ch5+ken+black

8/8/2019 Ch5+Ken+Black

1/53

Discrete Distributions

Learning Objectives

Distinguish between discrete randomvariables and continuous random variables.

Know how to determine the mean and

variance of a discrete distribution. Identify the type of statistical experiments

that can be described by the binomialdistribution, and know how to work such

problems.


2/53

Learning Objectives -- Continued

Decide when to use the Poisson distributionin analyzing statistical experiments, andknow how to work such problems.

Decide when binomial distributionproblems can be approximated by thePoisson distribution, and know how to worksuch problems.

Decide when to use the hypergeometricdistribution, and know how to work suchproblems.


3/53

Discrete vs. Continuous Distributions

Random Variable -- a variable which containsthe outcomes of a chance experiment

Discrete Random Variable -- the set of allpossible values is at most a finite or a countably

infinite number of possible valuesNumber of new subscribers to a magazineNumber of bad checks received by a restaurant

Number of absent employees on a given day

Continuous Random Variable -- takes on values

at every point over a given interval Current Ratio of a motorcycle distributorship

Elapsed time between arrivals of bank customers Percent of the labor force that is unemployed


4/53

Some Special Distributions

DiscretebinomialPoissonhypergeometric

Continuousuniformnormalexponential

tchi-squareF


5/53

Discrete Distribution -- Example

0

1

2

34

5

0.37

0.31

0.18

0.090.04

0.01

Number ofCrises Probability

Distribution of DailyCrises

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5

P

r

o

b

a

b

i

l

it

yNumber of Crises


6/53

Requirements for a

Discrete Probability Function

Probabilities are between 0 and 1,inclusively

Total of all probabilities equals 10 1

P X( ) for all X

P X( )over all x = 1


7/53

Requirements for a Discrete

Probability Function -- Examples

X P(X)

-1

0

1

2

3

.1

.2

.4

.2

.11.0

X P(X)

-1

0

1

2

3

-.1

.3

.4

.3

.1

1.0

X P(X)

-1

0

1

2

3

.1

.3

.4

.3

.11.2

PROBABILITY

DISTRIBUTION:: YESYES NONO NONO


8/53

Mean of a Discrete Distribution

( ) = = E X X P X ( )X

-1

0

1

23

P(X)

.1

.2

.4

.2

.1

-.1

.0

.4

.4

.3

1.0

X P X ( )

= 1.0= 1.0


9/53

Variance and Standard Deviation

of a Discrete Distribution

( ) 2.1)(22

= = XPX 10.12.12

==

X-1

0

12

3

P(X).1

.2

.4

.2

.1

-2

-1

01

2

X 4

1

01

4

.4

.2

.0

.2

.4

1.2

)(2

X2

( ) ( )X P X


10/53

Mean of the Crises Data Example

( ) = = = E X X P X ( ) .115X P(X) X P(X)

0 .37 .00

1 .31 .31

2 .18 .36

3 .09 .27

4 .04 .16

5 .01 .05

1.15

0

0.1

0.2

0.3

0.40.5

0 1 2 3 4 5

P

ro

b

a

b

i

l

i

t

yNumber of Crises


11/53

Variance and Standard Deviation

of Crises Data Example

( ) 41.1)(22

== XPX = = =2

1 41 119. .

X P(X) (X- ) (X- ) 2 (X- ) 2P(X)0 .37 -1.15 1.32 .49

1 .31 -0.15 0.02 .01

2 .18 0.85 0.72 .13

3 .09 1.85 3.42 .31

4 .04 2.85 8.12 .32

5 .01 3.85 14.82 .15

1.41


12/53

Binomial Distribution

Experiment involves n identical trials Each trial has exactly two possible outcomes:

success and failure Each trial is independent of the previous trials

p is the probability of a success on any one trial q = (1-p) is the probability of a failure on anyone trial

p and q are constant throughout the experiment Xis the number of successes in the n trials Applications

Sampling with replacementSampling without replacement -- n < 5% N


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Binomial Distribution

Probabilityfunction

Mean value

Varianceandstandarddeviation

( )P X

n

X n X X n

X n X

p q( )!

