Discrete

Distributions

1
 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 calculate probabilities based on the binomial
distribution.

2
Random Variable

A variable which contains the outcomes of a chance

experiment

Focusing our attention on the numerical features of the

elementary outcomes, we introduce the idea of a random
variable.

A random variable X associates numerical values with

each elementary outcome of an experiment.
A quantity resulting from an experiment that, by chance,
can assume different values.

3
The numerical values are determined by
some characteristic of the elementary
outcome, and typically it will vary from
outcome to outcome. The word random
serves to emphasize the fact that before the
experiment is performed, we do not know the
specific outcome and, consequently, its
associated value of X. The following examples
illustrate the concept of a random variable.

4
Experiment:

Outcome Value of X
HHH 3
HHT 2
HTH 2
HTT 1
THH 2
THT 1
TTH 1
TTT 0

5
For each elementary outcome, there is only
one value for X. However, several elementary
outcomes may yield the same value.
Scanning our list, we now identify the events
(the collections of elementary outcomes) that
correspond to distinct values of X.

6
Numerical Value of X as Composition of the
an Event Event
X =0 = {TTT}

X=1 = {HTT, THT, TTH}

X=2 = {HHT, HTH, THH}

X=3 = {HHH}

7
The random variable X, the number of heads in three tosses
of a coin, defines a correspondence between the collections of
elementary outcomes and the real numbers 0, 1, 2, and 3.

Guided by this example, we have two general facts:

The events corresponding to the distinct values of X are

incompatible; that is, any two of these events cannot occur
together.

The union of these events is the entire sample space.

8
 Discrete vs. Continuous Distributions

Random Variable - a variable which contains the

outcomes of a chance experiment
Discrete Random Variable – A random variable that
only takes on distinct values
ex: Number of heads on 10 flips, Number of defective items
in a random sample of 100, Number of times you check
your watch during class, etc.
Continuous Random Variable – A random variable
that takes on infinite values by increasing precision.
For each two values, there always exists a valid value
in between them.
ex: Time until a bulb goes out, height, etc.
9
 Describing a Distribution

A distribution can be described by constructing a

graph of the distribution
Measures of central tendency and variability can be
applied to distributions

10
 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)
Mean of a discrete distribution
µ = ∑ (Xi * P(Xi))
where µ is the long run average,
Xi = the ith outcome

11
 Describing a Discrete Distribution

Variance of a discrete distribution is obtained in a

manner similar to raw data, summing the squared
deviations from the mean and weighting them by
P(Xi) (rather than dividing by n):
Var(Xi) = ∑ ((Xi – m)2* P(Xi))
Standard Deviation is computed by taking the square
root of the variance

12
 Discrete Distribution -- Example

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 on the following slide 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.

13
 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

14
 Mean and Standard Deviation
of a Discrete Distribution
# Crises (X) Prob (P) XP X-m (X-m2 (X-m2P
0 0.37 0 -1.15 1.32 0.49
1 0.31 0.31 -0.15 0.02 0.01
2 0.18 0.36 0.85 0.72 0.13
3 0.09 0.27 1.85 3.42 0.31
4 0.04 0.16 2.85 8.12 0.32
5 0.01 0.05 3.85 14.82 0.15
SUM: 1 1.15 8.1 28.44 1.41
m   XP  1.15
  2    X  m 2
P X   1.41  1.19

15
 Requirements for a Discrete
Probability Function -- Examples
- Each probability must be between 0 and 1
- The sum of all probabilities must be equal to 1.
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
VALID NOT NOT
VALID VALID
16
 Binomial Distribution
The binomial distribution is a discrete distribution
where X is the number of “successes” and the
following four conditions are met:
There are n trials
The n trials are independent of each other
The outcome is dichotomous – only two outcomes possible
The probability of “success” is constant
Example, 10 coin flips, X = # of heads
X = the number of “successes” and we say X follows a
Binomial distribution with n trial and P(success) = p
If the data follow a binomial distribution, then we
can summarize P(Xi) for all values of Xi = 1, …, n
through the binomial probability distribution formula
17
 Binomial Distribution

