3
Chapter 6
Discrete
Probability
Distributions
© 2010 Pearson Prentice Hall. All rights reserved
Section 6.1 Probability Rules
© 2010 Pearson Prentice Hall. All rights
reserved
6-2
© 2010 Pearson Prentice Hall. All rights
reserved
6-3
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-4
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. See
the figure.
© 2010 Pearson Prentice Hall. All rights
reserved
6-5
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. See the figure.
© 2010 Pearson Prentice Hall. All rights
reserved
6-6
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
© 2010 Pearson Prentice Hall. All rights
reserved
6-7
Classify the following random variable according to whether it is discrete
or continuous.
The number of goals scored in a soccer game
(a) Discrete
(b) Continuous
© 2010 Pearson Prentice Hall. All rights
reserved
(c) Not sure
6-8
A random variable is
(a) A numerical measure of the outcome of a probability experiment.
(b) A qualitative attribute of a population.
(c) The variable for which an algebraic equation is solved.
(d) Generated by a random number table.
(e) Not sure
© 2010 Pearson Prentice Hall. All rights
reserved
6-9
© 2010 Pearson Prentice Hall. All rights
reserved
6-10
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-11
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
DVDs a person rents from
a video store during a
single visit.
x
0
1
P(x)
0.06
0.58
2
3
0.22
0.10
4
5
0.03
0.01
© 2010 Pearson Prentice Hall. All rights
reserved
6-12
© 2010 Pearson Prentice Hall. All rights
reserved
6-13
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
© 2010 Pearson Prentice Hall. All rights
reserved
6-14
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
© 2010 Pearson Prentice Hall. All rights
reserved
6-15
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
© 2010 Pearson Prentice Hall. All rights
reserved
6-16
Does the following table describe a discrete probability
distribution?
x
P(x)
-1
0
1
2
0.2
0.3
0.1
0.4
(a) No, because –1 is not allowed as a value for x
(b) No, because the probabilities do not sum to 1
(c) No, because the probabilities are not in order
(d) Yes, the probabilities are all positive and sum to 1
(e) Not sure
© 2010 Pearson Prentice Hall. All rights
reserved
6-17
© 2010 Pearson Prentice Hall. All rights
reserved
6-18
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-19
EXAMPLE
Drawing a Probability Histogram
Draw a probability histogram 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
P(x)
0
0.06
1
0.58
2
0.22
3
0.10
4
0.03
5
0.01
Probability
DVDs Rented at a Video Store
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
Number of DVDs Rented
© 2010 Pearson Prentice Hall. All rights
reserved
6-20
© 2010 Pearson Prentice Hall. All rights
reserved
6-21
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
P(x)
0
0.06
1
0.58
2
0.22
3
0.10
4
0.03
5
0.01
X   x  P( x)
 0(0.06)  1(0.58)  2(0.22)  3(0.10)  4(0.03)  5(0.01)
 1.49
© 2010 Pearson Prentice Hall. All rights
reserved
6-22
© 2010 Pearson Prentice Hall. All rights
reserved
6-23
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-24
x1  x2  ...  x100
X 
 1.49
100
© 2010 Pearson Prentice Hall. All rights
reserved
6-25
As the number of trials of the experiment increases, the mean number of rentals
approaches the mean of the probability distribution.
© 2010 Pearson Prentice Hall. All rights
reserved
6-26
© 2010 Pearson Prentice Hall. All rights
reserved
6-27
The mean of the discrete random variable
defined in the table below is approximately
x
-1
P(x)
0.2
0
1
2
0.3
0.1
0.4
© 2010 Pearson Prentice Hall. All rights
reserved
6-28
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-29
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
© 2010 Pearson Prentice Hall. All rights
reserved
6-30
© 2010 Pearson Prentice Hall. All rights
reserved
6-31
© 2010 Pearson Prentice Hall. All rights
reserved
6-32
EXAMPLE
Computing the Variance and Standard Deviation
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.
2
x


x


P( x)




