© 2012 McGraw-Hill
Ryerson Limited
© 2009 McGraw-Hill Ryerson Limited
1
Lind
Marchal
Wathen
Waite
© 2012 McGraw-Hill Ryerson Limited
2
Learning Objectives
LO 1 Define the terms probability distribution and random
variable.
LO 2 Distinguish between discrete and continuous
probability distributions.
LO 3 Calculate the mean, the variance, and the standard
deviation of a discrete probability distribution.
LO 4 Describe the characteristics of, and compute,
probabilities using the binomial probability
distribution.
© 2012 McGraw-Hill Ryerson Limited
3
Learning Objectives
LO 5 Describe the characteristics of, and compute,
probabilities using the hypergeometric probability
distribution.
LO 6 Describe the characteristics of, and compute,
probabilities using the Poisson probability
distribution.
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LO
1
What Is Probability Distribution?
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Probability Distribution
It gives the entire range of values that can occur based on
an experiment.
It is similar to a relative frequency distribution.
However, instead of describing the past, it describes a
possible future event and how likely that event is.
LO 1
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Example – Probability Distributions
Suppose we are now interested in getting a suit of cards
from a well-shuffled deck. What is the probability
distribution for the suit diamonds?
The possible results are: diamond, club, spade, heart
Probability of
Suit
Outcome, P(x)
diamond
club
heart
spade
Total
LO 1
1
 0.25
4
1
 0.25
4
1
 0.25
4
1
 0.25
4
4
1
4
© 2012 McGraw-Hill Ryerson Limited
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Solution – Probability Distributions
P(x)
Probability distribution
Probability
0.5
0.25
0
Spade
Club
Heart
Diamond
Suits
LO 1
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Solution – Probability Distributions
There are 52 possible outcomes.
LO 1
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Solution – Probability Distributions
Let E be the event “drawing a diamond”.
An examination of the sample space shows that there are
52 cards, of which 13 are diamonds, so
n(E) = 13 and n(S) = 52.
Hence the probability of event E occurring is
P(E) = 13 / 52 = 0.25.
LO 1
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Characteristics of a Probability Distribution
1. The probability of a particular outcome is between 0
and 1 inclusive.
2. The outcomes are mutually exclusive events.
3. The list is exhaustive. The sum of the probabilities of
the various events is equal to 1.
LO 1
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You Try It Out!
The possible outcomes of an experiment involving a card
drawn from a well-shuffled deck of 52 cards:
a) Develop a probability distribution for drawing a king.
b) Portray the probability distribution graphically.
c) What is the sum of the probabilities?
LO 1
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LO
2
Random Variables
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Random Variables
In any experiment of chance, the outcomes occur
randomly.
So the numerical value associated with an outcome is often
called a random variable.
LO 2
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Example – Probability Distributions
The following diagram illustrates the terms:
• experiment
• outcome
• event
• random variable
First, for the experiment of drawing a card from a wellshuffled deck of 52 cards, there are 52 possible outcomes.
In this experiment, we are interested in the event of
drawing a diamond.
LO 2
© 2012 McGraw-Hill Ryerson Limited
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Example – Probability Distributions
The random variable is a card drawn from a well-shuffled
deck of 52 cards. We want to know the probability of the
event given that the random variable equals 1.
The result is P (suit of diamond) = 0.25
LO 2
© 2012 McGraw-Hill Ryerson Limited
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Example – Probability Distributions
Discrete:
A variable that can assume only certain, clearly separated
values.
It is usually the result of counting something.
Examples:
1. The number of heads appearing when a coin is tossed
three times.
2. The number of students earning an A in this class.
LO 2
© 2012 McGraw-Hill Ryerson Limited
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Example – Probability Distributions
Continuous:
Can assume an infinite number of values within a given
range.
It is usually the result of some type of measurement.
Examples:
1. The length of each song on a CD.
2. The height of each student in the class.
LO 2
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LO
3
The Mean, Variance, and
Standard Deviation of a
Probability Distribution
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The Mean of a Probability Distribution
The mean is a typical value used to represent the central
location within a probability distribution.
It also is the long-run average value of the random variable
The mean of a probability distribution is also referred to as
its expected value.
