = 0.3 ?Solution:

( )6P X

= 0.9894 - 0.9527 = 0.0367

Trials Success 0.1 0.2 0.25 0.310 5 0.9999 0.9936 0.9803 0.952710 6 1.0000 0.9991 0.9965 0.989410 7 0.9999 0.9996 0.9984

20

Multinomial Experiments – not on exam

• More than 2 possible outcomes– Small, medium, large

• If given trial can result in any one of k outcomes, E1, E2, …, Ek with probabilities p1, p2, …, pk occurs in n independent trials.

1 21 2 1 2 1 2

1 2

( , ,..., ; , ,..., ) ..., ,...

kxx xk k k

k

nf x x x p p p p p p

x x x

1

k

ii

x n

1

1k

ii

p

21

Example Multinomials – not on exam

• Arrivals and departures from an airport• 3 runways, each with a probability of being

used for any random plane:

– R1: p1 = 2/9 R2: p2 = 1/6 R3: p3 = 11/18

• What is the probability that 6 planes will land such that:

– R1: 2 planes R2: 1 plane R3: 3 planes

1127.0

18

11

6

1

9

2

3,1,2

6)18/11,6/1,9/2;3,1,2(

312

f

22

Homework: 5ADo problemson pp. 150-151

Up to but not including 17

Artificial intelligence is no match for natural stupidity.

"This 'telephone' has too many shortcomings to be seriously considered as a means of communication. The device is inherently of no value to us." --Western Union internal memo, 1876.

"The wireless music box has no imaginable commercial value. Who would pay for a message sent to nobody in particular?" --David Sarnoff's associates in response to his urging for investment in the radio in the 1920s.

"We don't like their sound, and guitar music is on the way out." -- Decca Recording Co. rejecting the Beatles, 1962.

23

Hypergeometric Distribution

• Involves the way sampling is done• Binomial requires independence (e.g., replace card

after dealing)• Hypergeometric does NOT require independence.

Therefore, highly useful when testing destroys item.• Applies, also, when sample is a significant fraction of

the population • Used in acceptance sampling, electronics testing,

and quality assurance.

24

Hypergeometric Distribution

• Hypergeometric ExperimentRandom sample of size n selected without

replacement from N items

k of the N items may be classified as successes and N-k are classified as failure

• Hypergeometric random variable: number, X, of successes in hypergeometric experiment.

• Hypergeometric distribution:– Prob distr. of a hypergeometric variable– h( x; N, n, k )

25

Hypergeometric Distribution in Acceptance Sampling

• Acceptance sampling: parts are sampled to determine whether the entire lot should be sampled.

26

Hypergeometric Distribution

nx

n

Nxn

kN

x

k

knNxh ,...2,1,0,),,;(

x = assumed number of successes in samplek = number of success in population to be sampledn = size of random sampleN = size of population from which sample is taken

P(X= x) =

Proportion of successes in population = k/NProporton of successes in sample = x/n

27

Hypergeometric Distribution

! ( )!!( )! ! ( ) !

( ; , , )!

! !

k N kx k x n x N k n x

h x N n kN

n N n

( ; , , )

k N k

x n xh x N n k

N

n

No selections

No ways to failNo ways to succeed

28

Example Shipment SamplingSuppose our plan is to sample 3 out of a lot of 10 in each shipment. We will accept the lot if none of the 3 are bad.• How often will we accept the lot when 2 of 10 are actually bad?

• Solution: find P(accept | 2 of 10 in shipment are bad)– i.e., NONE of the 3 are unacceptable)

02

0

23-

103

0.467

bad good

all

( ; , , )

k

x nn

N

NN

x

n

x

k

h k

x = assumed number of “successes” (defectives) in sample = 0

23

10

k = number of success in population to be sampled =n = size of random sample =N = size of population from which sample is taken =

10-0( )P X

29

Example Lots of 40Lots of 40 components should be rejected if they contain 3 or more

defectives. However, we sample 5 at random from each lot and reject if even 1 defective is found. What is the prob that exactly 1 sample is bad if there are 3 defectives in the whole lot?

!!

! ! ! !( ; , , ) , 0,1,2,...

!

! !

N kk

x k x n x N k n xh x N n k x n

N

n N n

( ; , , )h

1!(3-1)!

3!

5

(40-

-1)!

405!(40-5)!

0.30113401

3)!

(5 (40- 3-5+1)!

