Chapter 5
Probability Distributions
5-1 Review and Preview
5-2 Probability Distributions
5-3 Binomial Probability Distributions
5-4 Parameters for Binomial Distributions
5-5 Poisson Probability Distributions
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 5.1-1
Review and Preview
This chapter describe and analyze probability
distributions by combining
•descriptive statistics presented in Chap. 2 and 3
•probability presented in Chapter 4
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Section 5.1-2
Preview
• Introduce random variables
• Focus on discrete probability distributions
• In particular, we discuss binomial and Poisson
probability distributions.
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Section 5.1-3
Combining Descriptive Methods
and Probabilities
In this chapter we will construct probability distributions by presenting
possible outcomes along with the relative frequencies we expect.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 5.1-4
Chapter 5
Probability Distributions
5-1 Review and Preview
5-2 Probability Distributions
5-3 Binomial Probability Distributions
5-4 Parameters for Binomial Distributions
5-5 Poisson Probability Distributions
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 5.1-5
Key Concept
This section introduces the important concept of a
probability distribution
Give consideration to distinguishing between
outcomes that are likely to occur by chance and
outcomes that are not likely to occur by chance
(“unusual”).
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Section 5.1-6
Random Variable
Probability Distribution
 Random Variable
a variable (typically represented by x) that has a
single numerical value, determined by chance,
for each outcome of a procedure
 Discrete Random Variable (ch.5)
 Continuous random variable (ch.6)
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Section 5.1-7
Probability Distribution:
Requirements
1. There is a numerical random variable x and its
values are associated with corresponding
probabilities.
2. The sum of all probabilities must be 1.
 P  x  1
3. Each probability value must be between 0 and
1 inclusive.
0  P  x  1
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Section 5.1-8
Example
Consider a simple example that involves two births.
Let x be the variable:
x = number of girls in two births.
The possible values for x is 0, 1, and 2; and the
probability associated with each value is 0.25, 0.5,
and 0.25. Summarize in the table:
x
0
P(x) 0.25
1
2
0.5
0.25
This is a probability distribution. (satisfies all three
conditions)
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Section 5.1-9
Graphs
Graph of the probability distribution is shown below.
probability
0.50
0.25
0.00
0
1
2
Number of girls in two births
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Section 5.1-10
Example
Let x be the response to the survey question:
should Marijuana use be legal? Determine whether
the following table is a probability distribution?
x
P(x)
yes
no
don’t know
0.41
0.52
0.07
Not a probability distribution, value is not numerical.
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 5.1-11
Example
Is the following table a probability distribution?
x
P(x)
1
2
3
0.3
0.26
0.1
Not a probability distribution, sum of P(x) is not 1.
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Section 5.1-12
Mean, Variance and
Standard Deviation of a
Probability Distribution
  [ x  P( x)]
Mean
 2  [( x   ) 2  P( x)]
Variance
 2  [( x 2  P( x)]   2
Variance (shortcut)
  [( x  P( x)]  
Standard Deviation
2
2
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Section 5.1-13
Expected Value
The expected value of a discrete random
variable is denoted by E, and it represents
the mean value of the outcomes.
E  [ x  P ( x )]
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Section 5.1-14
Example
Continue the number of girls in two births. Find
the mean, variance, and standard deviation.
x
P(x)
xP(x)
(x  )2P(x)
0
0.25
0
0.25
1
0.50
0.5
0
2
0.25
0.5
0.25
1
0.5
Total
=1
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
2 = 0.5
Section 5.1-15
Example
The following table describes the probability
distribution for the number of girls in two births.
Find the mean, variance, and standard
deviation.
   x  P( x)  1.0
   ( x   )  P( x)  0.5
2
2
  0.5  0.707
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Section 5.1-16
Identifying Unusual Results
Range Rule of Thumb
According to the range rule of thumb, most
values should lie within 2 standard deviations
of the mean.
We can therefore identify “unusual” values
by determining if they lie outside these limits:
Maximum usual value =   2
Minimum usual value =   2
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Section 5.1-17
Example – continued
We found for families with two children, the mean number of
girls is 1.0 and the standard deviation is 0.7 girls.
Use those values to find the maximum and minimum usual
values for the number of girls.
maximum usual value    2  1.0  2  0.7   2.4
minimum usual value    2  1.0  2  0.7   0.4
Two girls is not unusual.
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Section 5.1-18
Example – Expected Value
Expected value for the Texas Pick 3 Game. In the Texas
Pick 3 lottery, you can bet $1 by selecting three digits, each
between 0 and 9 inclusive. If the same three numbers are
drawn in the same order, you win and collect $500.
a.How many selections are possible?