It is a weighted average where the possible values of a
random variable are weighted by their corresponding
probabilities of occurrence.
LO 3
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The Variance and Standard Deviation of a
Probability Distribution
Measures the amount of spread in a distribution
The computational steps are:
1. Subtract the mean from each value, and square this
difference.
2. Multiply each squared difference by its probability.
3. Sum the resulting products to arrive at the variance.
Find the standard deviation by taking the positive square
root of the variance.
LO 3
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Example – Mean, Variance, and Standard Deviation
John Ragsdale sells new cars for Pelican Ford. John
usually sells the largest number of cars on a Saturday. He
has the following probability distribution for the number of
cars he expects to sell on a particular Saturday.
Number of
Cars Sold,
x
0
1
2
3
4
Total
LO 3
Probability,
P(x)
0.05
0.15
0.35
0.30
0.15
1.0
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Example – Probability Distributions
Continued
1. What type of distribution is this?
2. On a typical Saturday, how many cars does John expect
to sell?
3. What is the variance of the distribution?
LO 3
© 2012 McGraw-Hill Ryerson Limited
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Solution – Probability Distributions
1. This is a discrete probability distribution for the random
variable “number of cars sold”.
John expects to sell only within a certain range of cars.
Also, the outcomes are mutually exclusive. He cannot
sell a total of both three and four cars on the same
Saturday.
The sum of all possible outcomes totals 1.
Hence, these circumstances qualify as a probability
distribution.
LO 3
© 2012 McGraw-Hill Ryerson Limited
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Solution – Probability Distributions
Continued
2. The mean number of cars sold is computed by
weighting the number of cars sold by the probability of
selling that number and adding the product.
    xP( x)
 0(.05)  1(.15)  2(.35)  3(.3)  4(.15)
 2.35
LO 3
© 2012 McGraw-Hill Ryerson Limited
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Solution – Probability Distributions
Continued
The calculations are summarized in the following table.
Number of
Cars Sold,
x
0
1
2
3
4
Total
LO 3
Probability,
P(x)
0.05
0.15
0.35
0.30
0.15
1.0
xP(x)
0.00
0.15
0.70
0.90
0.60
  2.35
© 2012 McGraw-Hill Ryerson Limited
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Solution – Probability Distributions
Continued
Again, a table is useful for organizing the calculations for
the variance,
Number of
2
2
Probability,
x     x     x    P( x)
Cars Sold,

P(x)
x
0
0.05
0–2.35 5.5225 0.2761
1
0.15
1–2.35 1.8225 0.2734
2
0.35
2–2.35 0.1225 0.0429
3
0.30
3–2.35 0.4225 0.1268
4
0.15
4–2.35 2.7225 0.4084
 2  1.1275
   2  1.1275  1.0618
LO 3
© 2012 McGraw-Hill Ryerson Limited
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You Try It Out!
A soft drink seller offers three sizes of cola—small, medium,
and large—for wholesale. The colas are sold for $0.70, $1,
and $1.30, respectively. Twenty percent of the orders are for
small, 55 percent are for medium, and 25 percent are for the
large sizes. Organize the size of the colas and the
probability of a sale into a probability distribution.
a) Is this a discrete probability distribution? Indicate why or
why not.
b) Compute the mean amount charged for a cola.
c) What is the variance in the amount charged for a cola?
The standard deviation?
LO 3
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LO
4
The Binomial Probability Distribution
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The Binomial Probability Distribution
Characteristics of a Binomial Probability Distribution
1. An outcome on each trial of an experiment is classified
into one of two mutually exclusive categories—a
success or a failure.
2. The random variable counts the number of successes in
a fixed number of trials.
3. The probability of success and failure stays the same for
each trial.
4. The trials are independent, meaning that the outcome of
one trial does not affect the outcome of any other trial.
LO 4
© 2012 McGraw-Hill Ryerson Limited
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The Binomial Probability Distribution
Binomial Probability Distribution given as:
Where: C denotes a combination.
n is the number of trials.
x is a number of successes.
p is the probability of a success on each trial.
LO 4
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EXAMPLE – The Binomial Probability Distribution
Suppose a die is tossed six times.
a) What is the probability of getting exactly two 4s?
b) What is the probability of getting exactly three 6s?
