2.12. SOME TYPICAL PROBABILITY DISTRIBUTIONS 63
Table 2.8: Number of automobiles arriving at a tollbooth per minute
e number of automobiles, x
2 3 4 5 6 7 8 9
Probability mass function p(x)
0.06 0.09 0.15 0.22 0.18 0.16 0.10 0.04
Solution:
Per Equation (2.45), the mean of the number of automobiles arriving at a tollbooth per minute
X (a discrete random variable) is:
x
D E
.
X
/
D
X
All i
x
i
p
.
x
i
/
D 2 0:06 C 3 0:09 C 4 0:15 C 5 0:22
C 6 0:18 C7 0:16 C 8 0:10 C 9 0:04
D 5:45:
Per Equation (2.51), the standard deviation of a discrete random variable X is
x
D
p
Var
.
X
/
D
q
E
Œ
X
2
2
x
D
s
X
All i
x
2
i
p
.
x
i
/
2
x
D
s
2
2
0:06 C 3
2
0:09 C 4
2
0:15 C 5
2
0:22 C 6
2
0:18 C 7
2
0:16
C8
2
0:10 C 9
2
0:04 5:45
2
D
p
3:2075 D 1:7909:
Per Equation (2.28), the coefficient of variance of this discrete random variable X is
x
D
x
x
D
1:7909
5:45
D 0:3286:
2.12 SOME TYPICAL PROBABILITY DISTRIBUTIONS
When the PDF or PMF of a random variable is specified and known, every possible information
about it is fully defined and can be determined. Followings are several typical types of probability
distributions for describing various types of a discrete and continuous random variable.
2.12.1 BINOMIAL DISTRIBUTION
Suppose that a random variable in one trial has only two outcomes. One outcome is named as
“success” and represented by numerical number “1” with the probability p. Another outcome is
64 2. FUNDAMENTAL RELIABILITY MATHEMATICS
named as “failure” and represented by a numerical number “0” with a probability q. Since this
random variable has only two outcomes, we have:
p C q D 1: (2.53)
e repeated independent n trials of such a random variable are called n-Bernoulli trials. e
sample space of one such random variable has two sample points “1” as “success” and “0” as
“failure.” en, the sample space of n-Bernoulli trials will have 2
n
sampling points. For exam-
ple, 3-Bernoulli trials will have 2
3
D 8 sample points, which includes 000, 001, 010, 011, 100,
101, 110, and 111. Let X represents the number of successes in the n-Bernoulli trials. en, the
discrete random variable X of the n-Bernoulli trials can be described by the Binomial distribu-
tion.
Binomial distribution: For n-Bernoulli trials where a “success” is represented by a numerical
number “1” with a probability p, the PMF of the random variable X in the n-Bernoulli trials
with the number of “success” x follows the Binomial distribution expressed as:
p
.
X D x
/
D
n
x
!
p
x
.1 p/
nx
for x D 0; 1; 2; : : : ; n; (2.54)
where X is a discrete random variable which is equal to the number of “success” in the n-
Bernoulli trials. x is the realizing value of the random variable X.
n
x
is the number of possible
combinations of x objects from a set of n objects and is equal to
nŠ
xŠ.n x/Š
. p is the probability
of a “success” in one trial.
e CDF of the Binomial distribution per Equation (2.43) will be:
F
.
x
/
D P
.
X x
/
D
kDx
X
kD0
"
n
k
!
p
knk
#
: (2.55)
e Binomial distribution is fully defined by two distribution parameters n and p. e mean
X
and standard deviation
X
of a Binominal distribution X are determined by the following two
equations:
X
D E
.
X
/
D np (2.56)
X
D
p
var
.
X
/
D
p
np.1 p/: (2.57)
In Microsoft Excel, the functions for calculating PMF and the CDF of a Binomial distribution
are:
p
.
x
/
D BINOM:DIST
.
x; n; p; FALSE
/
(2.58)
F
.
x
/
D BINOM:DIST
.
x; n; p; TRUE
/
: (2.59)
2.12. SOME TYPICAL PROBABILITY DISTRIBUTIONS 65
In MATLAB, the commands for calculating the PMF and CDF of a binomial distribution are:
p
.
x
/
D binopdf
.
x; n; p
/
(2.60)
F
.
x
/
D binocdf
.
x; n; p
/
: (2.61)
Example 2.36
A thermal power plant buys three boilers. It is assumed that the probability of a boiler function-
ing without failure for a year is 0.9. (1) Calculate and plot the PMF and CDF of this Binomial
distribution. (2) Calculate the mean and standard deviation of it. (3) What will be the probability
of an event with at least one boiler functioning without failure?
Solution:
1. e PMF and CDF.
In this example, n D 3; p D 0:9. Per Equations (2.53), (2.57), or (2.59), we can calculate
the PMF. Let use the function in Excel to run the calculations:
p
.
x D 0
/
D BINOM:DIST
.
0; 3; 0:9; FALSE
/
D 0:001
p
.
x D 1
/
D BINOM:DIST
.
1; 3; 0:9; FALSE
/
D 0:027
p
.
x D 2
/
D BINOM:DIST
.
2; 3; 0:9; FALSE
/
D 0:243
p
.
x D 3
/
D BINOM:DIST
.
3; 3; 0:9; FALSE
/
D 0:729:
Per Equations (2.55), (2.59), or (2.61), we can calculate the CDF. Let use the command
in MATLAB to run the calculations:
F .0/ D binocdf
.
0; 3; 0:9
/
D 0:001
F .1/ D binocdf
.
1; 3; 0:9
/
D 0:028
F .2/ D binocdf
.
2; 3; 0:9
/
D 0:271
F .3/ D binocdf
.
3; 3; 0:9
/
D 1:
e PMF and CDF of this 3-Bernoulli trial are shown in Figures 2.16 and 2.17, respec-
tively.
2. e mean and standard deviation.
Per Equations (2.56) and (2.57), the mean and standard deviation are:
X
D E
.
X
/
D np D 3 0:9 D 2:7
X
D
p
var
.
X
/
D
p
np D
p
3 0:9 .1 0:9/ D 0:520:
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