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Chapter
6
Discrete
Probability
Distributions
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Section
6.1
Discrete Random
Variables
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
Objectives
1. Distinguish between discrete and continuous
random variables
2. Identify discrete probability distributions
3. Construct probability histograms
4. Compute and interpret the mean of a discrete
random variable
5. Interpret the mean of a discrete random
6. Compute the standard deviation of a discrete
random variable
5-3
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Objective 1
• Distinguish between Discrete and Continuous
Random Variables
5-4
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A random variable is a numerical
measure of the outcome from a probability
experiment, so its value is determined by
chance. Random variables are denoted
using letters such as X.
6-5
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A discrete random variable has either a
finite or countable number of values. The
values of a discrete random variable can be
plotted on a number line with space
between each point.
6-6
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A continuous random variable has
infinitely many values. The values of a
continuous random variable can be plotted
on a line in an uninterrupted fashion.
6-7
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EXAMPLE
Distinguishing Between Discrete and
Continuous Random Variables
Determine whether the following random variables are
discrete or continuous. State possible values for the
random variable.
(a) The number of light bulbs that burn out in a room of
10 light bulbs in the next year.
Discrete; x = 0, 1, 2, …, 10
(b) The number of leaves on a randomly selected oak
tree.
Discrete; x = 0, 1, 2, …
(c) The length of time between calls to 911.
Continuous; t > 0
6-8
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Objective 2
• Identify Discrete Probability Distributions
5-9
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A probability distribution provides the
possible values of the random variable X
and their corresponding probabilities. A
probability distribution can be in the form
of a table, graph or mathematical formula.
6-10
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EXAMPLE
A Discrete Probability Distribution
The table to the right shows
the probability distribution
for the random variable X,
where X represents the
number of movies streamed
on Netflix each month.
6-11
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x
0
P(x)
0.06
1
2
3
4
0.58
0.22
0.10
0.03
5
0.01
Rules for a Discrete Probability
Distribution
Let P(x) denote the probability that the random
variable X equals x; then
1. Σ P(x) = 1
2. 0 ≤ P(x) ≤ 1
5-12
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EXAMPLE
Identifying Probability Distributions
Is the following a probability distribution?
x
0
1
2
P(x)
0.16
0.18
0.22
3
4
5
0.10
0.30
0.01
No. Σ P(x) = 0.97
6-13
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EXAMPLE
Identifying Probability Distributions
Is the following a probability distribution?
x
0
1
2
P(x)
0.16
0.18
0.22
3
4
5
0.10
0.30
– 0.01
No. P(x = 5) = –0.01
6-14
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EXAMPLE
Identifying Probability Distributions
Is the following a probability distribution?
x
0
1
2
P(x)
0.16
0.18
0.22
3
4
5
0.10
0.30
0.04
Yes. Σ P(x) = 1
6-15
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Objective 3
• Construct Probability Histograms
5-16
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A probability histogram is a histogram
in which the horizontal axis corresponds
to the value of the random variable and
the vertical axis represents the probability
of that value of the random variable.
6-17
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EXAMPLE
Drawing a Probability Histogram
Draw a probability histogram of
the probability distribution to the
right, which represents the number
of movies streamed on Netflix each
month.
6-18
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x
0
1
2
3
4
5
P(x)
0.06
0.58
0.22
0.10
0.03
0.01
Objective 4
• Compute and Interpret the Mean of a Discrete
Random Variable
5-19
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The Mean of a Discrete Random Variable
The mean of a discrete random variable is
given by the formula
 x    x  P x 
where x is the value of the random variable
and P(x) is the probability of observing the
value x.
5-20
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EXAMPLE
Computing the Mean of a Discrete
Random Variable
Compute the mean of the
probability distribution to the
right, which represents the
number of DVDs a person
rents from a video store
during a single visit.
x    x  P x 
x
0
1
2
3
4
5
P(x)
0.06
0.58
0.22
0.10
0.03
0.01
 0(0.06)  1(0.58)  2(0.22)  3(0.10)  4(0.03)  5(0.01)
 1.49
6-21
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Interpretation of the Mean of a Discrete Random Variable
Suppose an experiment is repeated n independent times
and the value of the random variable X is recorded. As
the number of repetitions of the experiment increases,
the mean value of the n trials will approach μX, the mean
of the random variable X. In other words, let x1 be the
value of the random variable X after the first
experiment, x2 be the value of the random variable X
after the second experiment, and so on. Then
x1  x2 L  xn
x
n
The difference between x and μX gets closer to 0 as n
increases.
5-22
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The following data represent the number of DVDs
rented by 100 randomly selected customers in a single
visit. Compute the mean number of DVDs rented.
6-23
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x1  x2  ...  x100
X
 1.49
100
6-24
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As the number of trials of the experiment increases,
the mean number of rentals approaches the mean of
the probability distribution.
6-25
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Objective 5
• Compute and Interpret the Mean of a Discrete
Random Variable
5-26
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Because the mean of a random variable represents
what we would expect to happen in the long run,
it is also called the expected value, E(X), of the
random variable.
6-27
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EXAMPLE
Computing the Expected Value of a Discrete
Random Variable
A term life insurance policy will pay a beneficiary a certain sum of money upon
the death of the policy holder. These policies have premiums that must be paid
annually. Suppose a life insurance company sells a $250,000 one year term life
insurance policy to a 49-year-old female for $530. According to the National
Vital Statistics Report, Vol. 47, No. 28, the probability the female will survive
the year is 0.99791. Compute the expected value of this policy to the
insurance company.
x
P(x)
530
0.99791
530 – 250,000
= -249,470
0.00209
Survives
Does not survive
E(X) = 530(0.99791) + (-249,470)(0.00209)
= $7.50
6-28
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Objective 6
• Compute the Standard Deviation of a Discrete
Random Variable
5-29
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Standard Deviation of a Discrete Random Variable
The standard deviation of a discrete random
variable X is given by
X 

