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Chapter 07

OMIS 2010 Chapter Notes - Chapter 07: Random Variable, Countable Set, Standard Deviation


Department
Operations Management and Information System
Course Code
OMIS 2010
Professor
Alan Marshall
Chapter
07

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Chapter 7: Random Variables and Discrete Probability Distributions
7.1 Random Variables and Probability Distributions
This section introduced the concept of a random variable, which assigns a numerical value to each
simple event in a sample space, thereby enabling us to work with numerical-valued outcomes. A random
variable is discrete if the number of possible values that it can assume is finite or countably infinite;
otherwise, the random variable is continuous. The second important concept introduced was that of a
discrete probability distribution, which lists all possible values that a discrete random variable can take
on, together with the probability that each value will be assumed. When constructing a discrete probabil-
ity distribution, bear in mind that the probabilities must sum to 1.
Example 7.1
The local taxation office claims that 10% of the income tax forms processed this year contain errors.
Suppose that two of these forms are selected at random. Find the probability distribution of the random
variable X, defined as the number of forms selected that contain an error.
Solution
A sample space for this random experiment is
S =
{
}
21212121 ,,, FFFFFFFF
where, for example, 21FF indicates that the first form selected contains an error and the second form
does not. The random variable X assigns a value of 0, 1, or 2 to each of these simple events. To find the
probability that each of these values will be assumed, we must determine the simple events that result in
each of these values being assumed and the probabilities that the simple events will occur. For example,
X = 2 if and only if the event 21FF occurs. Since the occurrence of an error on one form is independent
of the occurrence of an error on the other form,
01.)1)(.1(.)()()()2()2( 2121
=
=
=
=== FPFPFFPpXP
Similarly,
81.)9)(.9(.)()()()0()0( 18.)1)(.9(.)9)(.1(. )()()()( )()()1()1(
2121
2121
2121
======
=+=
+=
+===
FPFPFFPpXP
FPFPFPFP
FFPFFPpXP
Summarizing the probability distribution of X in tabular form, we obtain
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x p(x)
0 .81
1 .18
1 .01
Note that we could have used a probability tree to identify the simple events and their associated
probabilities.
EXERCISES
7.1 For each of the following random variables, indicate whether the variable is discrete or continu-
ous and specify the possible values that it can assume.
a) X = the number of customers served by a restaurant on a given day
b) X = the time in minutes required to complete a particular assembly
c) X = the number of accidents at a particular intersection in a given week
d) X = the number of questions answered correctly by a randomly selected student who wrote
a quiz consisting of 15 multiple choice questions
e) X = the weight in ounces of a newborn baby chosen at random
7.2 The following table gives the probability distribution of X, the number of telephones in a ran-
domly selected home in a certain community.
x p(x)
0 .021
1 .412
2 .283
3 .188
73
d
c
d
d
c

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4 .096
If one home is selected at random, find the probability that it will have:
a) no telephone;
b) fewer than two telephones;
c) at least three telephones;
d) between one and three telephones (inclusive).
7.3 The security force on a particular campus estimates that 80% of the cars that are parked illegally
on campus are detected and issued a $10 parking ticket. Suppose your car will be parked ille-
gally on two occasions during the coming year. Determine the probability distribution of the
random variable X = the total dollar amount of parking tickets that you will receive.
7.2 Expected Value and Variance
This section dealt with the two main descriptive measures of a discrete random variable: its expected
value and its variance. As well as being able to calculate the numerical values of these measures, you are
expected to have a general understanding of their meanings and to be familiar with the laws of expected
value and of variance.
74
0.21
0.433
0.188
0.979
X
p(X)
0,0
1,0
0,1
1,1
(.2)(.2)
(.8)(.2)
(.2)(.8)
(.8)(.8)
= 0.04
= 0.16
= 0.16
=0.64
X
p(x)
$0
$10
$20
0.04
0.32
0.64
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