Probability P( X ) 
n! X
p q n  X
function X !n  X !
for 0  X  n, q  1  p
Mean value m  n p
Variance
and  2  n pq
Standard
   2  n pq
Deviation

18
 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
X is the number of families in the sample with
‘Children in Household’

Children in Number of Listing of Sample Space

Family Household Automobiles
(A,B), (A,C), (A,D), (A,A),
A Yes 3
(B,A), (B,B), (B,C), (B,D),
B Yes 2
(C,A), (C,B), (C,C), (C,D),
C No 1
(D,A), (D,B), (D,C), (D,D)
D Yes 2

19
 Binomial Distribution: Development
Continued
Listing of
Families A, B, and D have Sample Space
P(outcome)
X
children in the household;
family C does not (A,B), 1/16 2
Success is ‘Children in (A,C), 1/16 1
(A,D), 1/16 2
Household:’ p = 0.75 (A,A) 1/16 2
Failure is ‘No Children in (B,A),
(B,B),
1/16
1/16
2
2
Household:’ q = 1- p = 0.25 (B,C), 1/16 1
X is the number of families (B,D),
(C,A),
1/16
1/16
2
1
in the sample with ‘Children (C,B), 1/16 1
in Household’ (C,C), 1/16 0
(C,D), 1/16 1
(D,A), 1/16 2
(D,B), 1/16 2
(D,C), 1/16 1
(D,D) 1/16 2

20
Binomial Distribution: Development
Continued
Families A, B, and D have Possible
children in the household; Sequences
P(sequence)
X
family C does not
Success is ‘Children in
Household:’ p = 0.75 (F,F) (.25)(.25)  (.25)2 0

Failure is ‘No Children in (S,F) (.75)(.25) 1

Household:’ q = 1- p = 0.25
X is the number of families in (F,S) (.25)(.75) 1
the sample with ‘Children in
Household’ (S,S) (.75)(.75)  (.75)2 2

21
 Binomial Distribution: Development
Continued
Possible
Possible
P(sequence)
P(sequence) P(X)
P(X)
Sequences
Sequences X
X X
X

25)) (.25) 2 25)) (.25) 2 =0.0625

2 2
(F,F)
(F,F) (.(.25
25)(.
)(.25 00 00 (.(.25
25)(.
)(.25 =0.0625
(S,F)
(S,F) (.(.75
75)(.
)(.25
25)) 11 11 22(.(.25
25)(.
)(.75
75))=0.375
=0.375

75))(.75) 2 =0.5625
2
(F,S)
(F,S) (.(.25
25)(.
)(.75
75)) 11 22 (.(.75
75)(.
)(.75 =0.5625

75))(.75) 2
2
(S,S)
(S,S) (.(.75
75)(.
)(.75 22
nn!! xx nnxx
P X)) 
P(( X pq
X !! nn  X
X X  !!
22!! 2200 22!! 2211
X  00))   00..0625 X 11))  
00 11
P
P(( X .75 .25 0625 P
P(( X  .75 .25  00..375
375
 
0! 2  0 !
0! 2 0 ! 11!! 22 
1 !
1 !

22!! 2222
X  22))   00..5625
22
P
P(( X
 
22!! 22  22 !!.75 .25 5625

22
 Binomial Distribution:
Demonstration Problem
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?

23
 Binomial Distribution:
Demonstration Problem
In this 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? => P(X≤2)
The hard part of this problem is identifying p, n, and x –
emphasize this when studying the problems.

24
 Binomial Distribution:
Demonstration Problem
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)   .06  .94  (1)(1)(.2901)  .2901
0!(20  0)!

20! 1 20  1
P( X  1)    
.06 .94  (20)(.06)(.3086)  .3703
1!(20  1)!