 x     P x 
2
x
  x
2
 P x   
2
X
where x is the value of the random variable, μX is
the mean of the random variable, and P(x) is the
probability of observing a value of the random
variable.
5-30
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EXAMPLE
Deviation
Computing the Variance and Standard
of a Discrete Random Variable
Compute the variance and
standard deviation of the
following probability distribution
which represents the number of
DVDs a person rents from a
video store during a single visit.
6-31
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x
0
P(x)
0.06
1
2
3
4
0.58
0.22
0.10
0.03
5
0.01
EXAMPLE
x
0
1
2
3
4
5
Computing the Variance and Standard
Deviation of a Discrete Random Variable
P(x)
0.06
0.58
0.22
0.1
0.03
0.01
x  X
–1.43
–0.91
–1.27
–1.39
–1.46
–1.48
   x   X  P(x)
2
2
X
x   X  x   X  P(x)
2
2.0449
0.8281
1.6129
1.9321
2.1316
2.1904
0.122694
0.480298
0.354838
0.19321
0.063948
0.021904
 X  1.236892
 1.236892
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2
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 1.11
Section
6.2
The Binomial
Probability
Distribution
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Objectives
1. Determine whether a probability experiment
is a binomial experiment
2. Compute probabilities of binomial
experiments
3. Compute the mean and standard deviation of
a binomial random variable
4. Construct binomial probability histograms
5-34
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Criteria for a Binomial Probability Experiment
An experiment is said to be a binomial experiment if
1. The experiment is performed a fixed number of times.
Each repetition of the experiment is called a trial.
2. The trials are independent. This means the outcome of
one trial will not affect the outcome of the other trials.
3. For each trial, there are two mutually exclusive (or
disjoint) outcomes, success or failure.
4. The probability of success is fixed for each trial of the
experiment.
5-35
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Notation Used in the Binomial Probability
Distribution
•There are n independent trials of the
experiment.
•Let p denote the probability of success so
that 1 – p is the probability of failure.
•Let X be a binomial random variable that
denotes the number of successes in n
independent trials of the experiment.
So, 0 < x < n.
5-36
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EXAMPLE
Identifying Binomial Experiments
Which of the following are binomial experiments?
(a) A player rolls a pair of fair die 10 times. The
number X of 7’s rolled is recorded.
Binomial experiment
6-37
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EXAMPLE
Identifying Binomial Experiments
Which of the following are binomial experiments?
(b)
The 11 largest airlines had an on-time percentage of
84.7% in November, 2001 according to the Air
Travel Consumer Report. In order to assess reasons
for delays, an official with the FAA randomly selects
flights until she finds 10 that were not on time. The
number of flights X that need to be selected is
recorded.
Not a binomial experiment – not a fixed number of trials.
6-38
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EXAMPLE
Identifying Binomial Experiments
Which of the following are binomial experiments?
(c)
In a class of 30 students, 55% are female. The
instructor randomly selects 4 students. The
number X of females selected is recorded.
Not a binomial experiment – the trials are
not independent.
6-39
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Objective 2
• Compute Probabilities of Binomial
Experiments
5-40
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EXAMPLE
Constructing a Binomial Probability Distribution
According to the Air Travel Consumer Report, the 11
largest air carriers had an on-time percentage of 79.0%
in May, 2008. Suppose that 4 flights are randomly
selected from May, 2008 and the number of on-time
flights X is recorded.
Construct a probability distribution for the random
variable X using a tree diagram.
6-41
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Binomial Probability Distribution Function
The probability of obtaining x successes in
n independent trials of a binomial
experiment is given by
P x   n C x p 1 p 
x
n x
x  0,1, 2,..., n
where p is the probability of success.
5-42
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Phrase
Math Symbol
“at least” or “no less than”
or “greater than or equal to”
≥
“more than” or “greater than”
>
“fewer than” or “less than”
<
“no more than” or “at most”
or “less than or equal to
≤
“exactly” or “equals” or “is”
=
6-43
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EXAMPLE
Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all
car-owning households have three or more cars.
(a) In a random sample of 20 car-owning households,
what is the probability that exactly 5 have three or
more cars?
P(5)  20 C5 (0.35)5 (1 0.35)205
 0.1272
6-44
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EXAMPLE
Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all
car-owning households have three or more cars.
(b) In a random sample of 20 car-owning
households, what is the probability that less than 4
have three or more cars?
P(X  4)  P(X  3)
 P(0)  P(1)  P(2)  P(3)
 0.0444
6-45
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EXAMPLE
Using the Binomial Probability Distribution Function
According to the Experian Automotive, 35% of all
car-owning households have three or more cars.
(c) In a random sample of 20 car-owning households,
what is the probability that at least 4 have three or
more cars?
P(X  4)  1  P(X  3)
 1  0.0444
 0.9556
6-46
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Objective 3
• Compute the Mean and Standard Deviation of
a Binomial Random Variable
5-47
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Mean (or Expected Value) and Standard
