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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
an Event
Composition of the
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
Number of
Crises
0
1
2
3
4
5
Probability
0.37
0.31
0.18
0.09
0.04
0.01
P
r
o
b
a
b
i
l
i
t
y
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
Number of Crises
14
Mean and Standard Deviation
of a Discrete Distribution
# Crises (X) Prob (P)
0
0.37
1
0.31
2
0.18
3
0.09
4
0.04
5
0.01
SUM:
1
XP
0
0.31
0.36
0.27
0.16
0.05
1.15
X-m
-1.15
-0.15
0.85
1.85
2.85
3.85
8.1
(X-m2 (X-m2P
1.32
0.49
0.02
0.01
0.72
0.13
3.42
0.31
8.12
0.32
14.82
0.15
28.44
1.41
m   XP  1.15
  2 
2


X

m
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
0
1
2
3
.1
.2
.4
.2
.1
1.0
-1
0
1
2
3
-.1
.3
.4
.3
.1
1.0
-1
0
1
2
3
.1
.3
.4
.3
.1
1.2
VALID
NOT
VALID
NOT
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
function
n!
X
n

X
P( X ) 

p q
X !n  X !
for 0  X  n, q  1  p
Mean value
m  n p
Variance
and
Standard
Deviation
 2  n pq
   2  n pq
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’
Family
Children in
Household
Number of
Automobiles
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)
19
Binomial Distribution: Development
Continued
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’
Listing of
Sample Space
P(outcome)
(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)
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
20
Binomial Distribution: Development
Continued
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’
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
21
Binomial Distribution: Development
Continued
Possible
Possible
Sequences
Sequences
P(sequence)
P(sequence)
(F,F)
(F,F)
X
X
X
X
2
(.(.25
25)(.
)(.25
25)) (.25) 2
00
00
(S,F)
(S,F)
(.(.75
75)(.
)(.25
25))
11
11
(F,S)
(F,S)
(.(.25
25)(.
)(.75
75))
11
22
(S,S)
(S,S)
2
(.(.75
75)(.
)(.75
75))(.75) 2
22
P
P(( X
X  00)) 
22!!
00
2200
 00..0625
0625
.
75
.
25


0!
2

0
!
0! 2  0 !
P
P(( X
X  22)) 
22!!
22
2222
 00..5625
5625
.
75
.
25


22!! 22  22 !!
P(X)
P(X)
2
(.(.25
25)(.
)(.25
25)) (.25) 2 =0.0625
=0.0625
25)(.
)(.75
75))=0.375
=0.375
22(.(.25
2
(.(.75
75)(.
)(.75
75))(.75) 2 =0.5625
=0.5625
xx nnxx
nn!!
P
P(( X
X)) 
pq
X
X !! nn  X
X  !!
P
P(( X
X
 11)) 

22!!
11
2211

 00..375
375
.
75
.
25

11!! 22 
1
!
1 !
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
P( X  0) 
20!
0
20  0
 (1)(1)(.2901)  .2901
 .06  .94
0!(20  0)!
P( X  1) 
20!
1
20  1
.
06
.
94
 (20)(.06)(.3086)  .3703
  
1!(20  1)!
20!
2
20  2
P( X  2) 
 (190)(.0036)(.3283)  .2246
 .06  .94
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
Example:
Solve the binomial probability for n=20, p=.40, and
x=10 by using Binomial Probability Distribution.
27
Using the Binomial Table
Solution:
n = 20
PROBABILITY
X
0.1
0.2
0.3
0.4
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
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
n  20
p .40
P ( X  10)  20C10
.40 .60
10
10
 01171
.
28
Binomial Distribution using Table: U.S.
Census Bureau Problem
n = 20
X
0
1
2
3
4
5
6
7
8
…
20
PROBABILITY
0.05
0.06
0.3585 0.2901
0.3774 0.3703
0.1887 0.2246
0.0596 0.0860
0.0133 0.0233
0.0022 0.0048
0.0003 0.0008
0.0000 0.0001
0.0000 0.0000
…
…
0.0000 0.0000
0.07
0.2342
0.3526
0.2521
0.1139
0.0364
0.0088
0.0017
0.0002
0.0000
…
0.0000
n  20
p . 06
q . 94
P( X  2 )  P( X  0 )  P( X  1)  P( X  2 )
. 2901. 3703. 2246 . 8850
P( X  2)  1  P( X  2)  1. 8850 .1150
m  n  p  (20)(. 06)  1. 20


2
 n  p  q  ( 20)(. 06)(. 94)  1.128


2
 1.128  1. 062
29
Binomial Distribution Table:
Demonstration Problem
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
n  20
p . 06
q . 94
P( X  2 )  P( X  0 )  P( X  1)  P( X  2 )
. 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
30
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)
31
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
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
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
9
10
11
12
13
Prob
.084
.039
.014
.004
.001
.142
37
Solution:
p(3 < x < 6):
x
3
4
5
6
Prob
.047
.101
.162
.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
P( X  7) = 3.2 e
8!
9
3.2
 3.2 e
9!
10
3.2
 3.2 e
10!
 0.016333
50
Solution:
  3.2 customers/ 4 minutes
X = 10 customers/ 8 minutes
Adjusted 
 = 6.4 customers/ 8 minutes
 e
X
P(X) =

X!
10
P ( X = 10) =
 6. 4
6.4 e
10!
 0.0528
51
51
Excel’s Poisson Function
=
X
1.6
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
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
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
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
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
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
55
Poisson Distribution:
Using the Poisson Tables

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
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  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
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
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  2 )  1  P( X  2 )  1  P( X  0)  P( X  1)
 1. 2019 . 3230 . 4751
57
Poisson Distribution: Graphs
 1. 6
0.35
0.30
  6. 5
0.16
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
2
3
4
5
6
Prob.
.0988
.0198
.0030
.0004
.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
4
5
6
7
8
x > 3 .0336
x
Prob.
.0260
.0062
.0012
.0002
.0000
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,
a)
p = .01,
 = n(p) = 300(.01) = 3
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