1000
b.What is the probability of winning?
0.001
c.If you win, what is your net profit?
$499
d.Find the expected value.
0.499 – 0.999 = –$0.50
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Section 5.1-19
Identifying Unusual Results
Probabilities
Rare Event Rule for Inferential Statistics
If, under a given assumption (such as the
assumption that a coin is fair), the probability
of a particular observed event (such as 992
heads in 1000 tosses of a coin) is extremely
small, we conclude that the assumption is
probably not correct.
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Section 5.1-20
Identifying Unusual Results
Probabilities
Using Probabilities to Determine When
Results Are Unusual
 Unusually high: x successes among n
trials is an unusually high number of
successes if P( x or more)  0.05.
 Unusually low: x successes among n
trials is an unusually low number of
successes if P( x or fewer)  0.05 .
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Section 5.1-21
Chapter 5
Probability Distributions
5-1 Review and Preview
5-2 Probability Distributions
5-3 Binomial Probability Distributions
5-4 Parameters for Binomial Distributions
5-5 Poisson Probability Distributions
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Section 5.1-22
Key Concept
Presents a basic definition of a binomial probability
distribution.
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Section 5.1-23
Binomial Probability Distribution
A binomial probability distribution results from a
procedure that meets all the following requirements:
1. The procedure has a fixed number of trials.
2. The trials must be independent. (The outcome of
any trial doesn’t affect the probabilities in the other
trials.)
3. Each trial have two categories of outcomes
(commonly referred to as success and failure).
4. The probability of a success remains the same in
all trials.
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Section 5.1-24
Notation for Binomial
Probability Distributions
S and F (success and failure) denote the two
possible categories of all outcomes; p and q will
denote the probabilities of S and F, respectively, so
P(S)  p
(p = probability of success)
P(F)  1  p  q
(q = probability of failure)
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Section 5.1-25
Notation (continued)
n
denotes the fixed number of trials.
x
denotes a specific number of successes in n
trials, so x can be any whole number between
0 and n, inclusive.
p
denotes the probability of success in one of
the n trials.
q
denotes the probability of failure in one of the
n trials.
P(x) denotes the probability of getting exactly x
successes among the n trials.
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Section 5.1-26
Caution
 Be sure that x and p both refer to the same
category being called a success.
 When sampling without replacement,
consider events to be independent if
n  0.05 N .
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Section 5.1-27
Example
 When an adult is randomly selected, there is a
0.85 probability that this person knows what
Twitter is.
 Suppose we want to find the probability that
exactly three of five randomly selected adults
know of Twitter.
 Does this procedure result in a binomial
distribution?
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yes
Section 5.1-28
Methods for Finding
Probabilities
We will now discuss three methods for
finding the probabilities corresponding to
the random variable x in a binomial
distribution.
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Section 5.1-29
Method 1: Using the Binomial
Probability Formula
n!
P( x) 
 p x  q n x
(n  x)! x!
for x  0, 1, 2,
,n
where
n = number of trials
x = number of successes among n trials
p = probability of success in any one trial
q = probability of failure in any one trial (q = 1 – p)
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Section 5.1-30
Method 2: Using Technology
STATDISK, Minitab, Excel, SPSS, SAS and the TI-83/84 Plus calculator
can be used to find binomial probabilities.
STATDISK
MINITAB
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Section 5.1-31
Method 2: Using Technology
STATDISK, Minitab, Excel and the TI-83 Plus calculator can all be used
to find binomial probabilities.
EXCEL
TI-83 PLUS Calculator
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Section 5.1-32
Method 3: Using
Table A-1 in Appendix A
The following is part of Table A-1 with n = 5 and p = 0.6 in
the binomial distribution
n
x
5
0
1
2
3
4
5
0.01
…
0.60
…
0.010
0.077
0.230
0.346
0.259
0.078
It gives values P(x), for x = 0, 1, 2, 3, 4, 5, for n = 5, p = 0.6
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Section 5.1-33
Strategy for Finding
Binomial Probabilities
1. Use computer software or a TI-83/84 Plus
calculator.
2. If items in 1 are un available, use Table A-1.
3. If items in 1 and 2 are unavailable, use the
binomial probability formula.
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Section 5.1-34
Example
Given there is a 0.85 probability that any given adult
knows of Twitter, use the binomial probability formula to
find the probability of getting exactly three adults who
know of Twitter when five adults are randomly selected.
We have:
n  5, x  3, p  0.85, q  0.15
We want:
P  3
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Section 5.1-35
Example
We have:
n  5, x  3, p  0.85, q  0.15
5!
P(3) 
 (0.85)3  (0.15)53
(5  3)!3!
5!