30

Mean and Variance of Hypergeometric

• The mean and variance of a hypergeometric distribution, h(x; N, n, k) are:

kn

N

N

k

N

nk

N

nN1

12

31

Example: Hypergeometric Mean and Variance

• Find the mean and variance for the data given

n = 5 N = 40 k = 3 x = 1

N

k

N

nk

N

nN1

12

40

40 40 40

5 5 x 3 3

1 1 0.3113

375.040

)3(5nk

N

kn

N

32

Relationship to Binomial Distribution

• If n/N < 0.05 then doesn’t matter much if replace or not.

• k/N plays the role of the binomial value, p

• Variance differs by a factor of (N-n)/N-1negligible as n/N becomes very small

• Therefore, binomial is “large population edition” of the hypergeometric distribution.

np

npq2

11

N

nN

N

k

N

kn

N

k

N

kn 1

N

nk

qN

kn

1

n/N < 0.05

33

Example Tires

• Among 5000 tires shipped, 1000 are blemished. If one selects 10 at random, what is prob that exactly 3 are blemished?

• Solution:– n/N = 10/5000 << 0.05– Therefore, can use the binomial distribution– h(3:5000,10,1000) ~ b(3;10,0.2)

2

0

3

0

)2.0,10;()2.0,10;()3(xx

xbxbXP

=0.8791 – 0.6778 = 0.2013

• Compares well to h(3;5000,10,1000) = 0.2015

34

Multivariate Hypergeometric Distribution – not on exam

• If N items can partitioned in the k cells A1, A2, …. Ak with

a1, a2,…ak elements, then the prob distr of the random

variables X1, X2,…,Xk, representing the number of

elements selected from A1, A2, …. Ak in a random

sample of size n, is:

n

N

x

a

x

a

x

a

nNaaaxxxf k

k

kk

....

),,,...,,;,...,,( 2

2

1

1

2121

1

k

ii

x n

k

ii Na

1

35

Example Multivariate Hypergeometric – not on exam

• A group of 10 individuals are used for a biological case study with the following blood types:– Type O = 3 people– Type A = 4 people– Type B = 3 people

• What is the prob that a random sample of 5 will contain 1 Type O, 2 Type A, 2 Type B ?

• Solution:

– X1 = 1 x2 = 2 x3 = 2

– A1 = 3 a2 = 4 a3 = 3 N = 10 n = 5

14

3

5

102

3

2

4

1

3

)5,10,3,4,3;2,2,1(

f

36

HW: 5Bp. 157-158,

problems: 29, 30, 31, 33, 35, 41, 43, 47*, 49*

* Set up and substitute; do not calculate

37

Negative Binomial and Geometric Distributions

• Consider case where repeat experiments until a fixed number of successes occur.

• Called negative binomial experiments.• Example: Suppose testing drug we hope will provide some relief

to 60% of patients, and we want to know the probability that the 5th patient to experience relief will be the 7th patient to receive the drug. Let k= number of success

5 7 57 1 !( 7) 0.6 0.4 0.1866

5 1 ! 7 1 5 1 !P X

• Prob for a specific order = pk-1 qx-k p = pk qx-k

P = 0.65(1-0.6)2 for each of several orders of successes and failures

For the kth success to occur on the xth trial, there must have been (k-1) successes and (x-k) failures among the first (x-1) trials since the xth trial must be a success.

38

Negative Binomial Random Variable

• Negative binomial random (X)– Number of trials to produce k successes in a negative binomial

experiment

,...2,1,,1

1),;(*

kkkxqpk

xpkxb kxk

Probability distribution of X, the number of trials on which the kth success occurs, is:

1 !*( ; , ) , , 1, 2,...

1 ! !k x kx

b x k p p q x k k kk x k

39

Example NBA-a

• NBA team must win 4 of 7 games to be Champion. Suppose Team A has prob. of beating Team B of 55%.– A) what is the probabilty A will win in 6 games?– Solution:

1 !(1 )

1 !*( ; , )

1 ! !k x kx

b x k p p qk x k

),;(*b

x = 6k = 4p = 0.55

66 4 0.553!

0.55 0.554 6-4

4 25!0.55 0.45 10 0.01853 0.1853

3!2!

2!

40

Example NBA-b• NBA team must win 4 of 7 games to be Champion. Suppose

Team A has probability of beating Team B of 55% each game that is played.– B) what is the probability A will win the series?– Solution: (Hint: can win in 4, 5, 6, or 7 games)

*(4; ,4 0.55)b

*( ; , )b x k p

6083.01668.01853.01647.00915.0

)(teamAwinsP 4*(5; ,0.55)b 4*(6; ,0.55)b 4*(7; ,0.55)b

k = 4p = 0.55

1 !

