2.12. SOME TYPICAL PROBABILITY DISTRIBUTIONS 69
e distribution parameter for the number of defects in a cable with a length of 1500 m
in 5 years is:
D
0:007
100 1
.
1500 5
/
D 0:525 .defects per 1500 m per 5 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 poisspdf
.
0; 0:525
/
D 0:5916
p
.
1
/
D poisspdf
.
1; 0:525
/
D 0:3106
p
.
2
/
D poisspdf
.
2; 0:525
/
D 0:0815
p
.
3
/
D poisspdf
.
3; 0:525
/
D 0:0143
p
.
4
/
D poisspdf
.
4; 0:525
/
D 0:0019
p
.
5
/
D poisspdf
.
5; 0:525
/
D 0:0002:
e PMFs of these two Poisson distributions are tabulated in Table 2.9.
Table 2.9: e PMFs of two Poisson distributions
x
0 1 2 3 4 5
p(x) with λ = 0.056
0.9455 0.0530 0.00148 2.77E-5 3.87E-7 4.34E-9
p(x) with λ = 0.525
0.5916 0.3106 0.0815 0.0143 0.0019 0.0002
2.12.3 UNIFORM DISTRIBUTION
Uniform distribution: If a random variable X has the same occurrence likelihood between the
lower boundary a and the upper boundary b, it is a uniform distribution and often abbreviated
as X D U.a; b/. e PDF of such uniform distribution with distribution parameters a and b is:
f
.
x
/
D
8
<
:
1
b a
for a x b
0 elsewhere:
(2.70)
e CDF of a uniform distribution with distribution parameters a and b is:
F
.
x
/
D
8
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
:
0 for x < a
x a
b a
for a x b
1 for b < x:
(2.71)