This chapter summary discusses discrete probability distributions. It distinguishes between discrete and continuous random variables and distributions. It describes how to determine the mean and variance of discrete distributions. It introduces some common discrete distributions like the binomial and Poisson distributions. For the binomial distribution, it explains how to calculate the probability of a given number of successes in a given number of trials. For the Poisson distribution, it provides the probability formula and explains that it models independent events occurring continuously over an interval.

1. Business Statistics, 6th ed.
by Ken Black
Chapter 5
Discrete
Probability
Distributions
Copyright 2010 John Wiley & Sons, Inc. 1
Copyright 2010 John Wiley & Sons, Inc.

2. Learning Objectives
Distinguish between discrete random variables 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 binomial distribution, and know
how to work such problems.
Copyright 2010 John Wiley & Sons, Inc. 2

3. Discrete vs. Continuous Distributions
Discrete distributions – constructed from discrete
(individually distinct) random variables
Continuous distributions – based on continuous
random variables
Random Variable - a variable which contains the
outcomes of a chance experiment
Copyright 2010 John Wiley & Sons, Inc. 3

4. Discrete vs. Continuous Distributions
Categories of Random Variables
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
Copyright 2010 John Wiley & Sons, Inc. 4

5. Describing a Discrete Distribution
A discrete distribution can be described by
constructing a graph of the distribution
Measures of central tendency and variability can be
applied to discrete distributions
Discrete values of outcomes are used to represent
themselves
Copyright 2010 John Wiley & Sons, Inc. 5

6. Describing a Discrete Distribution
Mean of discrete distribution – is the long run
average
If the process is repeated long enough, the average of the
outcomes will approach the long run average (mean)
Requires the process to eventually have a number which is
the product of many processes
Mean of a discrete distribution
µ = ∑ (X * P(X))
where (X) is the long run average;
X = outcome, P = Probability of X
Copyright 2010 John Wiley & Sons, Inc. 6

7. Describing a Discrete Distribution
Variance and Standard Deviation of a discrete
distribution are solved by using the outcomes (X) and
probabilities of outcomes (P(X)) in a manner similar
to computing a mean
Standard Deviation is computed by taking the square
root of the variance
Copyright 2010 John Wiley & Sons, Inc. 7

8. Some Special Distributions
Discrete
binomial
Poisson
Hypergeometric
Continuous
normal
uniform
exponential
t
chi-square
F
Copyright 2010 John Wiley & Sons, Inc. 8

9. Discrete Distribution -- Example
Observe the discrete distribution in the following
table.
An executive is considering out-of-town business
travel for a given Friday. At least one crisis could occur
on the day that the executive is gone. The distribution
contains the number of crises that could occur during
the day the executive is gone and the probability that
each number will occur. For example, there is a .37
probability that no crisis will occur, a .31 probability of
one crisis, and so on.
Copyright 2010 John Wiley & Sons, Inc. 9

10. Discrete Distribution -- Example
Distribution of Daily Crises
P 0.5
Number of r
Probability o 0.4
Crises
b
a 0.3
0 0.37 b
1 0.31 i 0.2
2 0.18 l
i 0.1
3 0.09
t
4 0.04 y 0
5 0.01 0 1 2 3 4 5
Number of Crises
Copyright 2010 John Wiley & Sons, Inc. 10

11. Variance and Standard Deviation
of a Discrete Distribution
2 2 2
X P( X ) 1.2 12 110
. .
2
X P(X)
2
X (X ) (X ) P( X )
-1 .1 -2 4 .4
0 .2 -1 1 .2
1 .4 0 0 .0
2 .2 1 1 .2
3 .1 2 4 .4
1.2
Copyright 2010 John Wiley & Sons, Inc. 11

12. Requirements for a Discrete
Probability Function -- Examples
X P(X) X P(X) X P(X)
-1 .1 -1 -.1 -1 .1
0 .2 0 .3 0 .3
1 .4 1 .4 1 .4
2 .2 2 .3 2 .3
3 .1 3 .1 3 .1
1.0 1.0 1.2
: YES NO NO
Copyright 2010 John Wiley & Sons, Inc. 12

13. Mean of a Discrete Distribution
E X X P( X )
X P(X) X P( X)
-1 .1 -.1
0 .2 .0
1 .4 .4
2 .2 .4
3 .1 .3
1.0 = 1.0
Copyright 2010 John Wiley & Sons, Inc. 13

14. Mean of the Crises Data Example
E X X P( X ) 115
.
X P(X) X P(X) P
r 0.5
0 .37 .00 o 0.4
b
1 .31 .31 a 0.3
b
2 .18 .36 0.2
i
l 0.1
3 .09 .27 i
t
0
4 .04 .16 0 1 2 3 4 5
y
Number of Crises
5 .01 .05
1.15
Copyright 2010 John Wiley & Sons, Inc. 14