1 ! !k x kx

p qk x k

74 4

4 4 4

1 !0.55 0.45

1 ! !x

x

x

x

41

Geometric Distribution

• Special case of negative binomial with k=1• Example: toss coin until first occurrence of a

“head”• If repeated trials can result in success with

probability, p, and failure with q =1-p, then the probability distribution of the random variable, X, the number of the trial on which the first success occurs is:

g(x,p) = p qx-1, x = 1,2,3,….

1 !*( ; , )

1 ! !k x kx

b x k p p qk x k

42

Example: First Defective

• In a certain mfg process, on the average, 1/100 items is defective. What is the probability that the 5th item inspected is the 1st defective?

p = 0.01 k=1 x = 5

Solution: g(x,p) = p q

x-1

g(5; 0.01) == (0.01) (1-0.01)4 = 0.0096

43

Example: Telephone Exchanges

• When many calls are coming in at once, telephone exchanges can be overwhelmed, making callers have to call repeatedly to get a connection.– p = 0.05 for probability of connection– What is probability that 5 attempts are necessary?

• Solution:– x = 5– P(X=x) = g(x,p)

= p q x-1

= g(5; 0.05)

= 0.05(1-0.05)5-1 = 0.041

44

Example: Particles in Wafers

• The probability that a wafer contains a large particle of contamination is 0.01. If it is assumed that the wafers are independent, what is the probability that exactly 125 wafers need to be analyzed before a large particle is detected?– Let X denote the number of samples analyzed

until a large particle is detected– p = 0.01

= p q x-1

P(X=125) = g( ; )

= ( ) =

125 0.01

0.0029125-10.01 1- 0.01

45

Mean and Variance of Geometric

• Mean and variance of a random variable following the geometric distribution are:

p

1

22 1

p

p

46

Poisson Distribution and the Poisson Process

• “Poisson Experiments” yield numerical values of a random variable, X, the number of outcomes occurring during a given time interval or in a specified region.– Example: number of calls received per hour– Example: number of field mice per acre– Derived from Poisson process

• Used for quality control, quality assurance, and acceptance sampling.

47

Properties of Poisson Process

• Properties– Number of outcomes independent of number occurring in other

times or spaces– Probability that a single outcome will occur during a short

period or small space is proportional to the size of the region and independent of outcomes outside that region

– The probability that more than one outcome will occur in a short time interval or small place is negligible.

– Can sample only a small fraction of the total volume (“law of rare events”).

• X = Poisson random variable– Follows Poisson distribution– Mean = µ = t– t = specific time or region of interest = single unit– P(x; t )

Derivation from Binomial DistributionWikipedia: This limit is sometimes known as the law of rare events, since each of the individual Bernouilli event rarely triggers. The name may be misleading because the total count of success events in a Poisson process need not be rare if the parameter λ is not small. For example, the number of telephone calls to a busy switchboard in one hour follows a Poisson distribution with the events appearing frequent to the operator, but they are rare from the point of the average member of the population who is very unlikely to make a call to that switchboard in that hour. The proof may proceed as follows. First, recall from from calculus that:

48

If the binomial probability can be defined such that p = λ / n, we can evaluate the limit of P as n goes large:

and then note that, since k is fixed, this is a rational function of n with limit 1.

The F term can be written as:

and the definition of the Binomial distribution:

( )!

keP x k

k

49

Poisson Distribution

The probability distribution of a Poisson random variable, X, representing the number of outcomes occurring in a given time interval or specified region denoted by t, is:

,...,2,1,0,

!);(

xx

tetxp

xt

2.);();(0

ATableseetxptrPr

x

0

0.05

0.1

0.15

0.2

0.25

0.3

0 2 4 6 8 10 12

X

f(x

)

lt = 2

00.020.040.060.08

0.1

0.120.140.160.18

0.2

0 2 4 6 8 10 12

X

f(x

)

lt = 5

66.9 6.9(6;6.9) 0.151

6!

ep

50

Example: Flaws in Copper Wire• For thin copper wire the number of flaws follows a Poisson

distribution with a mean of 2.3 flaws per millimeter. Determine the probability of exactly 2 flaws in 1 millimeter of wire.

( ; )p x t 22.3 2.3

(2;2.3) 0.2652!

ep

lt =

!

xte t

x

x = 2 in 1 mm

• P(4 flaws in 2 mm of wire) = 44.6 4.6

(4;4.6) 0.1884!

ep

• P(6 flaws in 3 mm of wire) =

lt = 2.3/mm * 2 mm = 4.6

lt = 2.3/mm * 3 mm = 6.9

t = =l 2.3 flaws/mm 1 mm 2.3/mm * 1 mm = 2.3

51

Example: Flaws in Copper Wire - B

• For thin copper wire, the number of flaws follows a Poisson distribution with a mean of 2.3 flaws per millimeter.