LO 4
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Solution – The Binomial Probability Distribution
a)
P(2)  n Cx ( p) x (1  p) n  x
 5 C2 (0.167) 2 (1  0.167)52
 (10)(0.02789)(.57801)  0.161
b)
P(3)  n Cx ( p) x (1  p) n  x
 5 C3 (0.167)3 (1  0.167)53
 (10)(0.004657)(.6939)  0.03231
LO 4
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The Mean and Variance of a Binomial Distribution
LO 4
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Example: The Binomial Probability Distribution
Suppose a die is tossed five times.
a) What is the average number of getting exactly two 4s?
b) What is the variance of getting exactly two 4s?
LO 4
© 2012 McGraw-Hill Ryerson Limited
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Example – The Binomial Probability Distribution
SOLUTION – The Binomial Probability Distribution.)
a)
  np  (5)(0.167)  0.833
2

 np (1  p )  5(0.167)(1  0.167)
b)
 0.694
LO 4
© 2012 McGraw-Hill Ryerson Limited
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Solution – Mean and Variance: Another Solution
Number of
4s,
x
P(x)
xP(x)
x  
0
0.4019
0.0000
-0.8333
0.6944
0.2791
1
0.4019
0.4019
0.1667
0.0278
0.0112
2
0.1607
0.3215
1.1667
1.3612
0.2188
3
0.0321
0.0964
2.1667
4.6946
0.1509
4
0.0032
0.0129
3.1667
10.0280
0.0322
5
0.0001
0.0006
4.1667
17.3614
0.0022
x   x  
2
  0.8333
LO 4
© 2012 McGraw-Hill Ryerson Limited
2
P( x)
 2  0.6944
37
Example – Binomial Distributions: Using Tables
Ten percent of the worm gears produced by an automatic,
high-speed milling machine are defective. What is the
probability that out of five gears selected at random none
will be defective? Exactly one? Exactly two? Exactly three?
Exactly four? Exactly five out of five?
LO 4
© 2012 McGraw-Hill Ryerson Limited
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EXAMPLE – Binomial Distributions: Using Tables
Probability of 0, 1, 2, … Successes for a p of 0.05, 0.10,
0.20, 0.50, and 0.70 and n of 5
LO 4
x
0.05
0.1
0.2
0.3
0.4
0.5
0.6
0
0.774 0.590 0.328 0.168 0.078 0.031 0.010 0.002 0.000 0.000 0.000
1
0.204 0.328 0.410 0.360 0.259 0.156 0.077 0.028 0.006 0.000 0.000
2
0.021 0.073 0.205 0.309 0.346 0.313 0.230 0.132 0.051 0.008 0.001
3
0.001 0.008 0.051 0.132 0.230 0.313 0.346 0.309 0.205 0.073 0.021
4
0.000 0.000 0.006 0.028 0.077 0.156 0.259 0.360 0.410 0.328 0.204
5
0.000 0.000 0.000 0.002 0.010 0.031 0.078 0.168 0.328 0.590 0.774
© 2012 McGraw-Hill Ryerson Limited
0.7
0.8
0.9
0.95
39
Binomial Distributions In Excel – Megastat
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Binomial Distributions In Excel
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You Try It Out!
Hospital records show that 75% of the patients suffering
from a certain disease, will die of it.
Suppose we select a random sample of six patients.
a) Does this situation fit the assumptions of the binomial
distribution?
b) What is the probability that the six patients will recover?
c) Use formula (5-3) to determine the exact probability that
four of the six sampled patients will recover.
d) Use Appendix A to verify your answers to parts (b) and
(c) .