Deviation of a Binomial Random Variable
A binomial experiment with n independent
trials and probability of success p has a
mean and standard deviation given by the
formulas
 X  np and  X  np 1 p 
5-48
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EXAMPLE
Finding the Mean and Standard Deviation of a
Binomial Random Variable
According to the Experian Automotive, 35% of all carowning households have three or more cars. In a simple
random sample of 400 car-owning households,
determine the mean and standard deviation number of
car-owning households that will have three or more cars.
 X  np
 (400)(0.35)
 140
6-49
 X  np(1  p)
 (400)(0.35)(1  0.35)
 9.54
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Objective 4
• Construct Binomial Probability Histograms
5-50
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EXAMPLE
Constructing Binomial Probability
Histograms
(a) Construct a binomial probability histogram with
n = 8 and p = 0.15.
(b) Construct a binomial probability histogram with
n = 8 and p = 0. 5.
(c) Construct a binomial probability histogram with
n = 8 and p = 0.85.
For each histogram, comment on the shape of the
distribution.
6-51
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6-52
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6-53
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6-54
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EXAMPLE
Constructing Binomial Probability
Histograms
Construct a binomial probability histogram with
n = 25 and p = 0.8. Comment on the shape of
the distribution.
6-55
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6-56
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EXAMPLE
Constructing Binomial Probability
Histograms
Construct a binomial probability histogram with
n = 50 and p = 0.8. Comment on the shape of
the distribution.
6-57
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6-58
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EXAMPLE
Constructing Binomial Probability
Histograms
Construct a binomial probability histogram with
n = 70 and p = 0.8. Comment on the shape of the
distribution.
6-59
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6-60
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For a fixed probability of success, p, as the
number of trials n in a binomial experiment
increase, the probability distribution of the
random variable X becomes bell-shaped.
As a general rule of thumb, if np(1 – p) > 10,
then the probability distribution will be
approximately bell-shaped.
5-61
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Use the Empirical Rule to identify unusual
observations in a binomial experiment.
The Empirical Rule states that in a bellshaped distribution about 95% of all
observations lie within two standard
deviations of the mean.
5-62
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Use the Empirical Rule to identify unusual
observations in a binomial experiment.
95% of the observations lie between μ – 2σ
and μ + 2σ.
Any observation that lies outside this
interval may be considered unusual
because the observation occurs less than
5% of the time.
5-63
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EXAMPLE
Using the Mean, Standard Deviation and
Empirical Rule to Check for Unusual
Results in a Binomial Experiment
According to the Experian Automotive, 35% of all
car-owning households have three or more cars. A
researcher believes this percentage is higher than the
percentage reported by Experian Automotive. He
conducts a simple random sample of 400 car-owning
households and found that 162 had three or more
cars. Is this result unusual ?
 X  np
 (400)(0.35)
 140
6-64
 X  np(1  p)
 (400)(0.35)(1  0.35)
 9.54
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
Using the Mean, Standard Deviation and
Empirical Rule to Check for Unusual
Results in a Binomial Experiment
 X  np
 X  np(1  p)
 (400)(0.35)
 140
 X  2 X  140  2(9.54)
 (400)(0.35)(1  0.35)
 9.54
 X  2 X  140  2(9.54)
 120.9
 159.1
The result is unusual since 162 > 159.1
6-65
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Section
6.3
The Poisson
Probability
Distribution
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Objectives
1. Determine whether a probability experiment
follows a Poisson process
2. Compute probabilities of a Poisson random
variable
3. Find the mean and standard deviation of a
Poisson random variable
5-67
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Objective 1
• Determine if a Probability Experiment Follows
a Poisson Process
5-68
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A random variable X, the number of successes in a
fixed interval, follows a Poisson process provided
the following conditions are met.
1. The probability of two or more successes in any
sufficiently small subinterval is 0.
2. The probability of success is the same for any two
intervals of equal length.
3. The number of successes in any interval is
independent of the number of successes in any
other interval provided the intervals are not
overlapping.
6-69
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EXAMPLE
Illustrating a Poisson Process
The Food and Drug Administration sets a Food Defect
Action Level (FDAL) for various foreign substances
that inevitably end up in the food we eat and liquids we
drink.
For example, the FDAL level for insect filth in
chocolate is 0.6 insect fragments (larvae, eggs, body
parts, and so on) per 1 gram.
6-70
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Objective 2
• Compute Probabilities of a Poisson Random
Variable
5-71
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Poisson Probability Distribution Function
If X is the number of successes in an
interval of fixed length t, then the probability
of obtaining x successes in the interval is
P x 
t 