x  X
X
X
x
P(x)
0
0.06
1
0.58
2
0.22
3
0.10
4
0.03
5
0.01
2
x
0
1
2
3
4
5
P(x)
0.06
0.58
0.22
0.1
0.03
0.01
-1.43
-0.91
-1.27
-1.39
-1.46
-1.48
2.0449
0.8281
1.6129
1.9321
2.1316
2.1904
0.122694
0.480298
0.354838
0.19321
0.063948
0.021904
© 2010 Pearson Prentice Hall. All rights
reserved
 X2    x   X  P( x)
2
 1.236892
 X  1.236892
 1.11
6-33
Section 6.2 The Binomial Probability Distribution
© 2010 Pearson Prentice Hall. All rights
reserved
6-34
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-35
Notation Used in the
Binomial Probability Distribution
• There are n independent trials of the experiment
• Let p denote the probability of success so that
is the probability of failure.
1–p
• Let X be a binomial random variable that denotes the
number of successes in n independent trials of the
experiment. So, 0 < x < n.
© 2010 Pearson Prentice Hall. All rights
reserved
6-36
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
(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.
(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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-37
Decide whether the experiment is a binomial experiment. If it is not, explain why.
Testing a pain reliever using 400 people to determine if it is effective. The
random variable represents the number of people who find the pain reliever
to be effective.
(a) A binomial experiment
(b) Not a binomial experiment
(c) Not sure
© 2010 Pearson Prentice Hall. All rights
reserved
6-38
•
A random sample of 12 students reports how many
brothers they have. Is this a binomial experiment?
(a) Yes, because the students are independent
(b) Yes, because the probabilities are the same for
each trial
(c) No, because sisters are excluded
(d) No, because the successes are not limited to one
“success” or one “failure
(e) Not sure
© 2010 Pearson Prentice Hall. All rights
reserved
6-39
© 2010 Pearson Prentice Hall. All rights
reserved
6-40
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-41
© 2010 Pearson Prentice Hall. All rights
reserved
6-42
© 2010 Pearson Prentice Hall. All rights
reserved
6-43
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
(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
(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
© 2010 Pearson Prentice Hall. All rights
reserved
6-44
According to government data, the probability that an adult
was never married is 15%. In a random survey of 10
adults, what is the probability that two were never
married?
© 2010 Pearson Prentice Hall. All rights
reserved
6-45
According to government data, the probability that an adult
was never married is 15%. In a random survey of 10
adults, what is the probability that less than 2 were never
married?
© 2010 Pearson Prentice Hall. All rights
reserved
6-46
© 2010 Pearson Prentice Hall. All rights
reserved
6-47
© 2010 Pearson Prentice Hall. All rights
reserved
6-48
EXAMPLE
Finding the Mean and Standard Deviation of a Binomial
Random Variable
According to the Experian Automotive, 35% of all car-owning
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
 X  np (1  p )
 (400)(0.35)
 (400)(0.35)(1  0.35)
 140
 9.54
© 2010 Pearson Prentice Hall. All rights
reserved
6-49
According to government data, the probability that an adult
was never married is 15%. In a random survey of 500
adults, what is the expected number who were never
married?
© 2010 Pearson Prentice Hall. All rights
reserved
6-50
© 2010 Pearson Prentice Hall. All rights
reserved
6-51
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-52
© 2010 Pearson Prentice Hall. All rights
reserved
6-53
© 2010 Pearson Prentice Hall. All rights
reserved
6-54
© 2010 Pearson Prentice Hall. All rights
reserved
6-55
Construct a binomial probability histogram with
n = 25 and p = 0.8. Comment on the shape of
the distribution.
© 2010 Pearson Prentice Hall. All rights
reserved
6-56
© 2010 Pearson Prentice Hall. All rights
reserved
6-57
Construct a binomial probability histogram with
n = 50 and p = 0.8. Comment on the shape of
the distribution.
© 2010 Pearson Prentice Hall. All rights
reserved
6-58
© 2010 Pearson Prentice Hall. All rights
reserved
6-59
Construct a binomial probability histogram with
n = 70 and p = 0.8. Comment on the shape of
the distribution.
© 2010 Pearson Prentice Hall. All rights
reserved
6-60
© 2010 Pearson Prentice Hall. All rights
reserved
6-61
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.
© 2010 Pearson Prentice Hall. All rights
reserved
6-62
© 2010 Pearson Prentice Hall. All rights
reserved
6-63
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
 X  np (1  p )
 (400)(0.35)
 (400)(0.35)(1  0.35)
 140
 9.54
 X  2 X  140  2(9.54)
The result is unusual since 162 > 159.1
 120.9
 X  2 X  140  2(9.54)
 159.1
© 2010 Pearson Prentice Hall. All rights
reserved
6-64
According to government data, the probability that an adult
was never married is 15%. In a random survey of 500
adults, a sociologist found that 478 were married. Is this
result unusual?
(a) Yes
(b) No
(c) Not sure
© 2010 Pearson Prentice Hall. All rights
reserved
6-65