Probability distributions
Although there are a large number of probability or statistical distributions in published literature, this section presents just four such distributions considered useful for performing various types of reliability, maintainability, and safety analysis studies concerned with engineering systems [1416].
Exponential distribution
This is one of the simplest continuous random variable distributions frequently used in the industrial sector, particularly for performing reliability studies. Its probability density function is expressed by [9,17]
f(t) = ae “, for t > 0, a > 0
(2.36)
where
f(t) is the probability density function.
t is the time (i.e., a continuous random variable).
a is the distribution parameter.
By inserting Equation (2.36) into Equation (2.21), we get the equation for the cumulative distribution function:
F(f) = le"“'
(2.37)
With the aid of Equations (2.26) and (2.36), we obtain the following equation for the distribution expected value (i.e., mean value):
E(f) = 
 (2.38)
 7 ^{a}
Example 2.9
Assume that the mean time to failure of engineering system is 1000 hours. Calculate the failure probability of the engineering system during a 500hour mission by using Equations (2.37) and (2.38).
By inserting the specified data value into Equation (2.38), we obtain
a = —i— = 0.001 failures per hour 1000 ^{r}
By substituting the calculated and the given data values into Equation (2.37), we get
F(500) = e‘^{(0tx,1)(500)}
= 0.6065
Thus, the failure probability of the engineering system during the 500hour mission is 0.6065.
Rayleigh distribution
This continuous random variable probability distribution is named after its founder, John Rayleigh (18421919) [1], and its probability density function is defined by
for 0 > 0, t > 0
(2.39)
where
0 is the distribution parameter.
By substituting Equation (2.39) into Equation (2.21), we obtain the following equation for the cumulative distribution function:
(2.40)
By inserting Equation (2.39) into Equation (2.26), we get the following equation for the distribution expected value:
E(t) = 0f
 3Ï
 2/
 (2.41)
where
f(.) is the gamma function and is defined by
00
r(n) = forn>0
 (2.42)
 0
Weibull distribution
This continuous random variable distribution was developed by Waloddi Weibull, a Swedish professor in mechanical engineering in the early 1950s and its probability density function is defined [18]
/(1) = ^^ fort>0,b>0,e>0 (2A3)
0
where
b and 0 are the distribution shape and scale parameters, respectively.
By inserting Equation (2.43) into Equation (2.21), we obtain the following equation for the cumulative distribution function:
F(t) = le’^{(f/0)i}’ (2.44)
By substituting Equation (2.43) into Equation (2.26), we obtain the following equation for the distribution expected value:
E(t) = 0rl + ^ (2.45)
It is to be noted that for b = 1 and b = 2, the exponential and Rayleigh distributions are the special cases of this distribution, respectively.
Bathtub hazard rate curve distribution
This is another continuous random variable distribution, and it can represent bathtubshaped, increasing and decreasing hazard rates. This distribution was developed in 1981 [19], and in the published literature by other authors around the world, it is generally referred to as the Dhillon distribution/law/model [2039].
The probability density function of the distribution is expressed by [19]
/(t) = 170(et)^{!,}'^{1}e I > for t>0,6>0,b>0 (2.46)
where
b and 9 are the distribution shape and scale parameters, respectively.
By substituting Equation (2.46) into Equation (2.21), we obtain the following equation for cumulative distribution function:
Wil
F(f) = le^{l 1} (2.47)
It is to be noted that for b = 0.5, this probability distribution gives the bathtubshaped hazard rate curve, and for b = 1, it gives the extreme value probability distribution. In other words, the extreme value probability distribution is the special case of this probability distribution at b = 1.
Solving firstorder differential equations with Laplace transforms
Usually, Laplace transforms are used to find solutions to firstorder linear differential equations in reliability, maintainability, and safety analysisrelated studies of engineering systems. The example presented below demonstrates the finding of solutions to a set of linear firstorder differential equations, describing a engineering system in regard to reliability and safety, using Laplace transforms.
Example 2.10
Assume that an engineering system can be in any of the three states: operating normally, failed safely, or failed unsafely. The following three firstorder linear differential equations describe the engineering system under consideration:
^iû_{+}(x_{s}+x„)p_{o}(t)=o
(2.48)
^_{W}) = 0 at 
(2.49) 
(2.50) 
where
X_{s} is the engineering system constant safe failure rate, is the engineering system constant unsafe failure rate.
Pi (t) is the probability that the engineering system is in state i at time, for i = 0 (operating normally), i = 1 (failed safely), and i = 2 (failed unsafely).
At time t = 0, Po (0) = 1, Pi (0) = 0, and P_{2} (0) = 0.
Solve differential Equations (2.48)—(2.50) by using Laplace transforms.
Using Table 2.1, differential Equations (2.48)(2.50), and the given initial conditions, we obtain
sP_{0}(s)l + (X_{s}+X„)P_{o}(s) = O 
(2.51) 
sP_{!}(s)X_{s}Po(s)=0 
(2.52) 
sft(s)X„P_{u}(s) = 0 
(2.53) 
By solving Equations (2.51)—(2.53), we get
Po(s)=, ? . ' (s + A._{s} + X„) Pl(s)= z x A(s)= z + X_{M} J 

By taking the inverse Laplace transforms of Equations (2.54)(2.56), we obtain
P_{0}(t) = e‘<^{Xs+x}“)' 
(2.57) 

Thus, Equations (2.57)(2.59) are the solutions to differential Equations (2.48)(2.50).