x
x!
e
 t
x  0,1, 2, 3,...
where λ (the Greek letter lambda)
represents the average number of
occurrences of the event in some interval of
length 1 and e ≈ 2.71828.
5-72
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EXAMPLE
Illustrating a Poisson Process
The Food and Drug Administration sets a Food Defect
Action Level (FDAL) for various foreign substances
that inevitably end up in the food we eat and liquids we
drink. For example, the FDAL level for insect filth in
chocolate is 0.6 insect fragments (larvae, eggs, body
parts, and so on) per 1 gram. Suppose that a chocolate
bar has 0.6 insect fragments per gram. Compute the
probability that the number of insect fragments in a 10gram sample of chocolate is
(a) exactly three. Interpret the result.
(b) fewer than three. Interpret the result.
(c) at least three. Interpret the result.
6-73
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(a) λ = 0.6; t = 10
(0.6 10)3 0.6(10)
P(3) 
e
3!
 0.0892
(b) P(X < 3) = P(X < 2)
= P(0) + P(1) + P(2)
= 0.0620
(c) P(X > 3) = 1 – P(X < 2)
= 1 – 0.0620
= 0.938
6-74
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Objective 3
• Find the Mean and Standard Deviation of a
Poisson Random Variable
5-75
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Mean and Standard Deviation of a Poisson
Random Variable
A random variable X that follows a Poisson
process with parameter λ has mean (or
expected value) and standard deviation
given by the formulas
 X  t and  X  t   X
where t is the length of the interval.
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Poisson Probability Distribution Function
If X is the number of successes in an
interval of fixed length and X follows a
Poisson process with mean μ, the
probability distribution function for X is
P x  
5-77

