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Chapter 5
Discrete
Probability
Distributions
© 2002 Thomson / South-Western
Slide 5-1
Learning Objectives
• Distinguish between discrete random
variables and continuous random
variables.
• Identify the type of statistical
experiments that can be described by
the binomial distribution, and know how
to work such problems.
© 2002 Thomson / South-Western
Slide 5-2
Learning Objectives, continued
• Decide when to use the Poisson
distribution in analyzing statistical
experiments, and know how to work
such problems.
• Decide when to use the hypergeometric
distribution, and know how to work such
problems.
© 2002 Thomson / South-Western
Slide 5-3
Discrete vs Continuous Distributions
• Random Variable -- a variable which
contains the outcomes of a chance
experiment
• Discrete Random Variable -- the set of all
possible values is at most a finite or a
countable infinite number of possible values
• Continuous Random Variable -- takes on
values at every point over a given interval
© 2002 Thomson / South-Western
Slide 5-4
Some Special Distributions
• Discrete distributions are constructed
from discrete ransom variables.
• The binomial, Poisson, and hypergeometric
distributions are discrete distributions
• Continuous distributions are based on
continuous random variables.
• The normal, uniform, exponential, t, chisquare, and F distributions are continuous
distributions
© 2002 Thomson / South-Western
Slide 5-5
Binomial Distribution
A widely known discrete distribution
constructed by determining the
probabilities of X successes in n trials.
© 2002 Thomson / South-Western
Slide 5-6
Assumptions of the
Binomial Distribution
• The experiment involves n identical trials
• Each trial has only two possible outcomes:
success and failure
• Each trial is independent of the previous
trials
• The terms p and q remain constant
throughout the experiment
– p is the probability of a success on any one trial
– q = (1-p) is the probability of a failure on any one
trial
© 2002 Thomson / South-Western
Slide 5-7
Assumptions of the Binomial
Distribution, continued
• In the n trials X is the number of
successes possible where X is a whole
number between 0 and n.
• Applications
– Sampling with replacement
– Sampling without replacement causes p
to change but if the sample size n < 5%
N, the independence assumption is not
a great concern.
© 2002 Thomson / South-Western
Slide 5-8
Binomial Distribution: Development
• Experiment: randomly select, with
replacement, two families from the
residents of the four-family town
• Success is ‘Children in Household:’ p =
0.75
• Failure is ‘No Children in Household:’ q =
1- p = 0.25
• X is the number of families in the sample
with ‘Children in Household’
© 2002 Thomson / South-Western
Slide 5-9
Binomial Distribution:
Development continued (2)
Family
A
B
C
D
Children in
Household
Number of
Automobiles
Yes
Yes
No
Yes
3
2
1
2
© 2002 Thomson / South-Western
Listing of Sample Space
(A,B), (A,C), (A,D), (D,D),
(B,A), (B,B), (B,C), (B,D),
(C,A), (C,B), (C,C), (C,D),
(D,A), (D,B), (D,C), (D,D)
Slide 5-10
Binomial Distribution:
Development continued (3)
• Families A, B, and D
have children in the
household; family C
does not
• Success is ‘Children in
Household:’ p = 0.75
• Failure is ‘No Children in
Household:’ q = 1- p =
0.25
• X is the number of
families in the sample
with ‘Children in
Household’
© 2002 Thomson / South-Western
Listing of
Sample
Space
P(outcome)
(A,B),
(A,C),
(A,D),
(D,D),
(B,A),
(B,B),
(B,C),
(B,D),
(C,A),
(C,B),
(C,C),
(C,D),
(D,A),
(D,B),
(D,C),
(D,D)
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
X
2
1
2
2
2
2
1
2
1
1
0
1
2
2
1
2
Slide 5-11
Binomial Distribution:
Development continued (4)
Listing of
Sample
Space
P(outcome)
(A,B),
(A,C),
(A,D),
(D,D),
(B,A),
(B,B),
(B,C),
(B,D),
(C,A),
(C,B),
(C,C),
(C,D),
(D,A),
(D,B),
(D,C),
(D,D)
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
1/16
© 2002 Thomson / South-Western
X
P(X)
X
2
1
2
2
2
2
1
2
1
1
0
1
2
2
1
2
0
1
2
1/16
6/16
9/16
1
x
n x
n!
P( X ) 
pq
X ! n  X  !
2!
1
0
20
P( X  0) 
 0.0625 
0! 2  0 ! .75 .25
16
2!
3
1
2 1
P( X  1) 
 0.375 
1! 2  1 ! .75 .25
16
2!
9
2
22
P( X  2) 
 0.5625 
2! 2  2 ! .75 .25
16
Slide 5-12
Binomial Distribution:
Development continued (5)
• Families A, B, and D
have children in the
household; family C
does not
• Success is Children in
Household: p = 0.75
• Failure is No Children
in Household, q = 1p = 0.25
• X is the number of
families in the sample
with Children in
Household
© 2002 Thomson / South-Western
Possible
Sequences
P(sequence)
(F,F)
(.25)(.25)  (.25)2
0
(S,F)
(.75)(.25)
1
(F,S)
(.25)(.75)
1
(S,S)
(.75)(.75)  (.75)2
2
X
Slide 5-13
Binomial Distribution:
Development continued (6)
Possible
Sequences
P(sequence)
(F,F)
X
X
(.25)(. 25)  (.25)2
0
0
(S,F)
(.75)(.25)
1
1
(F,S)
(.25)(.75)
1
2
(S,S)
(.75)(.75)  (.75)2
2
P( X  0) 
2!
0
2 0
 0.0625
.
75
.
25