20! 2 20  2
P( X  2)   .06  .94  (190)(.0036)(.3283)  .2246
2!(20  2)!

25
 Binomial Distribution:
Demonstration Problem

What are the mean and standard deviation of this

distribution?
m  n  p  (20)(.06)  1.20
 2  n  p  q  (20)(.06)(.94)  1.128
   2  1.128  1.062

26
 Binomial Distribution using Table: U.S.
Census Bureau Problem

n  20
n = 20 PROBABILITY
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
2 0.1887 0.2246 0.2521
P( X  2 )  P( X  0 )  P( X  1)  P( X  2 )
3 0.0596 0.0860 0.1139 . 2901. 3703. 2246 . 8850
4 0.0133 0.0233 0.0364
5 0.0022 0.0048 0.0088 P( X  2)  1  P( X  2)  1. 8850 .1150
6 0.0003 0.0008 0.0017
7
8
0.0000 0.0001
0.0000 0.0000
0.0002
0.0000
m  n  p  (20)(. 06)  1. 20
…
20
… …
0.0000 0.0000
…
0.0000  2
 n  p  q  ( 20 )(. 06)(. 94)  1.128

   2
 1.128  1. 062

27
 Binomial Distribution Table:
Demonstration Problem

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
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

28
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)

29
 Example:

Solve the binomial probability for n=20, p=.40, and

x=10 by using Binomial Probability Distribution.

30
 Using the Binomial Table
Solution:
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 n  20
11 0.000 0.000 0.012 0.071 p .40
12 0.000 0.000 0.004 0.035
13
14
0.000
0.000
0.000
0.000
0.001
0.000
0.015
0.005
P ( X  10)  20C10 .40 .60
10 10
 01171
.
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 31
 Graphs of Selected Binomial
Distributions
n = 4 PROBABILITY 1.000
P = 0.5

X 0.1 0.5 0.9 0.900

0.800

0 0.656 0.063 0.000 0.700

0.600
0.500
1 0.292 0.250 0.004

P(X)
0.400
0.300
2 0.049 0.375 0.049 0.200
0.100
3 0.004 0.250 0.292 0.000
0 1 2 3 4
4 0.000 0.063 0.656 X

P = 0.1 P = 0.9
1.000 1.000
0.900 0.900
0.800 0.800
0.700 0.700
0.600 0.600
0.500 0.500
P(X)
P(X)

0.400 0.400
0.300 0.300
0.200 0.200
0.100 0.100
0.000 0.000
0 1 2 3 4 0 1 2 3 4
X X

32
Example:

Purchasing magazine reported the result of a survey in which

buyers were asked a series of questions with regard to
Internet usages. One question asked was how they would use
the Internet if security and other issue could be resolved.
78% said they would use it for pricing information, 75% said
they would use it to send purchase orders, and 70% said they
would use it for purchase order acknowledgements. Assume
that these percentages hold true for all buyers. A researcher
randomly samples 20 buyers and asks them how they would
use the Internet if security and other issues could be
resolved.

33
Questions:

What is the probability that exactly 14 of these buyers would

use the Internet for pricing information?

What is the probability that all of the buyers would-use the

Internet to send purchase orders?

What is probability that fewer than 12 would use the

Internet for purchase order acknowledgements?

34
Solution:
a) n = 20 p = .78 x = 14
20C (.78)14(.22)6 = 38,760(.030855)(.00011338)
14
= .1356

b) n = 20 p = .75 x = 20
20C (.75)20(.25)0 = (1)(.0031712)(1) = .0032
20

c) n = 20 p = .70 x < 12
Use table :
P(x=0) + P(x=1) + . . . + P(x=11) =

0.000 + 0.000 + 0.000 + 0.000 + 0.000 + 0.000 +

0.000 + 0.001 + 0.004 + 0.012 +0 .031 + 0.065
= .113
35
Example:
The Wall Street Journal reported some interesting
statistics on the job market. One statistic is that
40% of all workers say they would change jobs for
“slightly higher pay”. In addition, 88% of
companies say that there is a shortage of qualified
job candidates. Suppose 16 workers are randomly
selected and asked if they would change jobs for
“slightly higher pay”. What is the probability that
nine or more say yes? What is the probability that
three, four, five, or six say yes? If 13 companies
are contacted, what is the probability that exactly
10 say there is a shortage of qualified job
candidates? What is the probability that all of the
companies say there is a shortage of qualified job
candidates? What is the expected number of
companies that would say there is a shortage of
qualified job candidates?

