2.12. SOME TYPICAL PROBABILITY DISTRIBUTIONS 71
Example 2.38
e temperature in a device is uniformly distributed between 200
ı
C and 221
ı
C. (1) Build the
PDF and CDF of this uniform distribution. (2) Calculate the probability P .T 215
ı
C/.
Solution:
1. e PDF and CDF of this uniform distribution.
In this example, the distribution parameters are a D 200 and b D 221. Per Equa-
tions (2.69) and (2.70), the PDF and CDF are:
f
.
t
/
D
8
<
:
1
221 200
D
1
21
for 200 t 221
0 elsewhere:
F
.
t
/
D
8
ˆ
ˆ
ˆ
<
ˆ
ˆ
ˆ
:
0 for t < 200
t 200
21
for 200 t 221
1 for 221 < t:
2. Calculate the probability P .T 215
ı
C/.
We can use the definition of the PDF to calculate the probability P .T 215
ı
C/.
P
.
T 215
/
D
Z
215
1
f
.
t
/
dt D
Z
200
1
f
.
t
/
dt C
Z
215
200
f
.
t
/
dt
D
Z
200
1
0dt C
Z
215
200
1
21
dt D 0 C
t
21
ˇ
ˇ
ˇ
ˇ
221
200
D
215 200
21
D 0:7143:
P .T 215/ can also directly calculated by using the CDF.
P
.
T 215
/
D F
.
215
/
D
215 200
21
D 0:7143:
2.12.4 NORMAL DISTRIBUTION
e normal distribution is a very common continuous probability distribution. It often describes
the distribution of real-valued random variables whose distributions are not known and that are
affected by lots of uncertainty.
72 2. FUNDAMENTAL RELIABILITY MATHEMATICS
Normal distribution: e PDF of a normally distributed random variable X , also known as
a normal distribution or Gaussian distribution is:
f
X
.
x
/
D
1
p
2
x
exp
"
1
2
x
x
x
2
#
1 < x < 1; (2.74)
where
x
and
x
are the mean and standard deviation of a normally distributed random variable,
respectively.
e CDF of a normally distributed random variable X is
F
X
.
x
/
D P
.
X x
/
D
Z
x
1
1
p
2
x
exp
"
1
2
x
x
x
2
#
dx 1 < x < 1: (2.75)
If X follows a normal distribution with distribution parameters
x
and
x
, we have the following
equations:
E
.
X
/
D
x
(2.76)
x
D E
h
.
X
x
/
2
i
D
p
Var.X/D
x
: (2.77)
When two distribution parameters of a normal distribution
x
and
x
are known, the PDF of
this normal distribution is fully specified, as shown in Equation (2.74). A normal distribution
of a random variable with distribution parameters
x
and
x
can be expressed as
X D N
.
x
;
x
/
: (2.78)
e PDFs of a normal distribution with different distribution parameters are plotted in Fig-
ure 2.20. From Figure 2.20, the PDF of normal distribution has a bell-shaped curve with the
symmetrical line x D
x
. e mean
x
of normal distribution will control the horizontal loca-
tion of the bell-shaped curve and the standard deviation
x
will control the shape. When the
means are the same, the bell-shaped curve will be thinner with a smaller standard deviation.
e CDF in Equation (2.75) of a normal distribution does not have an explicit theoretical
solution. However, it can be easily calculated by using Excel and MATLAB.
In Microsoft Excel, the functions for calculating the PDF and the CDF of a normal
distribution are:
f
.
x
/
D NORM:DIST
.
x;
x
;
x
; false
/
(2.79)
F
.
x
/
D P
.
X x
/
D NORM:DIST
.
x;
x
;
x
; true
/
: (2.80)
In MATLAB, the commands for calculating the PDF and CDF of a normal distribution are:
f
.
x
/
D normpdf
.
x;
x
;
x
/
(2.81)
F
.
x
/
D P
.
X x
/
D normcdf
.
x;
x
;
x
/
: (2.82)
2.12. SOME TYPICAL PROBABILITY DISTRIBUTIONS 73
0-2-4 2 4 6 8 10 12 14 16 18 20 22
0.25
0.2
0.15
0.1
0.05
0
x
µ
x
= 6
σ
x
= 2
µ
x
= 14
σ
x
= 2
f (x)-Probability Density Function
µ
x
= 6
σ
x
= 4
Figure 2.20: Several PDFs of normal distributions.
Example 2.39
If the tensile yield strength X of a ductile material follows a normal distribution with the mean
x
D 61:2 (ksi) and the standard deviation
x
D 4:25 (ksi). Use Excel and MATLAB to calcu-
late probability P .50 X 70/.