0! 2  0 !
P(X)
(.25)(. 25)  (.25)2 =0.0625
2 (.25)(.75) =0.375
(.75)(.75)  (.75)2 =0.5625
x
n x
n!
P( X ) 
pq
X ! n  X  !
P( X  1) 
2!
1
2 1
 0.375
.
75
.
25


1! 2  1 !
2!
2
22
P( X  2) 
 0.5625
2! 2  2 ! .75 .25
© 2002 Thomson / South-Western
Slide 5-14
Binomial Distribution:
Demonstration Problem 5.2
n  20
p . 06
q . 94
P( X  2 )  P( X  0 )  P( X  1)  P( X  2 )
. 2901. 3703. 2246 . 8850
20!
P( X  0) 
0!(20  0)!
20!
P( X  1) 
1!(20  1)!
.06 .94
20 0
0
.06 .94
20!
P ( X  2) 
2!(20  2)!
201
1
.06 .94
© 2002 Thomson / South-Western
2
 (1)(1)(.2901) .2901
 (20)(.06)(.3086) .3703
20 2
 (190)(.0036)(.3283) .2246
Slide 5-15
Binomial
Table
© 2002 Thomson / South-Western
n = 20
X
0.1
0.2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
0.122
0.270
0.285
0.190
0.090
0.032
0.009
0.002
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.012
0.058
0.137
0.205
0.218
0.175
0.109
0.055
0.022
0.007
0.002
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
PROBABILITY
0.3
0.4
0.5
0.001
0.007
0.028
0.072
0.130
0.179
0.192
0.164
0.114
0.065
0.031
0.012
0.004
0.001
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.003
0.012
0.035
0.075
0.124
0.166
0.180
0.160
0.117
0.071
0.035
0.015
0.005
0.001
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.005
0.015
0.037
0.074
0.120
0.160
0.176
0.160
0.120
0.074
0.037
0.015
0.005
0.001
0.000
0.000
0.000
0.6
0.7
0.8
0.9
0.000
0.000
0.000
0.000
0.000
0.001
0.005
0.015
0.035
0.071
0.117
0.160
0.180
0.166
0.124
0.075
0.035
0.012
0.003
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.004
0.012
0.031
0.065
0.114
0.164
0.192
0.179
0.130
0.072
0.028
0.007
0.001
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.002
0.007
0.022
0.055
0.109
0.175
0.218
0.205
0.137
0.058
0.012
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.002
0.009
0.032
0.090
0.190
0.285
0.270
0.122
Slide 5-16
n = 20
X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
PROBABILITY
0.1
0.2
0.3
0.122
0.270
0.285
0.190
0.090
0.032
0.009
0.002
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.012
0.058
0.137
0.205
0.218
0.175
0.109
0.055
0.022
0.007
0.002
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.001
0.007
0.028
0.072
0.130
0.179
0.192
0.164
0.114
0.065
0.031
0.012
0.004
0.001
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.4
0.000
0.000
0.003
0.012
0.035
0.075
0.124
0.166
0.180
0.160
0.117
0.071
0.035
0.015
0.005
0.001
0.000
0.000
0.000
0.000
0.000
© 2002 Thomson / South-Western
Using the
Binomial Table:
Demonstration
Problem 5.3
n  20
p .40
P ( X  10)  20C10
.40 .60
10
10
 01171
.
Slide 5-17
Binomial Distribution
• Probability
function
P( X ) 
• Mean
value
• Variance
and
standard
deviation
X
n X
n!
q
for 0  X  n
p
X ! n  X  !
  n p