36
Solution:

n = 16 p = .40
P(x > 9): from Table

x Prob
9 .084
10 .039
11 .014
12 .004
13 .001
.142

37
Solution:
p(3 < x < 6):
x Prob
3 .047
4 .101
5 .162
6 .198
.508
n = 13 p = .88

P(x = 10) = 13C10(.88)10(.12)3 = 286(.278500976)(.001728)

= .1376

P(x = 13) = 13C13(.88)13(.12)0 = (1)(.1897906171)(1) = .1898

Expected Value = µ = n p = 13(.88) = 11.44

38
Question 5.22
Harley Davidson, director of quality control for the
Kyoto motor company is conducting his monthly spot
check of automatic transmissions. In this procedure,
10 transmissions are removed from the pool of
components and are checked for manufacturing
defects. Historically, only 2 percent of the
transmissions have such flows. (Assume that flaws
occur independently in different transmissions.)

(a) What is the probability that Harley’s sample

contains more than two transmissions with
manufacturing flaws? (Do not use the tables.)

(b) What is the probability that none of the selected

transmissions has any manufacturing flaws? (Do
not use the tables.)
39
Solution:
P (more than 2 flaws)
= 1 – P(0 flaws) – P(1 flaw) – P(2 flaws)
= 1  (0.02)0 (0.98)10  (0.02)1(0.98)9
 (0.02)2(0.98)8
= 1  0.8171  0.1667  0.0153 = 0.0009

P (0 flaws) = (.02)0 (.98)10 = 0.8171

40
Question 5.23
Diane Burns is the mayor of a large city. Lately, she has
become concerned about the possibility that large numbers of
people who are drawing unemployment checks are secretly
employed. Her assistants estimate that 40 percent of
unemployment beneficiaries fall into this category, but Ms.
Bruns is not convinced. She asks one of her aides to conduct a
quite investigation of 10 randomly selected unemployment
beneficiaries.

(a) If the mayer’s assistants are correct, what is the probability

that more than eight of individuals investigated have jobs?
(b) If the mayor’s assistants are correct, what is the probability
that only three of the investigated have jobs?

41
Solution:

a) P (more than 8 have jobs)

= P(r = 9) + P(r = 10)
= (.40)9 (.60)1 + (.40)10 (.60)0
= .0016 + .0001 = .0017.

b) P (3 have jobs) = (.40)3(.60)7 =.2150.

42
 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

43
 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 distribution contains a lambda value from
which the probabilities are determined
A Poisson distribution can be described by λ alone

44
 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.

45
 Poisson Distribution: Applications

Arrivals at queuing systems

airports -- people, airplanes, automobiles,
baggage
banks -- people, automobiles, loan applications
computer file servers -- read and write operations
Defects in manufactured goods
number of defects per 1,000 feet of extruded
copper wire
number of blemishes per square foot of painted
surface
number of errors per typed page

46
 Poisson Distribution

Probability function

 X e 
P( X )  for X  0,1,2,3,...
X!
where :
  longrun average
e  2.718282... (the base of natural logarithms )
 Mean value  Variance  Standard deviation

  
47
 Poisson Distribution:
Demonstration Problem
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?

A bank has an average random arrival rate of

3.2 customers every 4 minutes. What is the
probability of getting exactly 10 customers
during on 8-minute interval?

