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3
Chapter 6
Discrete
Probability
Distributions
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Section 6.1 Probability Rules
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reserved
6-2
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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.
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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.
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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.
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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
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6-7
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reserved
6-8
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.
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6-9
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
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reserved
6-10
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reserved
6-11
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
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reserved
6-12
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
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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
0.04
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reserved
6-14
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reserved
6-15
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.
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reserved
6-16
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
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reserved
6-17
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reserved
6-18
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
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6-19
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reserved
6-20
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.
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6-21
x1  x2  ...  x100
X
 1.49
100
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6-22
As the number of trials of the experiment increases, the mean number of rentals
approaches the mean of the probability distribution.
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6-23
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6-24
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.
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6-25
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
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6-26
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6-27
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reserved
6-28
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
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reserved
 X2    x   X  P( x)
2
 1.236892
 X  1.236892
 1.11
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