(0.614125)(0.0225)
2!3!
 (10)(0.614125)(0.0225)
 0.138
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Section 5.1-36
Rationale for the Binomial
Probability Formula
n!
x
n x
P( x) 
 p q
(n  x)! x!
The number of
outcomes with
exactly x successes
among n trials
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Section 5.1-37
Rationale for the Binomial
Probability Formula
n!
x
n x
P( x) 
 p q
(n  x)! x!
The probability of x
Number of
successes among n
outcomes with
trials for any one
exactly x successes
particular order
among n trials
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Section 5.1-38
Chapter 5
Probability Distributions
5-1 Review and Preview
5-2 Probability Distributions
5-3 Binomial Probability Distributions
5-4 Parameters for Binomial Distributions
5-5 Poisson Probability Distributions
Copyright © 2014, 2012, 2010 Pearson Education, Inc.
Section 5.1-39
Key Concept
In this section we consider important
characteristics of a binomial distribution:
• Center
• variation
• standard deviation.
A strong emphasis is placed on interpreting
and understanding those values.
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Section 5.1-40
Binomial Distribution: Formulas
Mean
  n p
Variance
  n pq
Std. Dev.
  n pq
2
Where
n = number of fixed trials
p = probability of success in one of the n trials
q = probability of failure in one of the n trials
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Section 5.1-41
Interpretation of Results
It is especially important to interpret results. The
range rule of thumb suggests that values are
unusual if they lie outside of these limits:
maximum usual value =   2
minimum usual value =   2
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Section 5.1-42
Example
McDonald’s has a 95% recognition rate. A special focus
group consists of 12 randomly selected adults.
For such a group, find the mean and standard deviation.
  np  12  0.95   11.4
  npq  12  0.95  0.05   0.754983  0.8 (rounded)
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Section 5.1-43
Example - continued
Use the range rule of thumb to find the minimum and
maximum usual number of people who would recognize
McDonald’s.
  2  11.4  2  0.8   13 people
  2  11.4  2  0.8   9.8 people
If a particular group of 12 people had all 12 recognize the
brand name of McDonald’s, that would not be unusual.
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Section 5.1-44
Chapter 5
Probability Distributions
5-1 Review and Preview
5-2 Probability Distributions
5-3 Binomial Probability Distributions
5-4 Parameters for Binomial Distributions
5-5 Poisson Probability Distributions
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Section 5.1-45
Key Concept
The Poisson distribution is another
discrete probability distribution which is
important because it is often used for
describing the behavior of rare events
(with small probabilities).
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Section 5.1-46
Poisson Distribution
The Poisson distribution is a discrete probability
distribution that applies to occurrences of some
event over a specified interval. The random
variable x is the number of occurrences of the event
in an interval. The interval can be time, distance,
area, volume, or some similar unit.
Formula
P( x) 
 x  e 
x!
where e  2.71828
  mean number of occurrences of the event over the interval
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Section 5.1-47
Requirements of the
Poisson Distribution
The random variable x is the number of occurrences
of an event over some interval.
The occurrences must be random.
The occurrences must be independent of each other.
The occurrences must be uniformly distributed over
the interval being used.
Parameters
The mean is  .
 The standard deviation is  
.
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Section 5.1-48
Differences from a
Binomial Distribution
The Poisson distribution differs from the binomial
distribution in these fundamental ways:
 Binomial distribution: n, p,
Poisson distribution: 
 Binomial: x takes value 0,1,. . .,n
Poisson: x takes values 0, 1, 2, . . . , with no
upper limit.
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Section 5.1-49
Example
For a recent period of 100 years, there were 530
Atlantic hurricanes. Assume the Poisson
distribution is a suitable model.
a.Find μ, the mean number of hurricanes per
year.
number of hurricanes 530

number of years

100
 5.3
b.If P(x) is the probability of x hurricanes in a
randomly selected year, find P(2).
P  2 
 e
x
x!

5.3  2.71828 
2

2!
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5.3
 0.0701
Section 5.1-50
Poisson as an Approximation
to the Binomial Distribution
Use the Poisson distribution to approximate the
binomial distribution when n is large and p is small.
Rule of Thumb to use the Possion to approximate
the binomial
n  100
n  10
Value for for 
 = np
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Section 5.1-51
Example
In the Maine Pick 4 game, you pay $0.50 to select a
sequence of four digits, such as 2449.
If you play the game once every day, find the probability
of winning at least once in a year with 365 days.
1
The chance of winning is p 
10, 000
1
 0.0365
Then, we need μ:   np  365
10,000
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Section 5.1-52
Example - continued
Because we want the probability of winning “at least”
once, we will first find P(0).
P  0 
0.0365  2.71828 
0
0!
0.0365
 0.9642
P(“at least”) = 1 – P(“none”) = 1 – 0.9642 = 0.0358
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Section 5.1-53