48
 Poisson Distribution:
Demonstration Problem - Solution
λ = 3.2 customers/4 minutes
We want to calculate P(X > 7 customers/4 minutes)
The problem can either be solved as:
P(X>7) = P(X=8) + P(X=9) + …, or
P(X>7) = 1 – P(X≤7) = 1 – [P(X=7) + P(X=6) + … + P(X=0)]
The answer can be obtained directly or through software
The answer you get is 1.7% of the time.
Bank officers could use these results to help them make
staffing decisions.

49
Solution:

  3.2 customers/ 4 minutes

X = 7 customers/ 4 minutes

 e
X 

P(X) =
X!
8 3.2 9 3.2 10 3.2

P( X  7) = 3.2 e  3.2 e  3.2 e  0.016333

8! 9! 10!

50
Solution:

  3.2 customers/ 4 minutes

X = 10 customers/ 8 minutes
Adjusted 
 = 6.4 customers/ 8 minutes
 e
X 

P(X) =
X!
10  6.4

P ( X = 10) = 6.4 e  0.0528

10!

51 51
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)

52
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 0.0169
15 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

53
Poisson Distribution: Using the Poisson
Tables
If a real estate office sells 1.6 houses on an
average weekday and sales of houses on
weekdays are Poisson distributed, what is
the probability of selling exactly four houses
in one day?
What is the probability of selling no houses
in one day?
What is the probability of selling more than
five houses in one day?
What is the probability of selling 2 or more
houses in one day?
54
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

  1. 6
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
7
0.0000
0.0000
0.0035
0.0008
0.0047
0.0011
0.0504
0.0216
P( X  4 )  0. 0551
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

55
 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

56
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
57
 Poisson Distribution: Graphs

 1. 6   6. 5
0.35
0.16
0.30 0.14
0.25 0.12

0.20 0.10
0.08
0.15
0.06
0.10
0.04
0.05 0.02
0.00 0.00
0 1 2 3 4 5 6 7 8 0 2 4 6 8 10 12 14 16

58
Problem:
According to the United National Environmental Program and
World Health Organization, in Bombay, India, air pollution
standards for particulate matter are exceeded an average of
5.6 days in every three-week period. Assume that the
distribution of number of days exceeding the standards per
three-week period is Poisson distributed.

What is the probability that the standard is not exceeded on

any day during a three-week period?

What is the probability that the standard is exceeded exactly

6 days of a three-week period?

What is the probability that the standard is exceeded exactly

15 or more days during a three-week period? If this outcome
actually occurred, what might you conclude?

59
Solution:

 = 5.6 days3 weeks

a) Prob(x=0   = 5.6):from Table = .0037

b) Prob(x=6   = 5.6):from Table A.3 = .1584

c) Prob(x > 15   = 5.6):

x Prob.
15 .0005
16 .0002
17 .0001
x > 15 .0008
Because this probability is so low, if it actually occurred,
the researcher would actually have to question the Lambda
value as too low for this period.
60
Problem:
A high percentage of people who fracture or
dislocate a bone see a doctor for that condition.
Suppose the percentage is 99%. Consider a sample
in which 300 people are randomly selected who
have fractured or dislocated a bone.

What is the probability that exactly five of them

did not see a doctor?

What is the probability that fewer than four of

them did not see a doctor?

What is the expected number of people who would

not see a doctor?

61
 Solution:

n = 300, p = .01,  = n(p) = 300(.01) = 3

a) Prob(x = 5):
Using  = 3 and Table = .1008

b) Prob (x < 4)
= Prob.(x = 0) + Prob.(x = 1) + Prob.(x = 2)
+ Prob.(x = 3)
= .0498 + .1494 + .2240 + .2240 = .6472

c) The expected number = µ =  = np = 3

62
 Problem:

The average number of annual trips per family to amusement parks

in the United State is Poisson distributed, with a mean of 0.6 trips per
year. What is the probability of randomly selecting an American
family and finding the following:

a) The family did not make a trip to an amusement park last year?
b) The family took exactly one trip to an amusement park last year?
c) The family took two or more trips to amusement parks last year?
d) The family took three or fewer trips to amusement parks over a
three-year period?
e) The family took exactly four trips to amusement parks during a
six-year period?