LO 4
© 2012 McGraw-Hill Ryerson Limited
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Example – Probability Distributions
x
0.05
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.95
0
0.599 0.349 0.107 0.028 0.006 0.001 0.000 0.000 0.000 0.000 0.000
1
0.315 0.387 0.268 0.121 0.04 0.010 0.002 0.000 0.000 0.000 0.000
2
0.075 0.194 0.302 0.233 0.121 0.044 0.011 0.001 0.000 0.000 0.000
3
0.01 0.057 0.201 0.267 0.215 0.117 0.042 0.009 0.001 0.000 0.000
4
0.001 0.011 0.088 0.200 0.251 0.205 0.111 0.037 0.006 0.000 0.000
5
0.000 0.001 0.026 0.103 0.201 0.246 0.201 0.103 0.026 0.001 0.000
6
0.000 0.000 0.006 0.037 0.111 0.205 0.251 0.200 0.088 0.011 0.001
7
0.000 0.000 0.001 0.009 0.042 0.117 0.215 0.267 0.201 0.057 0.010
8
0.000 0.000 0.000 0.001 0.011 0.044 0.121 0.233 0.302 0.194 0.075
9
0.000 0.000 0.000 0.000 0.002 0.010 0.040 0.121 0.268 0.387 0.315
10 0.000 0.000 0.000 0.000 0.000 0.001 0.006 0.028 0.107 0.349 0.599
LO 4
© 2012 McGraw-Hill Ryerson Limited
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Binomial – Shapes for Varying P (n constant)
0.70
0.70
0.60
0.60
0.50
0.50
0.40
0.40
0.30
0.30
0.20
0.20
0.10
0.10
0.00
0
1
2
3
4
5
6
7
8
9 10
0.00
0
P = 0.05
n = 10
1
2
3
4
5
6
7
8
9 10
P = 0.1
n = 10
0.70
0.60
0.50
0.40
0.30
0.20
P = 0.2
n = 10
LO 4
0.10
0.00
0
1
2
3
4
5
6
7
8
9
10
© 2012 McGraw-Hill Ryerson Limited
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Binomial – Shapes for Varying P (n constant)
0.70
0.70
0.60
0.60
0.50
0.50
0.40
0.40
0.30
0.30
0.20
0.20
0.10
0.10
0.00
0.00
0
1
2
3
4
5
6
7
8
9 10
0
1
3
4
5
6
7
8
9 10
P = 0.7
n = 10
P = 0.5
n = 10
LO 4
2
© 2012 McGraw-Hill Ryerson Limited
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Binomial – Shapes for Varying n
0.50
0.50
0.40
0.40
0.30
0.30
0.20
0.20
0.10
0.10
0.00
0.00
0
1
2
3
4
0
1
P = 0.1, n = 7
3
4
5
6
7
9
10 11
P = 0.7, n = 12
0.50
0.50
0.40
0.40
0.30
0.30
0.20
0.20
0.10
0.10
0.00
0.00
0
1
2
3
4
5
P = 0.1, n = 20
LO 4
2
6
7
8
0
1
2
3
4
5
6
7
8
P = 0.1, n = 40
© 2012 McGraw-Hill Ryerson Limited
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Cumulative Binomial Distributions
We might be interested in the cumulative binomial probability
of obtaining 45 or fewer heads in 100 tosses of a coin.
Or we may be interested in the probability of selecting at
most two defectives at random from production during the
previous hour.
In these cases we need cumulative frequency distributions
similar to the ones developed in Chapter 2.
LO 4
© 2012 McGraw-Hill Ryerson Limited
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Example – Probability Distributions
The probability that a student is accepted to a prestigious
college is 0.6. If 10 students from the same school apply:
a) What is the probability that exactly 7 are accepted?
b) What is the probability that at most 2 are accepted?
LO 4
© 2012 McGraw-Hill Ryerson Limited
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Solution – Probability Distributions
a) P (7)  C ( p) x (1  p) n  x
n x
 10 C7 (0.6) 7 (1  0.6)107
 120(0.0279936)(0.064)
 0.214990848
 0.215
b)
P ( x  2 | n  10 and p  0.60)
 P ( x  0)  P( x  1)  P ( x  2)
 0.1681  0.3601  0.3087
 0.8369
LO 4
© 2012 McGraw-Hill Ryerson Limited
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Cumulative Binomial Distributions in Excel
LO 4
© 2012 McGraw-Hill Ryerson Limited
50
You Try It Out!