! !=

for 0

= n p

2

2

=

= =

n p q

n p q


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Binomial Distribution: Development

Experiment: randomly select, with replacement,two families from the residents of Tiny Town Success is Children in Household: p = 0.75 Failure is No Children in Household: q = 1-p =

0.25

Xis the number of families in the sample withChildren in Household

FamilyChildren inHousehold

Number ofAutomobiles

A

B

C

D

Yes

Yes

No

Yes

3

2

1

2

Listing of Sample Space

(A,B), (A,C), (A,D), (A,A),

(B,A), (B,B), (B,C), (B,D),

(C,A), (C,B), (C,C), (C,D),

(D,A), (D,B), (D,C), (D,D)


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Continued

Families A, B, and D havechildren in the household;family C does not

Success is Children inHousehold: p = 0.75

Failure is No Children inHousehold: q = 1- p = 0.25

Xis the number of families

in the sample withChildren in Household

(A,B),

(A,C),(A,D),

(A,A),

(B,A),

(B,B),

(B,C),

(B,D),

(C,A),

(C,B),(C,C),

(C,D),

(D,A),

(D,B),

(D,C),

(D,D)

Listing ofSampleSpace

2

12

2

2

2

1

2

1

10

1

2

2

1

2

X

1/16

1/161/16

1/16

1/16

1/16

1/16

1/16

1/16

1/161/16

1/16

1/16

1/16

1/16

1/16

P(outcome)


16/53


Continued

(A,B),

(A,C),

(A,D),

(A,A),

(B,A),

(B,B),

(B,C),

(B,D),

(C,A),

(C,B),(C,C),

(C,D),

(D,A),

(D,B),

(D,C),

(D,D)

Listing ofSampleSpace

2

1

2

2

2

2

1

2

1

10

1

2

2

1

2

X

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/16

1/161/16

1/16

1/16

1/16

1/16

1/16

P(outcome) X

0

1

2

1/16

6/16

9/16

1

P(X)

( )P X

n

X n X

x n x

p q( )!

! !=

( )

P X( )!

!

.. .= =

= =

02

0! 2 0

0 06251

16

0 2 0

75 25

( )P X( )

!

! !.. .= =

= =

1

2

1 2 10 375

3

16

1 2 1

75 25

( )P X( )

!

! !.. .= =

= =

2

2

2 2 205625

9

16

2 2 2

75 25


17/53


Continued Families A, B, and D

have children in thehousehold; family Cdoes not

Success is Children inHousehold: p = 0.75

Failure is No Childrenin Household: q = 1- p= 0.25

X is the number offamilies in the samplewith Children inHousehold

XPossible

Sequences

0

1

1

2

(F,F)

(S,F)

(F,S)

(S,S)

P(sequence)

(. )(. )25 25

(. )(. )25 75

(. )(. )75 25

(. )(. )75 75


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Continued

XPossible

Sequences

0

1

1

2

(F,F)

(S,F)

(F,S)

(S,S)

P(sequence)

(. )(. ) (. )25 25 225

(. )(. )25 75

(. )(. )75 25

(. )(. ) (. )75 75 275

X

0

1

2

P(X)

(. )(. )25 752 =0.375

(. )(. ) (. )75 75 275 =0.5625

(. )(. ) (. )25 25 225 =0.0625

( )P X

n

X n X

x n x

p q( )!

! !=

( )P X( ) !

!.. .= =

=

0 2

0! 2 00 0625

0 2 0

75 25 ( )P X( )!

! !.. .= =

=

1 2

1 2 10 375

1 2 1

75 25

( )P X( )

!

! !.. .= =

=

2

2

2 2 20 5625

2 2 2

75 25


19/53

Binomial Distribution:

Demonstration Problem 5.3

According to the US Census Bureau,approximately 6% of all workers inJackson, Mississippi, are unemployed. In

conducting a random telephone survey inJackson, what is the probability of gettingtwo or lesser unemployed workers in asample of 20?


20/53

Binomial Distribution:

Demonstration Problem 5.3n

p

q

P X P X P X P X

=

=

=

= = + = + =

= + + =

20

06

94

2 0 1 2

2901 3703 2246 8850

.

.

( ) ( ) ( ) ( )

. . . .

( ) ( )P X( ))!

( )( )(. ) .. .= =

= =

020!

0!(20 01 1 2901 2901

0 20 0

06 94

( ) ( )P X( ) !( )! ( )(. )(. ) .. .= = = =

1

20!

1 20 1 20 06 3086 3703

1 20 1

06 94

( ) ( )P X( )!( )!

( )(. )(. ) .. .= =

= =

220!