x
x!
e

x  0,1, 2, 3,...
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
Mean and Standard Deviation of a
Poisson Random Variable
The Food and Drug Administration sets a Food
Defect Action Level (FDAL) for various foreign
substances that inevitably end up in the food we eat
and liquids we drink. For example, the FDAL level
for insect filth in chocolate is 0.6 insect fragments
(larvae, eggs, body parts, and so on) per 1 gram.
(a)Determine the mean number of insect fragments
in a 5 gram sample of chocolate.
(b) What is the standard deviation?
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EXAMPLE
Poisson
Mean and Standard Deviation of a
Random Variable
(a)  X  t
 (0.6)(5)
3
We would expect 3 insect fragments in a
5-gram sample of chocolate.
(b)  X  t
 (0.6)(5)
 3
6-79
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EXAMPLE
A Poisson Process?
In 1910, Ernest Rutherford and
Hans Geiger recorded the
number of α-particles emitted
from a polonium source in
eighth-minute (7.5 second)
intervals. The results are
reported in the table to the
right. Does a Poisson
probability function accurately
describe the number of αparticles emitted?
Source: Rutherford, Sir Ernest; Chadwick, James; and Ellis, C.D.. Radiations from
Radioactive Substances. London, Cambridge University Press, 1951, p. 172.
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6-81
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Section
6.4
The
Hypergeometric
Probability
Distribution
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Objectives
1. Determine whether a probability experiment
is a hypergeometric experiment
2. Compute probabilities of hypergeometric
experiments
3. Find the mean and standard deviation of a
hypergeometric random variable
5-83
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Objective 1
• Determine if a Probability Experiment is a
Hypergeometric Experiment
5-84
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Criteria for a Hypergeometric Probability
Experiment
A probability experiment is said to be a
hypergeometric experiment provided
1. The finite population to be sampled has n
elements.
2. For each trial of the experiment, there are two
possible outcomes, success or failure. There
are exactly k successes in the population.
3. A sample size of n is obtained from the
population of size N without replacement.
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Notation Used in the Hypergeometric
Probability Distribution
•The population is size N.
•The sample is size n.
•There are k successes in the population.
•Let the random variable X denote the
number of successes in the sample of size n,
so x must be greater than or equal to the
larger of 0 or n – (N – k), and x must be less
than or equal to the smaller of n or k.
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EXAMPLE
A Hypergeometric Probability Distribution
The Dow Jones Industrial Average (DJIA) is a
collection of thirty publicly traded companies that are
meant to be representative of the United States
economy. In one certain month 18 of the 30 stocks in
the DJIA increased in value. If an investor randomly
invests in four stocks at the beginning of this month
and records X, the number of stocks that increased in
value during the month, is this a hypergeometric
probability experiment? List the possible values of
the random variable X.
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Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
A Hypergeometric Probability Distribution
This is a hypergeometric probability experiment
because
1. The population consists of N = 30 stocks.
2. Two outcomes are possible: either the stock
increases in value, or it does not increase in value.
There are k = 18 successes.
3. The sample size is n = 4.
The possible values of the random variable are
x = 0, 1, 2, 3, 4.
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Objective 2
• Construct the Probabilities of Hypergeometric
Experiments
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Hypergeometric Probability Distribution
The probability of obtaining x success
based on a random sample of size n from a
population of size N is given by

k C x  N k Cn x 
P x  
N
Cn
where k is the number of successes in the
population.
5-90
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
A Hypergeometric Probability Distribution
The Dow Jones Industrial Average (DJIA) is a
collection of thirty publicly traded companies that are
meant to be representative of the United States
economy. In one certain month 18 of the 30 stocks in
the DJIA increased in value.
What is the probability that an investor randomly
invests in four stocks at the beginning of this month
and three of the stocks increased in value?
6-91
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
A Hypergeometric Probability Distribution
N = 30; n = 4; k = 18; x = 3

k C x  N k Cn x 
P(x) 
N
Cn

18 C 3  3018 C 43 

30
C4
816 12 


27, 405
 0.3573
6-92
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Objective 3
• Compute the Mean and Standard Deviation of
a Hypergeometric Random Variable
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Mean and Standard Deviation of a
Hypergeometric Random Variable
A hypergeometric random variable has mean and
standard deviation given by the formulas
k
k Nk
 N  n
X  n  ;  X  
n 

 N 1
N
N
N
where n is the sample size
k is the number of successes in the population
N is the size of the population
5-94
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
A Hypergeometric Probability Distribution
The Dow Jones Industrial Average (DJIA) is a
collection of thirty publicly traded companies that are
meant to be representative of the United States
economy. In one certain month 18 of the 30 stocks in
the DJIA increased in value. In a random sample of
four stocks, determine the mean and standard
deviation of the number of stocks that will increase in
value.
6-95
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.
EXAMPLE
A Hypergeometric Probability Distribution
N = 30; n = 4; k = 18
18
X  4 
30
 2.4
30  4
18 30  18
X 
4 
30  1
30
30
 0.9277
We would expect 2.4 stocks out of 4 stocks to increase in
value during the month.
6-96
Copyright © 2013, 2010 and 2007 Pearson Education, Inc.