Ken Black QA ch05

8/3/2019 Ken Black QA ch05

1/49

Business Statistics, 5th ed.

by Ken Black

Chapter 5

DiscreteDistributions

Discrete Distributions

PowerPoint presentations prepared by Lloyd Jaisingh,

Morehead State University


2/49

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.


3/49

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.


4/49

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 values Number of new subscribers to a magazine

Number 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


5/49

Some Special Distributions

Discrete binomial Poisson hypergeometric

Continuous uniform normal exponential

t chi-square F


6/49

Discrete Distribution -- Example

012

345

0.370.310.18

0.090.040.01

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


7/49

Requirements for a

Discrete Probability Function

Probabilities are between 0 and 1,inclusively

Total of all probabilities equals 10 1

P X( ) for all

P X( )over all x 1


8/49

Requirements for a Discrete

Probability Function -- Examples

X P(X)

-1

01

2

3

.1

.2

.4

.2

.1

1.0

X P(X)

-1

01

2

3

-.1

.3

.4

.3

.1

1.0

X P(X)

-1

01

2

3

.1

.3

.4

.3

.1

1.2

PROBABILITYDISTRIBUTION

: YES NO NO


9/49

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


10/49

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


11/49

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


12/49

Variance and Standard Deviation

of Crises Data Example

41.1)(22

XPX 2

1 41 119. .

X P(X) (X-) (X-)2

(X-)2

P(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


13/49

Binomial Distribution Experiment involves n identical trials

Each trial has exactly two possible outcomes: successand 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 replacement Sampling without replacement --n < 5% N


14/49

Binomial Distribution

Probabilityfunction

Meanvalue

Variance

andstandarddeviation

P X

n

X n XX n

X n X

p q( )!

! !

for 0

n p

2

2

n p q

n p q


15/49

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 with

Children 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)


16/49


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

12222121

1012212

X

1/16

1/161/161/161/161/161/161/161/16

1/161/161/161/161/161/161/16

P(outcome)


17/49


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

212222121

1012212

X

1/161/161/161/161/161/161/161/161/16

1/161/161/161/161/161/161/16

P(outcome) X

012

1/166/169/16

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


18/49


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

.252


19/49


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 205625

2 2 2

75 25


20/49

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


21/49

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


22/49

Using the

Binomial TableDemonstration

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

.

( ) .. .


23/49

Binomial Distribution using Table:

Demonstration Problem 5.3

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 1 128

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


24/49

Excels Binomial Function

n = 20

p = 0.06

X P(X)

0 =BINOMDIST(A5,B$1,B$2,FALSE)1 =BINOMDIST(A6,B$1,B$2,FALSE)

2 =BINOMDIST(A7,B$1,B$2,FALSE)









25/49

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


26/49

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

)


27/49

Poisson Distribution

Describes discrete occurrences over acontinuum or interval

A discrete distribution

Describes rare events Each occurrence is independent of any other

occurrences. The number of occurrences in each interval

can vary from zero to infinity. The expected number of occurrences must

hold constant throughout the experiment.


28/49

Poisson Distribution: Applications

Arrivals at queuing systems airports -- people, airplanes, automobiles,

baggage banks -- 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 paintedsurface number of errors per typed page


29/49

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

Mean value

Standard deviation Variance


30/49

Poisson Distribution: Demonstration

Problem 5.7

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


31/49

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


32/49

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


33/49

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 0.22403 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


34/49

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


35/49

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.


36/49

Excels Poisson Function

= 1.6

X P(X)

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











37/49

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


38/49

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 acceptabl.n n p 20 7,

Use n p.


39/49

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

,

.


40/49

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 Xis the number of successes in the n trials The binomial is an acceptable approximation, if

n < 5%N. Otherwise it is not.


41/49

Hypergeometric Distribution

Probability functionNis population size

n is sample size

A is number of successes in population

x is 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

( )

Meanvalue

Variance and standard deviation


42/49

Hypergeometric Distribution:

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


43/49

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


44/49


Demonstration Problem 5.11

X P(X)0 0.02451 0.2206

2 0.48533 0.2696

N= 18n= 3A = 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


45/49


Binomial Approximation (largen)

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)


46/49


Binomial Approximation (smalln)

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)


47/49

Excels Hypergeometric Function

N = 24

A = 8

n = 5

X P(X)

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



3 =HYPGEOMDIST(A9,B$3,B$2,B$1)4 =HYPGEOMDIST(A10,B$3,B$2,B$1)


=SUM(B6:B11)


48/49

Minitabs Hypergeometric Function

X P(X =x)

0 0.102767

1 0.342556

2 0.368906

3 0.158103

4 0.026350

5 0.001318

Hypergeometric with N = 24, A = 8, n = 5


49/49

Copyright 2008 John Wiley & Sons, Inc.All rights reserved. Reproduction or translation

of this work beyond that permitted in section 117of the 1976 United States Copyright Act withoutexpress permission of the copyright owner isunlawful. Request for further information shouldbe addressed to the Permissions Department, JohnWiley & Sons, Inc. The purchaser may makeback-up copies for his/her own use only and notfor distribution or resale. The Publisher assumes

no responsibility for errors, omissions, or damagescaused by the use of these programs or from theuse of the information herein.

Ken Black QA ch05

Documents

Transcript of Ken Black QA ch05