2 20 2190 0036 3283 2246

2 20 2

06 94


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BinomialTable

n = 20 PROBABILITY

X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

0 0.122 0.012 0.001 0.000 0.000 0.000 0.000 0.000 0.000

1 0.270 0.058 0.007 0.000 0.000 0.000 0.000 0.000 0.0002 0.285 0.137 0.028 0.003 0.000 0.000 0.000 0.000 0.000

3 0.190 0.205 0.072 0.012 0.001 0.000 0.000 0.000 0.000

4 0.090 0.218 0.130 0.035 0.005 0.000 0.000 0.000 0.000

5 0.032 0.175 0.179 0.075 0.015 0.001 0.000 0.000 0.000

6 0.009 0.109 0.192 0.124 0.037 0.005 0.000 0.000 0.000

7 0.002 0.055 0.164 0.166 0.074 0.015 0.001 0.000 0.000

8 0.000 0.022 0.114 0.180 0.120 0.035 0.004 0.000 0.000

9 0.000 0.007 0.065 0.160 0.160 0.071 0.012 0.000 0.00010 0.000 0.002 0.031 0.117 0.176 0.117 0.031 0.002 0.000

11 0.000 0.000 0.012 0.071 0.160 0.160 0.065 0.007 0.000

12 0.000 0.000 0.004 0.035 0.120 0.180 0.114 0.022 0.000

13 0.000 0.000 0.001 0.015 0.074 0.166 0.164 0.055 0.002

14 0.000 0.000 0.000 0.005 0.037 0.124 0.192 0.109 0.009

15 0.000 0.000 0.000 0.001 0.015 0.075 0.179 0.175 0.032

16 0.000 0.000 0.000 0.000 0.005 0.035 0.130 0.218 0.090

17 0.000 0.000 0.000 0.000 0.001 0.012 0.072 0.205 0.19018 0.000 0.000 0.000 0.000 0.000 0.003 0.028 0.137 0.285

19 0.000 0.000 0.000 0.000 0.000 0.000 0.007 0.058 0.270

20 0.000 0.000 0.000 0.000 0.000 0.000 0.001 0.012 0.122


22/53

Using the

Binomial Table

Demonstration

Problem 5.4

n = 20 PROBABILITY

X 0.1 0.2 0.3 0.4

0 0.122 0.012 0.001 0.000

1 0.270 0.058 0.007 0.000

2 0.285 0.137 0.028 0.003

3 0.190 0.205 0.072 0.012

4 0.090 0.218 0.130 0.035

5 0.032 0.175 0.179 0.075

6 0.009 0.109 0.192 0.124

7 0.002 0.055 0.164 0.166

8 0.000 0.022 0.114 0.180

9 0.000 0.007 0.065 0.160

10 0.000 0.002 0.031 0.117

11 0.000 0.000 0.012 0.071

12 0.000 0.000 0.004 0.035

13 0.000 0.000 0.001 0.015

14 0.000 0.000 0.000 0.005

15 0.000 0.000 0.000 0.001

16 0.000 0.000 0.000 0.000

17 0.000 0.000 0.000 0.000

18 0.000 0.000 0.000 0.000

19 0.000 0.000 0.000 0.000

20 0.000 0.000 0.000 0.000

( ) ( )

n

p

P X C

=

=

= = =

20

40

10 0117120 1010 10

40 60

.

( ) .. .


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Binomial Distribution using Table:


n

p

q

P X P X P X P X

=

=

=

= = + = + =

= + + =

20

06

94

2 0 1 2

2901 3703 2246 8850

.

.

( ) ( ) ( ) ( )

. . . .

P X P X ( ) ( ) . .> = = =2 1 2 1 8850 1150

= = =n p ( )(. ) .20 06 1 20

2

2

20 06 94 1128

1 128 1 062

= = =

= = =

n p q ( )(. )(. ) .

. .

n = 20 PROBABILITY

X 0.05 0.06 0.07

0 0.3585 0.2901 0.2342

1 0.3774 0.3703 0.3526

2 0.1887 0.2246 0.2521

3 0.0596 0.0860 0.1139

4 0.0133 0.0233 0.0364

5 0.0022 0.0048 0.0088

6 0.0003 0.0008 0.0017

7 0.0000 0.0001 0.0002

8 0.0000 0.0000 0.0000

20 0.0000 0.0000 0.0000


24/53

Excels Binomial Function

n =20

p =0.06

X P(X)

0 =BINOMDIST(A5,B$1,B$2,FALSE)











25/53

Minitabs Binomial FunctionX P(X =x)

