ken black qa ch05
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Business Statistics, 5th ed.
by Ken Black
Chapter 5
DiscreteDistributions
Discrete Distributions
PowerPoint presentations prepared by Lloyd Jaisingh,
Morehead State University
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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.
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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.
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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
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Some Special Distributions
Discrete binomial Poisson hypergeometric
Continuous uniform normal exponential
t chi-square F
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 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)
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Binomial Distribution: Development
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)
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Binomial Distribution: Development
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
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Binomial Distribution: Development
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
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Binomial Distribution: Development
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
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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
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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
.
( ) .. .
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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
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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)
3 =BINOMDIST(A8,B$1,B$2,FALSE)
4 =BINOMDIST(A9,B$1,B$2,FALSE)
5 =BINOMDIST(A10,B$1,B$2,FALSE)
6 =BINOMDIST(A11,B$1,B$2,FALSE)
7 =BINOMDIST(A12,B$1,B$2,FALSE)
8 =BINOMDIST(A13,B$1,B$2,FALSE)
9 =BINOMDIST(A14,B$1,B$2,FALSE)
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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 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.
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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
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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
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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
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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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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
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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
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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: 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.
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Excels Poisson Function
= 1.6
X P(X)
0 =POISSON(D5,E$1,FALSE)
1 =POISSON(D6,E$1,FALSE)
2 =POISSON(D7,E$1,FALSE)
3 =POISSON(D8,E$1,FALSE)
4 =POISSON(D9,E$1,FALSE)
5 =POISSON(D10,E$1,FALSE)
6 =POISSON(D11,E$1,FALSE)
7 =POISSON(D12,E$1,FALSE)
8 =POISSON(D13,E$1,FALSE)
9 =POISSON(D14,E$1,FALSE)
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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
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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.
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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 Xis the number of successes in the n trials The binomial is an acceptable approximation, if
n < 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
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
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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
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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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Hypergeometric Distribution:
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
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Hypergeometric Distribution:
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)
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Hypergeometric Distribution:
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)
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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)
1 =HYPGEOMDIST(A7,B$3,B$2,B$1)
2 =HYPGEOMDIST(A8,B$3,B$2,B$1)
3 =HYPGEOMDIST(A9,B$3,B$2,B$1)4 =HYPGEOMDIST(A10,B$3,B$2,B$1)
5 =HYPGEOMDIST(A11,B$3,B$2,B$1)
=SUM(B6:B11)
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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
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