66 2. FUNDAMENTAL RELIABILITY MATHEMATICS
0.800
0.600
0.400
0.200
0.000
x-Number of Successes in the 3-Beroulii Trials
p (x)-Probability Mass
Function
0 1 2 3 4
0.001
0.027
0.243
0.729
Figure 2.16: e PMFs of the Binomial distribution.
1.000
0.750
0.500
0.250
0.000
x-Number of Successes in the 3-Beroulii Trials
F (x)-Cumulative Distribution
Function
0 1 2 3 4
Figure 2.17: e CDF of the Binomial distribution.
3. e probability of an event with at least one boiler functioning without failure.
e complimentary event of the event with at least one boiler functioning without failure
is that all three boilers are in failure. erefore,
P
.
X 1
/
D 1 p
.
x < 1
/
D 1 P
.
x D 0
/
D 1 0:001 D 0:999:
2.12. SOME TYPICAL PROBABILITY DISTRIBUTIONS 67
2.12.2 POISSON DISTRIBUTION
In many engineering problems, the occurrence of an event is only affected by chance. Some
examples are: (a) fatigue cracks that may occur in an automobile transmission shaft; (b) the
number of telephone calls that may be received at any time over a specified period of time; and
(c) the number of automobiles that may arrive at a tollbooth at any time over a specified period
of time. e occurrence of such an event can be described by Poisson distribution.
Poisson Distribution: If an event occurs randomly and independently at any time or any
point in space with the same likelihood at any subinterval, the random variable X , that denotes
the number of events in an interval, can be described by the Poisson distribution. e PMF of
Poisson distribution is
p
.
X D x
/
D
e
x
xŠ
I x D 0; 1; 2; : : : ; (2.62)
where X is the number of occurrence events in the interval and x is the realizing value of the
discrete random variable X. is the mean rate of occurrence of the event per the specified
interval.
e Poisson distribution is fully specified by one distribution parameter . ere are some
notes about . For example, if the number of defects of a manufacturing bar per inch is 0.0001,
the for a 20
00
-long bar will be:
D 0:0001 20 D 0:002 .defects per the 20
00
bar/:
In this example, the event is an occurrence of defects in a 20
00
-long bar, and the interval is not
specified and is “present,” that is, the present after the bars has been manufactured. e for a
5
00
-long bar will be
D 0:0001 5 D 0:0005 .defects per the 5
00
bar/:
For another example, if the average number of defects in a long underground cable per 1000-m
length per year is 0.07, the for a 500-m length in a four year will be
D
0:07
1000 1
.
500 4
/
D 0:14 .defects per the 500-m length per 4 years/:
In this example, the event is defects of the 500-m length cable in 4 years.
e CDF of a Poisson distribution per Equation (
2.43) will be:
F
.
x
/
D P
.
X x
/
D
kDx
X
kD1
e
k
kŠ
I x D 0; 1; 2; : : : : (2.63)
e mean and standard deviation of the Poisson distribution with a distribution parameter is
X
D E
.
X
/
D (2.64)
68 2. FUNDAMENTAL RELIABILITY MATHEMATICS
X
D
p
var
.
X
/
D
p
: (2.65)
In Microsoft Excel, the functions for calculating the PMF and CDF of a Poisson distribution
are:
p
.
x
/
D POISSON:DIST
.
x; ; FALSE
/
(2.66)
F
.
x
/
D POISSON:DIST
.
x; ; TRUE
/
: (2.67)
In MATLAB, the commands for calculating the PMF and CDF of a Poisson distribution are:
p
.
x
/
D poisspdf
.
x;
/
(2.68)
F
.
x
/
D poisscdf
.
x;
/
: (2.69)
Example 2.37
e average number of defects in a long underground cable per 100-m length per year is 0.007.
Calculate and tabulate the PMFs (x D 0; 1; 2; 3; 4; 5/ of following two cases: (1) the number of
defects in a cable with a length of 400 m in two years and (2) the number of defects in a cable
with a length of 1500 m in 5 years.
Solution:
1. p.x/ for the number of defects in a cable with a length of 400 m in 2 years.
e distribution parameter for the number of defects in a cable with a length of 400 m
in a period two years is:
D
0:007
100 1
.
400 2
/
D 0:056 .defects per 400 m per 2 years/:
Per Equations (2.62), (2.65), or (2.67), we can calculate the PMF. Let us use Equa-
tion (2.65) to run the calculations:
p
.
0
/
D POISSON:DIST
.
0; 0:056; FALSE
/
D 0:9455
p
.
1
/
D POISSON:DIST
.
1; 0:056; FALSE
/
D 0:0530
p
.
2
/
D POISSON:DIST
.
2; 0:056; FALSE
/
D 0:00148
p
.
3
/
D POISSON:DIST
.
3; 0:056; FALSE
/
D 2:77 10
5
p
.
4
/
D POISSON:DIST
.
4; 0:056; FALSE
/
D 3:87 10
7
p
.
5
/
D POISSON:DIST
.
5; 0:056; FALSE
/
D 4:34 10
9
:
2. p.x/ for the number of defects in a cable with a length of 1500 m in 5 years.
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