0 0.000000

1 0.000000

2 0.000000

3 0.000001

4 0.000006

5 0.000037

6 0.000199

7 0.000858

8 0.0030519 0.009040

10 0.022500

11 0.047273

12 0.084041

13 0.126420

14 0.160533

15 0.171236

16 0.152209

17 0.111421

18 0.066027

19 0.030890

20 0.010983

21 0.002789

22 0.000451

23 0.000035

Binomial with n = 23 and p = 0.64


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Graphs of Selected Binomial Distributions

n = 4 PROBABILITY

X 0.1 0.5 0.90 0.656 0.063 0.000

1 0.292 0.250 0.004

2 0.049 0.375 0.049

3 0.004 0.250 0.292

4 0.000 0.063 0.656

P = 0.1

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0 1 2 3 4X

P(X

)

P = 0.5

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0 1 2 3 4

X

P(X)

P = 0.9

0.000

0.100

0.200

0.300

0.400

0.500

0.600

0.700

0.800

0.900

1.000

0 1 2 3 4X

P(X

)


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Poisson Distribution

Describes discrete occurrences over acontinuum or interval

A discrete distribution

Describes rare events Each occurrence is independent of anyother occurrences.

The number of occurrences in each intervalcan vary from zero to infinity.

The expected number of occurrences musthold constant throughout the experiment.


28/53

Poisson Distribution: Applications

Arrivals at queuing systemsairports -- people, airplanes, automobiles,

baggagebanks -- people, automobiles, loan applications

computer file servers -- read and writeoperations Defects in manufactured goods

number of defects per 1,000 feet of extrudedcopper wire

number of blemishes per square foot of paintedsurfacenumber of errors per typed page


29/53

Poisson Distribution

Probability function

P X

X

X

where

long run average

e

X

e( )!

, , , , ...

:

. ...

= =

=

=

for

(the base of natural logarithms)

0 1 2 3

2 718282

s Mean valueMean value

s Standard deviationStandard deviations VarianceVariance


30/53


Bank customers arrive randomly onweekday afternoons at an average rate of3.2 customers every 4 minutes. What is the

probability of getting exactly 10 customersduring an 8 minute interval? What is theprobability of getting exactly 6 customersduring an 8 minute interval?


31/53


=

=

3 2

6 4

1010

0 0528

6 4

.

!

!.

.

customers / 4 minutes

X = 10 customers / 8 minutes

Adjusted

= . customers / 8 minutes

P(X) =

( = ) =

X

10

6.4

e

e

X

P X

=

=

3 2

6 4

66

0 1586

6 4

.

!

!.

.

customers / 4 minutes

X = 6 customers / 8 minutes

Adjusted

= . customers / 8 minutes

P(X) =

( = ) =

X

6

6.4

e

e

X

P X


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Poisson Distribution: Probability Table

X 0.5 1.5 1.6 3.0 3.2 6.4 6.5 7.0 8.0

0 0.6065 0.2231 0.2019 0.0498 0.0408 0.0017 0.0015 0.0009 0.0003

1 0.3033 0.3347 0.3230 0.1494 0.1304 0.0106 0.0098 0.0064 0.0027

2 0.0758 0.2510 0.2584 0.2240 0.2087 0.0340 0.0318 0.0223 0.0107

3 0.0126 0.1255 0.1378 0.2240 0.2226 0.0726 0.0688 0.0521 0.0286

4 0.0016 0.0471 0.0551 0.1680 0.1781 0.1162 0.1118 0.0912 0.0573

5 0.0002 0.0141 0.0176 0.1008 0.1140 0.1487 0.1454 0.1277 0.0916

6 0.0000 0.0035 0.0047 0.0504 0.0608 0.1586 0.1575 0.1490 0.1221

7 0.0000 0.0008 0.0011 0.0216 0.0278 0.1450 0.1462 0.1490 0.1396

8 0.0000 0.0001 0.0002 0.0081 0.0111 0.1160 0.1188 0.1304 0.1396

9 0.0000 0.0000 0.0000 0.0027 0.0040 0.0825 0.0858 0.1014 0.1241

10 0.0000 0.0000 0.0000 0.0008 0.0013 0.0528 0.0558 0.0710 0.0993

11 0.0000 0.0000 0.0000 0.0002 0.0004 0.0307 0.0330 0.0452 0.0722

12 0.0000 0.0000 0.0000 0.0001 0.0001 0.0164 0.0179 0.0263 0.0481

13 0.0000 0.0000 0.0000 0.0000 0.0000 0.0081 0.0089 0.0142 0.0296

14 0.0000 0.0000 0.0000 0.0000 0.0000 0.0037 0.0041 0.0071 0.016915 0.0000 0.0000 0.0000 0.0000 0.0000 0.0016 0.0018 0.0033 0.0090