© 2002 Thomson / South-Western
2
 n pq


2
 n pq
Slide 5-18
Graphs of Selected
Binomial Distributions
n = 4 PROBABILITY
X
0.1
0.5
0
0.656
0.063
1
0.292
0.250
2
0.049
0.375
3
0.004
0.250
4
0.000
0.063
0.9
0.000
0.004
0.049
0.292
0.656
P(X)
P = 0.5
1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
0
0
1
2
3
2
3
X
4
P = 0.9
1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
P(X)
P(X)
P = 0.1
1
2
© 2002 Thomson / South-Western
3
X
4
1.000
0.900
0.800
0.700
0.600
0.500
0.400
0.300
0.200
0.100
0.000
0
1
X
4
Slide 5-19
Assumptions of the
Poisson Distribution
• Describes discrete occurrences over a
continuum or interval
• A discrete distribution
• Describes rare events
• Each occurrence is independent 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.
© 2002 Thomson / South-Western
Slide 5-20
Poisson Distribution
• Probability function
e
X
P( X ) 

X!
for X  0,1, 2, 3,...
where:
  long  run average
e  2. 718282... (the base of natural logarithms )
 Mean value

 Variance
© 2002 Thomson / South-Western

 Standard deviation

Slide 5-21
Poisson Distribution:
Demonstration Problem 5.7
  3. 2 customers / 4 minutes
  3. 2 customers / 4 minutes
X = 10 customers / 8 minutes
X = 6 customers / 8 minutes
Adjusted 
Adjusted 
 = 6. 4 customers / 8 minutes
 = 6. 4 customers / 8 minutes
P(X) =  e
P(X) =  e
X

X
X!

X!
P( X = 10 ) = 6.4 e
10 !
10
6 . 4
© 2002 Thomson / South-Western
P( X = 6) = 6.4 e
6!
6
 0. 0528
6.4
 0.1586
Slide 5-22
Poisson Distribution: Probability Table

X
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
0.5
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
1.5
0.2231
0.3347
0.2510
0.1255
0.0471
0.0141
0.0035
0.0008
0.0001
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
© 2002 Thomson / South-Western
1.6
0.2019
0.3230
0.2584
0.1378
0.0551
0.0176
0.0047
0.0011
0.0002
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
3.0
0.0498
0.1494
0.2240
0.2240
0.1680
0.1008
0.0504
0.0216
0.0081
0.0027
0.0008
0.0002
0.0001
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
3.2
0.0408
0.1304
0.2087
0.2226
0.1781
0.1140
0.0608
0.0278
0.0111
0.0040
0.0013
0.0004
0.0001
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
6.4
0.0017
0.0106
0.0340
0.0726
0.1162
0.1487
0.1586
0.1450
0.1160
0.0825
0.0528
0.0307
0.0164
0.0081
0.0037
0.0016
0.0006
0.0002
0.0001
6.5
0.0015
0.0098
0.0318
0.0688
0.1118
0.1454
0.1575
0.1462
0.1188
0.0858
0.0558
0.0330
0.0179
0.0089
0.0041
0.0018
0.0007
0.0003
0.0001
7.0
0.0009
0.0064
0.0223
0.0521
0.0912
0.1277
0.1490
0.1490
0.1304
0.1014
0.0710
0.0452
0.0263
0.0142
0.0071
0.0033
0.0014
0.0006
0.0002
8.0
0.0003
0.0027
0.0107
0.0286
0.0573
0.0916
0.1221
0.1396
0.1396
0.1241
0.0993
0.0722
0.0481
0.0296
0.0169
0.0090
0.0045
0.0021
0.0009
Slide 5-23
Using the Poisson Tables:
Demonstration Problem 5.7