63
Solution:

 = 0.6 trips 1 year

a) Prob (x=0   = 0.6):

from Table = .5488

b) Prob (x=1   = 0.6):

from Table = .3293

64
c) Prob(x > 2   = 0.6):
from Table A.3
x Prob.
2 .0988
3 .0198
4 .0030
5 .0004
6 .0000

x > 2 .1220

65
Prob(x < 3  3 year period):
The interval length has been increased (3 times)
New Lambda =  = 1.8 trips3 years

Prob(x < 3   = 1.8):

from Table A.3 x Prob.
0 .1653
1 .2975
2 .2678
3 .1607
x < 3 .8913

66
Prob(x=4  6 years):
The interval has been increased (6 times)

New Lambda =  = 3.6 trips6 years

Prob(x=4   = 3.6):

from Table A.3 = .1912

67
Problem:
Ship collisions in the Houston Ship Channel are rare. Suppose the
numbers of collisions are Poisson distributed, with the mean of 1.2
collisions every four months.

What is probability of having no collisions occur over a four-month

period?
What is probability of having exactly two collisions in a two month
period?
What is probability of having one or fewer collisions in a six month
period? If this outcome occurred, what might you conclude about
ship channel conditions during this period? What might you
conclude about ship channel safety awareness during this period?
What might you conclude about weather conditions during this
period? What might you conclude about lambda?

68
Solution:
 = 1.2 collisions4 months
a) Prob(x=0   = 1.2):

from Table = .3012

b) Prob(x=2 2 months):

The interval has been decreased (by ½)

New Lambda =  = 0.6 collisions2 months

Prob(x=2   = 0.6):

from Table = .0988

69
Prob (x < 1 collision6 months):
The interval length has been increased (by 1.5)

New Lambda =  = 1.8 collisions6 months

Prob(x < 1  = 1.8):
from Table A.3 x Prob.
0 .1653
1 .2975

x < 1 .4628

The result is likely to happen almost half the time (46.26%).

Ship channel and weather conditions are about normal for this
period. Safety awareness is about normal for this period.
There is no compelling reason to reject the lambda value of 0.6
collisions per 4 months based on an outcome of 0 or 1 collision
per 6 months.
70
Problem:
A pen company arranges 1.2 defective pens per carton produced
(200 pens). The number of defects per carton is Poisson
distributed.

What is the probability of selecting a carton and finding no

defective pens?

What is the probability of finding eight or more defective pens

in a carton?

Suppose a purchaser of these pens will quit buying from the

company if a carton contains more than three defective pens.
What is the probability that a carton contains more than three
defective pens?

71
Solution:
 = 1.2 penscarton
a) Prob(x=0   = 1.2):

from Table A.3 = .3012

b) Prob(x > 8   = 1.2):

from Table A.3 = .0000

72
Prob(x > 3   = 1.2):
from Table A.3 x Prob.
4 .0260
5 .0062
6 .0012
7 .0002
8 .0000
x > 3 .0336

73
Question:
A high percentage of people who fracture or dislocate a bone see
a doctor for that condition. Suppose the percentage is 99%.
Consider a sample in which 300 people are randomly selected
who have fractured or dislocated a bone.
What is the probability that exactly five of them did not see a
doctor?

What is the probability that fewer than four of them did not see
a doctor?

What is the expected number of people who would not see a

doctor?

74
Solution:
n = 300, p = .01,  = n(p) = 300(.01) = 3

a) Prob(x = 5):
Using  = 3 and from Table = .1008

b) Prob (x < 4) = Prob.(x = 0) + Prob.(x = 1) +

Prob.(x = 2) + Prob.(x = 3)
= .0498 + .1494 + .2240 + .2240 = .6472

c) The expected number = µ =  = 3

75