16 0.0000 0.0000 0.0000 0.0000 0.0000 0.0006 0.0007 0.0014 0.0045

17 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0003 0.0006 0.0021

18 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0002 0.0009


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34/53

Poisson Distribution: Using the

Poisson Tables

X 0.5 1.5 1.6 3.0

0 0.6065 0.2231 0.2019 0.0498

1 0.3033 0.3347 0.3230 0.1494

2 0.0758 0.2510 0.2584 0.2240

3 0.0126 0.1255 0.1378 0.22404 0.0016 0.0471 0.0551 0.1680

5 0.0002 0.0141 0.0176 0.1008

6 0.0000 0.0035 0.0047 0.0504

7 0.0000 0.0008 0.0011 0.0216

8 0.0000 0.0001 0.0002 0.0081

9 0.0000 0.0000 0.0000 0.0027

10 0.0000 0.0000 0.0000 0.0008

11 0.0000 0.0000 0.0000 0.0002

12 0.0000 0.0000 0.0000 0.0001

=

= =

1 6

4 0 0551

.

( ) .P X


35/53

Poisson

Distribution:Using the

Poisson

Tables

X 0.5 1.5 1.6 3.0

0 0.6065 0.2231 0.2019 0.0498

1 0.3033 0.3347 0.3230 0.1494

2 0.0758 0.2510 0.2584 0.2240

3 0.0126 0.1255 0.1378 0.2240

4 0.0016 0.0471 0.0551 0.1680

5 0.0002 0.0141 0.0176 0.1008

6 0.0000 0.0035 0.0047 0.0504

7 0.0000 0.0008 0.0011 0.0216

8 0.0000 0.0001 0.0002 0.0081

9 0.0000 0.0000 0.0000 0.002710 0.0000 0.0000 0.0000 0.0008

11 0.0000 0.0000 0.0000 0.0002

12 0.0000 0.0000 0.0000 0.0001

=

> = = + = + = + =

= + + + =

1 6

5 6 7 8 9

0047 0011 0002 0000 0060

.

( ) ( ) ( ) ( ) ( )

. . . . .

P X P X P X P X P X


36/53

Poisson

Distribution:

Using the

PoissonTables

=

= < = = =

= =

1 6

2 1 2 1 0 1

1 2019 3230 4751

.

( ) ( ) ( ) ( )

. . .

P X P X P X P X

X 0.5 1.5 1.6 3.0

0 0.6065 0.2231 0.2019 0.0498

1 0.3033 0.3347 0.3230 0.14942 0.0758 0.2510 0.2584 0.2240

3 0.0126 0.1255 0.1378 0.2240

4 0.0016 0.0471 0.0551 0.1680

5 0.0002 0.0141 0.0176 0.1008

6 0.0000 0.0035 0.0047 0.0504

7 0.0000 0.0008 0.0011 0.0216

8 0.0000 0.0001 0.0002 0.0081

9 0.0000 0.0000 0.0000 0.002710 0.0000 0.0000 0.0000 0.0008

11 0.0000 0.0000 0.0000 0.0002

12 0.0000 0.0000 0.0000 0.0001


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Poisson Distribution: Using the

Poisson Tables

What is the probability of selling exactly 4houses in 2 days?

The interval has been changed from 1 dayto 2 days. A lambda of 1.6 for 1 dayconverts to a lambda of 3.2 for 2 days. Theprevious table no longer applies and the

table for lambda = 3.2 must be used.Looking up x = 4 gives the probability as0.1781.


38/53

Poisson Distribution: Graphs

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0 1 2 3 4 5 6 7 8

= 1 6.

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.16

0 2 4 6 8 10 12 14 16

= 6 5.


39/53

Excels Poisson Function

=1.6

X P(X)

0 =POISSON(D5,E$1,FALSE)











40/53

Minitabs Poisson Function

X P(X =x)

0 0.149569

1 0.284180

2 0.269971

3 0.170982

4 0.0812165 0.030862

6 0.009773

7 0.002653

8 0.000630

9 0.000133

10 0.000025

Poisson with mean = 1.9


41/53

Poisson Approximation

of the Binomial Distribution

Binomial probabilities are difficult tocalculate when n is large.

Under certain conditions binomial

probabilities may be approximated byPoisson probabilities.

Poisson approximation

If and the approximation is acceptable .n n p> 20 7,

Use = n p.