For a case in which n = 10 and p = 0.45, determine the
probability that:
a) x = 6
b) x ≤ 6
c) x > 6
LO 4
© 2012 McGraw-Hill Ryerson Limited
51
LO
5
Hypergeometric Probability
Distribution
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Hypergeometric Probability Distribution
A probability function f(x) that gives the probability of
obtaining exactly x elements of one kind and n − x
elements of another if n elements are chosen at random
without replacement from a finite population containing N
elements of which S are of the first kind and N − S are of
the second kind and that has the form.
LO 5
© 2012 McGraw-Hill Ryerson Limited
53
Hypergeometric Probability Distribution
Hypergeometric Distribution is:
Where:
N
is the size of the population.
S
is the number of successes in the population.
x
is the number of successes in the sample. It may be
0, 1, 2, 3, . . . .
n
is the size of the sample or the number of trials.
C is the symbol for a combination.
LO 5
© 2012 McGraw-Hill Ryerson Limited
54
Example – Probability Distributions
Assume that a jar contains 5 white marbles and 45 black
marbles. You close your eyes and draw 10 marbles without
replacement. What is the probability that exactly 4 of the 10
marbles are white?
LO 5
© 2012 McGraw-Hill Ryerson Limited
55
Solution – Probability Distributions
( 5 C4 )( 45 C6 )
P(x = 4) =
C
50 10
5.8145060
=
= 0.003 965
10 272 278 170
LO 5
© 2012 McGraw-Hill Ryerson Limited
56
Example – Probability Distributions
When the binomial requirement of a constant probability of
success cannot be met, then use the hypergeometric
distribution.
A rule of thumb: If the selected items are not returned to the
population and the sample size is less than 5 percent of the
population, the binomial distribution can be used to
approximate the hypergeometric distribution.
LO 5
© 2012 McGraw-Hill Ryerson Limited
57
Hypergeometric vs. Binomial
PlayTime Toys Inc. employs 50 people in the assembly
department. Forty of the employees belong to a union and 10
do not. Five employees are selected at random to form a
committee to meet with management regarding shift starting
times.
LO 5
Number of Union
Members on Committee
Hypergeometric
Probability, P(x)
Binomial Probability
(n = 5 and p = 0.80)
0
1
2
3
4
5
0.000
0.004
0.044
0.210
0.431
0.311
1.000
0.000
0.006
0.051
0.205
0.410
0.328
1.000
© 2012 McGraw-Hill Ryerson Limited
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Hypergeometric Distribution in Excel
LO 5
© 2012 McGraw-Hill Ryerson Limited
59
You Try It Out!
Kolzak Appliance Outlet just received a shipment of 15 TV
sets. Shortly after they were received, the manufacturer
called to report that he had inadvertently shipped five
defective sets. Ms. Kolzak, the owner of the outlet, decided
to test four of the 10 sets she received. What is the
probability that none of the four sets tested is defective?
LO 5
© 2012 McGraw-Hill Ryerson Limited
60
LO
6
POISSON PROBABILITY
DISTRIBUTION
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The Poisson Distribution
The Poisson probability distribution describes the
number of times some event occurs during a specified
interval. The interval may be time, distance, area, or
volume.
Characteristics:
1. The random variable is the number of times some event
occurs during a defined period.
2. The probability of the event is proportional to the size of
the interval.
3. The intervals do not overlap and are independent.
LO 6
© 2012 McGraw-Hill Ryerson Limited
62
Example – Probability Distributions
Poisson Distribution is:
Where:
µ
is the mean number of occurrences (successes) in a
particular interval.
e
is the constant 2.718 28… (base of the Napierian
logarithmic system).
x
is the number of occurrences (successes).
P(x) is the probability for a specified value of x.
LO 6
© 2012 McGraw-Hill Ryerson Limited
63
The Mean and Variance of the Poisson Probability
Distribution
The mean number of successes, s 2, can be determined in
binomial situations by np, where n is the number of trials
and p the probability of a success.
The variance of the Poisson distribution is also equal to np.
LO 6
© 2012 McGraw-Hill Ryerson Limited
64
Example – Poisson Probability Distribution
Suppose a random sample of 1000 salespeople sold a total
of 3000 life insurance policies per week. Thus, the average
number of policies sold per week by a salesperson is
3(3000/1000). Use Poisson’s law to calculate the
probability that a salesperson will not sell a policy in a given
week.
 x eu 30 e3
P( x  0) 

 0.0498
x!