X
0
1
2
3
4
5
6
7
8
9
10
11
12
0.5
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
1.5
0.2231
0.3347
0.2510
0.1255
0.0471
0.0141
0.0035
0.0008
0.0001
0.0000
0.0000
0.0000
0.0000
© 2002 Thomson / South-Western
1.6
0.2019
0.3230
0.2584
0.1378
0.0551
0.0176
0.0047
0.0011
0.0002
0.0000
0.0000
0.0000
0.0000
3.0
0.0498
0.1494
0.2240
0.2240
0.1680
0.1008
0.0504
0.0216
0.0081
0.0027
0.0008
0.0002
0.0001
  1. 6
P( X  4 )  0. 0551
Slide 5-24
Poisson Distribution: Graphs
 1. 6
0.35
0.30
0.14
0.25
0.12
0.20
0.10
0.08
0.15
0.06
0.10
0.04
0.05
0.00
0
  6. 5
0.16
0.02
1
2
3
4
5
© 2002 Thomson / South-Western
6
7
8
0.00
0
2
4
6
8
10
12
14
Slide 5-25
16
Assumptions of the
Hypergeometric Distribution
• It is a discrete distribution.
• Sampling is done without replacement.
• The number of objects in the population, N, is
finite and known.
• Each trial has exactly two possible outcomes:
success and failure.
• Trials are not independent
• X is the number of successes in the n trials
© 2002 Thomson / South-Western
Slide 5-26
Hypergeometric Distribution
• Probability function
– N is population size
P( x ) 
– n is sample size
– A is number of successes in
population
– x is number of successes in sample
• Mean
value
 ACx  N  ACn  x 
Cn
N
An

N
• Variance and standard deviation


© 2002 Thomson / South-Western
2


A( N  A) n( N  n)

N
2
( N  1)
2
Slide 5-27
Hypergeometric Distribution:
Probability Computations
N = 24
P( x  3) 
X=8
 ACx  N  ACn  x 
Cn
N
n=5
x
P(x)
0 0.1028
1 0.3426
2 0.3689
3 0.1581
4 0.0264

 8C 3 24  8C5  3
C5
 56120

42,504
.1581
24
5 0.0013
© 2002 Thomson / South-Western
Slide 5-28
Hypergeometric Distribution: Graph
N = 24
0.40
X=8
0.35
n=5
0.30
0.25
x
P(x)
0
0.1028
1
0.3426
2
0.3689
3
0.1581
4
0.0264
5
0.0013
© 2002 Thomson / South-Western
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
Slide 5-29
5
Hypergeometric Distributio:
Demonstration Problem 5.8
N = 18
n=3
A = 12
X
0
1
2
3
P(X)
0.0245
0.2206
0.4853
0.2696
P ( x  1)  P ( x  1)  P ( x  2)  P ( x  3)

 12C1 18  12C 3  1

C3
.2206.4853.2696
.9755
18
© 2002 Thomson / South-Western
 12C 2 18  12C 3  2
18
C3

 12C 3 18  12C 3  3
18
C3
Slide 5-30
The Hypergeometric Distribution
and the Binomial Distribution
• Because the hypergeometric distribution is
described by three parameters N, A and n,
it is practically impossible to create tables
for easy use.
• The binomial (which has tables) is an
acceptable approximation, if n < 5% N.
Otherwise it is not.
• Excel has eliminated all the tedious
calculations and allows the user to
compute the exact probabilities for the
hypergeometric.
© 2002 Thomson / South-Western
Slide 5-31