Solution:
By a Microsoft Excel function, that is, per Equation (2.79), we have:
P
.
50 X 70
/
D P
.
X 70
/
P
.
X 50
/
D F
.
70
/
F.50/
D NORM:DIST
.
70; 61:2; 4:25; true
/
NORM:DIST
.
50; 61:2; 4:25; true
/
D 0:9808 0:0042 D 0:9766:
By MATLAB, that is, per Equation (2.81), we have:
P
.
50 X 70
/
D P
.
X 70
/
P
.
X 50
/
D F
.
70
/
F
.
50
/
D normcdf
.
70; 61:2; 4:25
/
normcdf
.
50; 61:2; 4:25
/
D 0:9808 0:0042 D 0:9766:
e normal distribution is one of the frequently-used distribution function. e probabil-
ity theory proves that if
X
1
; : : : ; X
n
and are mutually independent normally distributed random
variables, the linear function of these X
1
; : : : ; X
n
and will also be a normal distributed random
74 2. FUNDAMENTAL RELIABILITY MATHEMATICS
variable, that is,
Y D
n
X
iD1
a
i
X
i
(2.83)
X
i
D N
.
i
;
i
/
i D 1; : : : ; n (2.84)
Y D N
y
;
y
(2.85)
Y
D
n
X
iD1
.
a
i
i
/
(2.86)
Y
D
v
u
u
t
n
X
iD1
.
a
i
i
/
2
; (2.87)
where a
i
is the constant coefficient.
i
and
i
are the mean and standard deviation of a normal
distributed random variable X
i
.
y
and
y
are the mean and standard deviation of a normal
distributed random variable Y .
Example 2.40
e combined normal stress on a critical point of a component is the sum of normal stresses
caused by mutually independent bending moment and axial loading. e normal stress
B
caused
by the bending moment follows a normal distribution with a mean
B
D 37:56 (ksi) and a
standard deviation
B
D 4:56 (ksi). e normal stress
A
caused by the axial loading is also a
normally distributed random variable with a mean
A
D 6:23 (ksi) and a standard deviation
A
D
1:02
(ksi). Calculate the distribution parameters of the combined normal stress
S
C
and
the probability of P .S
C
> 50/.
Solution:
In this problem, the combined normal stress is the sum of the normal stresses caused by
mutually-independent bending moment and axial loading, so we have:
S
C
D
B
C
A
:
Per Equations (2.84), (2.85), and (2.86), we have:
S
C
D
B
C
A
D 37:56 C 6:23 D 43:79 .ksi/
S
C
D
q
B
2
C
A
2
D
p
4:56
2
C 1:02
2
D 4:67 .ksi/:
So, the combined normal stress, which is also a normal distributed random variable, can be
expressed as:
S
C
D N
S
C
;
S
C
D N.43:79; 4:67/ .ksi/:
2.12. SOME TYPICAL PROBABILITY DISTRIBUTIONS 75
By Excel function per Equation (2.79), the probability of P .S
C
> 50/ will be:
P
.
S
C
> 50
/
D 1 P
.
50 S
C
/
D 1 NORM:DIST
.
50; 43:79; 4:67; true
/
D 1 0:9082 D 0:0918:
By MATLAB command per Equation (2.81), the probability of P .S
C
> 50/ will be:
P
.
S
C
> 50
/
D 1 P
.
50 S
C
/
D 1 normcdf .50; 43:79; 4:67/
D 1 0:9082 D 0:0918:
Standard normal distribution: It is a special normal distribution with a mean D 0 and
standard deviation D 1. e PDF of the standard normal distribution, typically expressed by
.
z
/
, as plotted in Figure 2.21, is:
f
.
z
/
D
.
z
/
D
1
p
2
exp
1
2
z
2
1 < z < 1: (2.88)
0-3-4 -2 -1 1 2 3 4
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
z -e Standard Normal Variable
f (z)-Probability Density
Function
Figure 2.21: e plot of the PDF of a standard normal distribution.
e CDF of a standard normal distribution, typically expressed by ˆ
.
z
/
, is
F
.
z
/
D ˆ
.
z
/
D P
.
Z z
/
D
Z
x
1
1
p
2
exp
1
2
z
2
dz 1 < z < 1: (2.89)
In Microsoft Excel, the functions for calculating the PDF and CDF of a standard normal
distribution are:
.z/ D NORM:S:DIST.z; false/ (2.90)
ˆ
.
z
/
D P
.
Z z
/
D NORM:S:DIST.z; true/: (2.91)
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