15. Variance and Standard Deviation
of Crises Data Example
2
2 2
X P( X ) 141
. 141 119
. .
X P(X) (X- ) (X- (X- 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
Copyright 2010 John Wiley & Sons, Inc. 15

16. Binomial Distribution
Probability n! X n X
function P( X )
X! n X !
p q for 0 X n
Mean value
n p
Variance
and 2
n p q
Standard
2
Deviation n p q
Copyright 2010 John Wiley & Sons, Inc. 16

17. Binomial Distribution:
Demonstration Problem 5.3
According to the U.S. Census Bureau,
approximately 6% of all workers in Jackson,
Mississippi, are unemployed. In conducting a
random telephone survey in Jackson, what is the
probability of getting two or fewer unemployed
workers in a sample of 20?
Copyright 2010 John Wiley & Sons, Inc. 17

18. Binomial Distribution:
Demonstration Problem 5.3
In the following example,
6% are unemployed => p
The sample size is 20 => n
94% are employed => q
x is the number of successes desired
What is the probability of getting 2 or fewer unemployed
workers in the sample of 20?
The hard part of this problem is identifying p, n, and x –
emphasis this when studying the problems.
Copyright 2010 John Wiley & Sons, Inc. 18

19. Binomial Distribution:
Demonstration Problem 5.3
n 20
p . 06
q . 94
P( X 2 ) P( X 0) P( X 1) P( X 2)
. 2901 . 3703 . 2246 . 8850
20! 0 20 0
P( X 0)
0!(20 0)!
.06 .94 (1)(1)(. 2901 ) .2901
20! 1 20 1
P( X 1)
1!(20 1)!
.06 .94 (20)(.06)(.3086) .3703
20! 2 20 2
P( X 2)
2!(20 2)!
.06 .94 (190)(.0036)(.3283) .2246
Copyright 2010 John Wiley & Sons, Inc. 19

20. Binomial Distribution Table:
Demonstration Problem 5.3
n = 20 PROBABILITY
n 20
X 0.05 0.06 0.07 p . 06
0 0.3585 0.2901 0.2342 q . 94
1 0.3774 0.3703 0.3526 P( X 2 ) P( X 0) P( X 1) P( X 2)
2 0.1887 0.2246 0.2521 . 2901 . 3703 . 2246 . 8850
3 0.0596 0.0860 0.1139
4 0.0133 0.0233 0.0364
P( X 2) 1 P( X 2) 1 . 8850 .1150
5 0.0022 0.0048 0.0088
6 0.0003 0.0008 0.0017
n p (20)(. 06) 1. 20
2
7 0.0000 0.0001 0.0002
n p q ( 20 )(. 06 )(. 94 ) 1.128
8 0.0000 0.0000 0.0000
2
… … … … 1.128 1. 062
20 0.0000 0.0000 0.0000
Copyright 2010 John Wiley & Sons, Inc. 20

21. Excel’s 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)
Copyright 2010 John Wiley & Sons, Inc. 21

22. Minitab’s Binomial Function
X 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 Binomial with n = 23 and p = 0.64
8 0.003051
9 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
Copyright 2010 John Wiley & Sons, Inc. 22

23. Mean and Std Dev of Binomial Distribution
Binomial distribution has an expected value or a long
run average denoted by µ (mu)
If n items are sampled over and over for a long time and if
p is the probability of success in one trial, the average long
run of successes per sample is expected to be np
=> Mean µ = np
=> Std Dev = √(npq)
Copyright 2010 John Wiley & Sons, Inc. 23

24. Poisson Distribution
The Poisson distribution focuses only on the number
of discrete occurrences over some interval or
continuum
Poisson does not have a given number of trials (n)
as a binomial experiment does
Occurrences are independent of other occurrences
Occurrences occur over an interval
Copyright 2010 John Wiley & Sons, Inc. 24

25. Poisson Distribution
If Poisson distribution is studied over a long period
of time, a long run average can be determined
The average is denoted by lambda (λ)
Each Poisson problem contains a lambda value from which
the probabilities are determined
A Poisson distribution can be described by λ alone
Copyright 2010 John Wiley & Sons, Inc. 25

26. Poisson Distribution
Probability function
X
P( X ) e for X 0,1,2,3,...
X!
where :
long run average
e 2.718282... (the base of natural logarithms)
 Mean value  Variance  Standard deviation
Copyright 2010 John Wiley & Sons, Inc. 26

27. Poisson Distribution:
Demonstration Problem 5.7
Bank customers arrive randomly on weekday
afternoons at an average of 3.2 customers
every 4 minutes. What is the probability of
having more than 7 customers in a 4-minute
interval on a weekday afternoon?
Copyright 2010 John Wiley & Sons, Inc. 27