• Determine the probability that there are < 2 flaws in 1 millimeter of wire.

2

0

( ; )x

p x t

0 1 22.3 2.3 2.32.3 2.3 2.3

0! 1! 2!

e e e

lt = 2.3 flaws

2

0 !

xt

x

e t

x

X ≤ 2 flaws

= 0.100 + 0.231 + 0.265 = 0.596

6 6 1

0 0

(6;4) ( ;4) ( ;4) 0.8893 0.7851 0.1042x x

P p x p x

52

Example: Particle Counter

In a lab experiment, average number of particles in 1 ms is 4. What is the probability that 6 particles enter the counter in a given milli-second?

);();();1();();(1

00

txptxptrPtrPtrpr

x

r

x

);( txp )4;6(p !x

te xt

1042.0!6

4 64

e

Solution: Using the Poisson distribution with:

x = t = 46

53

Example: Particle Counter (Using Table A.2)

In lab experiment, the average number of particles counted in 1 ms is 4. What is the prob that exactly 6 particles enter the counter in a given

ms? Poisson distribution with: x = 6 t = 4

);();();(1

00

txptxptrpr

x

r

x

6 6 1

0 0

(6;4) ( ;4) ( ;4)x x

p p x p x

x p(x,lt) P(x,lt)

0 0.0183 0.0183

1 0.0733 0.0916

2 0.1465 0.2381

3 0.1954 0.4335

4 0.1954 0.6288

5 0.1563 0.7851

6 0.1042 0.8893

0.8893 - 0.7851 = 0.1042

54

Example Tankers

• On the average, 10 tankers arrive each day. We can handle 15 tankers per day. What is the probability that on a given day some tankers will be turned away?

• Solution:– Let X = number tankers arriving– Using Table A.2, we have:

15

0x

p(x;10) - 1 15) P(X - 1 15) P(X

0.04870.9513- 1

55

Mean and Variance of Poisson

The mean and variance of the Poisson distribution p(x, t) both have

the value t.

56

Poisson as a limiting form of the binomial

• If n is large and p approaches 0, conditions are nearer to the continuous space or time region of the Poisson process.

• Note that if p approaches 1, we can just exchange the definition of success and failure and use the Poisson distribution.

• Accurate if:– n ≥ 20 and p ≤ 0.05,

or – n ≥ 100 and np ≤ 10

57

Theorem

• Let X be a binomial random variable with probability distribution, b(x;n,p). When:

n ∞ p 0 µ = np remains constant,

• Since– binomial: µ = np– Poisson: µ = t

• Then: t = np and b(x;n,p) = p(x, t) = p(x;µ)

58

Example: Accidents

• Accidents occur infrequently in an particular factory. Assume accidents are independent and p = 0.005 on any given day. What is the probability of an accident on one day of a period of 400 days?

• Solution:– Since p 0 and n is large, use Poisson distribution.– t = np = 400(0.005) = 2– X = 1

( 1)P X

12

2

2 20.271

1!

e

e

!x

npe xnp

!x

te xt

Example : Accidents - B

• Accidents occur infrequently in an particular factory. Assume accidents are independent and p = 0.005. What is the prob that there are at most 3 days out of 400 with an accident?

• Solution:– Since p 0 and n is large, use Poisson distribution.

( )P X

2 2 0 2 1 2 2 2 33

0

2 2 2 2 20.857

! 0! 1! 2! 3!

x

x

e e e e e

x

3

0 !x

xnp

x

npe

3

0 !x

xt

x

te

– t = np = 400(0.005) = 2

3<

59

60

Example: Glass Making• In glass-making, one or more defects occur on the avg on 1/1000

pieces. What is the prob that 8000 pieces will contain < 7 with defects?

6

0

( ;8000,0.001)x

b x

x p(x,lt) P(x,lt)

0 0.0003 0.0003

1 0.0027 0.0030

2 0.0107 0.0138

3 0.0286 0.0424

4 0.0573 0.0996

5 0.0916 0.1912

6 0.1221 0.3134

)7(XP

6

0

8

!

8

x

x

x

e

6

0 !x

xnp

x

npe

6

0

)8;(x

xp

3134.0

• Solution:– Is a binomial experiment, but since n=8000 and p=0.0001, can

approximate with the Poisson distr:

µ =(8000)(0.001) = 8

61

HW: 5COdd problems on p. 165-167,

omitting nos. 63 and 67

62