0!
LO 6
© 2012 McGraw-Hill Ryerson Limited
65
EXAMPLE – Poisson Probability Distribution Table

x
LO 6
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0.9048 0.8187 0.7408 0.6703 0.6065 0.5488 0.4966 0.4493 0.4066
1
0.0905 0.1637 0.2222 0.2681 0.3033 0.3293 0.3476 0.3595 0.3659
2
0.0045 0.0164 0.0333 0.0536 0.0758 0.0988 0.1217 0.1438 0.1647
3
0.0002 0.0011 0.0033 0.0072 0.0126 0.0198 0.0284 0.0383 0.0494
4
0.0000 0.0001 0.0003 0.0007 0.0016 0.0030 0.0050 0.0077 0.0111
5
0.0000 0.0000 0.0000 0.0001 0.0002 0.0004 0.0007 0.0012 0.0020
6
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0002 0.0003
7
0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000
© 2012 McGraw-Hill Ryerson Limited
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Poisson Probability Distribution in Excel
LO 6
© 2012 McGraw-Hill Ryerson Limited
67
Poisson Probability Distribution Shape
Always positively skewed.
Has no specific upper limit.
As μ becomes larger, the Poisson distribution becomes
more symmetrical.
LO 6
© 2012 McGraw-Hill Ryerson Limited
68
Shape of Poisson Probability Distribution
0.50
0.50
0.40
0.40
0.30
0.30
0.20
0.20
0.10
0.10
0.00
0.00
0
1
2
3
4
0
1
  0.7
2
3
4
5
6
7
  2.0
0.50
0.40
0.30
Series1
0.20
0.10
  6.0
LO 6
0.00
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You Try It Out!
If electricity power failures occur according to a Poisson
distribution with an average of three failures every twenty
weeks, calculate the probability that there will not be more
than one failure during a particular week.
LO 6
© 2012 McGraw-Hill Ryerson Limited
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Chapter Summary
I.
A random variable is a numerical value determined by the
outcome of a random experiment.
II. A probability distribution is a listing of all possible outcomes
of an experiment and the probability associated with each
outcome.
A. A discrete probability distribution can assume only certain
values. The main features are:
1. The sum of the probabilities is 1.
2. The probability of a particular outcome is between 0 and
1.
3. The outcomes are mutually exclusive.
B. A continuous distribution can assume an infinite number
of values within a specific range.
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Chapter Summary
III. The mean and variance of a discrete probability
distribution are computed as follows:
A. The mean is equal to:
   xP( x)
[5–1]
B. The variance is equal to:
 2   ( x   ) 2 P( x) 
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[5–2]
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Chapter Summary
IV. The binomial distribution has the following
characteristics:
A. Each outcome is classified into one of two mutually
exclusive categories.
B. The distribution results from a count of the number
of successes in a fixed number of trails.
C. The probability of a success remains the same from
trial to trial.
D. Each trial is independent.
E. A binomial probability is determined as follows:
P( x)  nCx p x (1  p)n  x
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[5–3]
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Chapter Summary
F. The mean is computed as
  np
G. The variance is:
 2  np(1  p)
© 2012 McGraw-Hill Ryerson Limited
[5–4]
[5–5]
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Chapter Summary
V. The hypergeometric distribution has the following
characteristics:
A. There are only two possible outcomes.
B. The probability of a success is not the same on each
trial.
C. The distribution results from a count of the number
of successes in a fixed number of trials.
D. A hypergeometric probability is computed from the
following equation.
P( x) 
( S Cx )( N  S Cn  x )
[5–6]
( N Cn )
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Chapter Summary
VI. The Poisson distribution has the following
characteristics:
A. It describes the number of times an event occurs
during a specified interval.
B. The probability of a “success” is proportional to the
length of the interval.
C. Non-overlapping intervals are independent.
D. It is a limiting form of the binomial distribution when
n is large and p is small.
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Chapter Summary
E. A Poisson probability is determined from the
following equation:
P( x) 
 x e 
[5–7]
x!
F. The mean and the variance are:
s 2 = np
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