28. Poisson Distribution:
Demonstration Problem 5.7
Solution
λ = 3.2 customers>minutes X > 7 customers/4 minutes
The solution requires obtaining the values of x = 8, 9, 10, 11, 12,
13, 14, . . . . Each x value is determined until the values are so
far away from λ = 3.2 that the probabilities approach zero. The
exact probabilities are summed to find x 7. If the bank has been
averaging 3.2 customers every 4 minutes on weekday
afternoons, it is unlikely that more than 7 people would
randomly arrive in any one 4-minute period. This answer
indicates that more than 7 people would randomly arrive in a
4-minute period only 1.69% of the time. Bank officers could use
these results to help them make staffing decisions.
Copyright 2010 John Wiley & Sons, Inc. 28

29. Poisson Distribution:
Demonstration Problem 5.7
3.2 customers/ minutes
4 3. 2 customers/ 4 minutes
X = 10 customers/ minutes
8 X = 6 customers/ 8 minutes
Adjusted Adjusted
= 6.4 customers/ minutes
8 = 6. 4 customers/ 8 minutes
X X
P(X) = e P(X) = e
X! X!
P( X = 6) = 6.4 e
10 6.4 6 6. 4
P( X = 10) = 6.4 e 0.0528 0.1586
10! 6!
Copyright 2010 John Wiley & Sons, Inc. 29

30. 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 0.0027
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
P( X 5) P ( X 6) P ( X 7) P ( X 8) P ( X 9)
.0047 .0011 .0002 .0000 .0060
Copyright 2010 John Wiley & Sons, Inc. 30

31. 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 0.0027
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
P( X 2) 1 P ( X 2) 1 P( X 0) P ( X 1)
1 .2019 .3230 .4751
Copyright 2010 John Wiley & Sons, Inc. 31

32. Excel’s 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)
Copyright 2010 John Wiley & Sons, Inc. 32

33. Minitab’s Poisson Function
X P(X =x)
0 0.149569
1 0.284180
2 0.269971 Poisson with mean = 1.9
3 0.170982
4 0.081216
5 0.030862
6 0.009773
7 0.002653
8 0.000630
9 0.000133
10 0.000025
Copyright 2010 John Wiley & Sons, Inc. 33

34. Mean and Std Dev of a Poisson Distribution
Mean of a Poisson Distribution is λ
Understanding the mean of a Poisson distribution
gives a feel for the actual occurrences that are likely
to happen
Variance of a Poisson distribution is also λ
Std Dev = Square root of λ
Copyright 2010 John Wiley & Sons, Inc. 34

35. Poisson Approximation
of the Binomial Distribution
Binomial problems with large sample sizes and small
values of p, which then generate rare events, are
potential candidates for use of the Poisson
Distribution
Rule of thumb, if n > 20 and np < 7, the
approximation is close enough to use the Poisson
distribution for binomial problems
Copyright 2010 John Wiley & Sons, Inc. 35

36. Poisson Approximation
of the Binomial Distribution
Procedure for Approximating binomial with Poisson
Begin with the computation of the binomial mean
distribution µ = np
Because µ is the expected value of the binomial, it becomes
λ for Poisson distribution
Use µ as the λ, and using the x from the binomial problem
allows for the approximation of the probabilities from the
Poisson table or Poisson formula
Copyright 2010 John Wiley & Sons, Inc. 36

37. Poisson Approximation
of the Binomial Distribution
Binomial probabilities are difficult to calculate when
n is large.
Under certain conditions binomial probabilities may
be approximated by Poisson probabilities.
If n 20 and n p 7, the approximation is acceptable
.
Poisson approximation Use n p.
Copyright 2010 John Wiley & Sons, Inc. 37

38. Hypergeometric Distribution
Sampling without replacement from a finite
population
The number of objects in the population is denoted N.
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,
if n < 5% N. Otherwise it is not.
Copyright 2010 John Wiley & Sons, Inc. 38

39. Hypergeometric Distribution
Probability function ACx N A Cn x
N is population size P( x )
n is sample size
N Cn
A is number of successes in population
x is number of successes in sample
Mean A n
Value N
Variance and standard 2 A( N A)n( N n)
deviation N
2
(N 1)
2
Copyright 2010 John Wiley & Sons, Inc. 39

40. Hypergeometric Distribution:
Probability Computations
N = 24 ACx N A Cn x
X=8
P( x 3)
N Cn
n=5
8 C3 24 8 C5 3
P(x)
x C5 24
0 0.1028
1 0.3426 56 120
2 0.3689 42,504
3 0.1581
4 0.0264 .1581
5 0.0013
Copyright 2010 John Wiley & Sons, Inc. 40

41. Excel’s 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)
Copyright 2010 John Wiley & Sons, Inc. 41

42. Minitab’s Hypergeometric Function
X P(X =x)
0 0.102767
1 0.342556
Hypergeometric with N = 24, A = 8, n = 5
2 0.368906
3 0.158103
4 0.026350
5 0.001318
Copyright 2010 John Wiley & Sons, Inc. 42