42/53

Poisson Approximation

of the Binomial Distribution

X Error

0 0.2231 0.2181 -0.0051

1 0.3347 0.3372 0.0025

2 0.2510 0.2555 0.00453 0.1255 0.1264 0.0009

4 0.0471 0.0459 -0.0011

5 0.0141 0.0131 -0.0010

6 0.0035 0.0030 -0.0005

7 0.0008 0.0006 -0.0002

8 0.0001 0.0001 0.0000

9 0.0000 0.0000 0.0000

Poisson

=1 5.

Binomial

n

p

=

=

50

03.X Error

0 0.0498 0.0498 0.0000

1 0.1494 0.1493 0.0000

2 0.2240 0.2241 0.0000

3 0.2240 0.2241 0.0000

4 0.1680 0.1681 0.0000

5 0.1008 0.1008 0.0000

6 0.0504 0.0504 0.0000

7 0.0216 0.0216 0.0000

8 0.0081 0.0081 0.0000

9 0.0027 0.0027 0.000010 0.0008 0.0008 0.0000

11 0.0002 0.0002 0.0000

12 0.0001 0.0001 0.0000

13 0.0000 0.0000 0.0000

Poisson

= 3 0.

Binomial

n

p

=

=

10 000

0003

,

.


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Hypergeometric Distribution

Sampling without replacementfrom a finitepopulation

The number of objects in the population isdenotedN.

Each trial has exactly two possible outcomes,success and failure.

Trials are not independent x is the number of successes in the n trials

The binomial is an acceptable approximation, ifn < 5%N. Otherwise it is not.


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Hypergeometric Distribution

Probability functionNis population size

n is sample size

A is number of successes in population

xis number of successes in sample

=A n

N

2

2

2

1

=

=

A N A n N n

NN

( ) ( )

( )

( ) ( )P x

C C

C

A x N A n x

N n

( ) =

Mean value

Variance and standard

deviation


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Hypergeometric Distribution:

Probability Computations

Twenty-four people, of whom 8 arewomen, apply for a job. If 5 of theapplicants are sampled randomly, what is

the probability that exactly 3 of thosesampled are women?


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Probability Computations

N = 24

X = 8

n = 5

x

0 0.1028

1 0.3426

2 0.3689

3 0.1581

4 0.0264

5 0.0013

P(x)

( ) ( )

( ) ( )

( ) ( )

P xC C

C

C C

C

A x N A n x

N n

( )

,

.

= =

=

=

=

3

56 120

42 504

1581

8 3 24 8 5 3

24 5


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Hypergeometric Distribution: Graph

N = 24

X = 8

n = 5

x

0 0.1028

1 0.3426

2 0.3689

3 0.1581

4 0.0264

5 0.0013

P(x)

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0 1 2 3 4 5


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Suppose 18 major computer companiesoperate in the United States and that 12 arelocated in Californias Silicon Valley. If 3

computer companies are selected randomlyfrom the entire list, what is the probabilitythat 1 or more of the selected companiesare located in the Silicon Valley?


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X P(X)0 0.02451 0.2206

2 0.48533 0.2696

N = 18

n = 3

A = 12

( ) ( ) ( ) ( ) ( ) ( )

P x P x P x P x

C C

C

C C

C

C C

C

( ) ( ) ( ) ( )

. . .

.

= = + = + =

= + +

= + +

=

1 1 2 3

2206 4853 2696

9755

12 1 18 12 3 1

18 3

12 2 18 12 3 2

18 3

12 3 18 12 3 3

18 3


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Binomial Approximation (large n)

Hypergeometric

N = 24

X = 8n = 5

Binomial

n = 5

p = 8/24 =1/3

x Error

0 0.1028 0.1317 -0.0289

1 0.3426 0.3292 0.0133

2 0.3689 0.32920.0397

3 0.1581 0.1646 -0.0065

4 0.0264 0.0412 -0.0148

5 0.0013 0.0041 -0.0028

P(x)P(x)


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Binomial Approximation (small n)

Hypergeometric

N = 240

X = 80

n = 5

Binomial

n = 5

p = 80/240 =1/3

x P(x) Error

0 0.1289 0.1317 -0.0028

1 0.3306 0.3292 0.0014

2 0.3327 0.3292 0.0035

3 0.1642 0.1646 -0.0004

4 0.0398 0.0412 -0.0014

5 0.0038 0.0041 -0.0003

P(x)


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Excels Hypergeometric Function

N =24

A =8

n =5

X P(X)

0 =HYPGEOMDIST(A6,B$3,B$2,B$1)






=SUM(B6:B11)


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