Preventative maintenance tests. The optimal schedulingofpreventative maintenance tests of some (but not all) ofnindependently operating components was developed in Reliability Engineering and System Safety(January$\mathbf{2006}$).The time (in hours) between failures of a component wasapproximated by an exponential distribution with mean$\mathit{\theta }$.

a. Suppose$\mathit{\theta }\mathbf{=}\mathbf{1000}$ hours. Find the probability that the time between component failures ranges between $\mathbf{1200}$and$\mathbf{1500}$hours.

b. Again, assume$\mathit{\theta }\mathbf{=}\mathbf{1000}$hours. Find the probability that the time between component failures is at least$\mathbf{1200}$hours.

c. Given that the time between failures is at leastrole="math" localid="1658214710824" $\mathbf{1200}$ hours, what is the probability that the time between failures is less than$\mathbf{1500}$hours?

The time between failures of a component was approximated by an exponential distribution with mean $\theta$.

Let, $X$denote the time between failures of a component.

Therefore, X follows an exponential distribution with mean$\theta$

The p.d.f of $X$is

$f\left(x\right)=\frac{1}{\theta }{e}^{-\frac{x}{\theta }};x\ge 0$

a.

The probability that the time between component failure ranges between 1200 and 1500 is$P(1200\le X\le 1500)$.

$\begin{array}{c}P(1200\le X\le 1500)=\underset{1200}{\overset{1500}{\int }}f\left(x\right)dx\\ =\underset{1200}{\overset{1500}{\int }}\frac{1}{1000}{e}^{-\frac{x}{1000}}dx\\ =\frac{1}{1000}\underset{1200}{\overset{1500}{\int }}{e}^{-\frac{x}{1000}}dx\\ =\frac{1}{1000}(1000e^{-\frac{6}{5}}-1000e^{-\frac{3}{2}})\\ =e^{-\frac{6}{5}}-e^{-\frac{3}{2}}\\ \approx 0.078\end{array}$

Hence, the probability that the time between component failure ranges between 1200 and 1500 is 0.078.

b.

The probability thatthe time between component failures is at least 1200 hours is $P(X\ge 1200)$.

Hence, the probability that the time between component failures is at least 1200 hours is 0.301.

c.

If the time between component failures is at least 1200 hours, then the probability that the time between component failures is less than 1500 hours is .

$\begin{array}{c}P(X<1500|X\ge 1200)=\frac{P(1200\le X<1500)}{P(X\ge 1200)}\\ =\frac{{\displaystyle \underset{1200}{\overset{1500}{\int }}f\left(x\right)dx}}{{\displaystyle \underset{1200}{\overset{\infty }{\int }}f\left(x\right)dx}}\\ =\frac{{\displaystyle \underset{1200}{\overset{1500}{\int }}\frac{1}{100}{e}^{\frac{-x}{1000}}dx}}{{\displaystyle \underset{1200}{\overset{\infty }{\int }}\frac{1}{100}{e}^{\frac{-x}{1000}}dx}}\\ =\frac{-1000{\left[e^{\frac{-x}{1000}}\right]}_{1200}^{1500}}{-1000{\left[e^{\frac{-x}{1000}}\right]}_{1200}^{\infty }}\\ =\frac{e^{\frac{-1500}{1000}}-e^{\frac{-1200}{1000}}}{e^{\frac{-\infty }{1000}}-e^{\frac{-1200}{1000}}}\\ =\frac{e^{-1.5}-e^{-1.2}}{-e^{-1.2}}\\ =\frac{e^{-1.5}}{-e^{-1.2}}+\frac{-e^{-1.2}}{-e^{-1.2}}\\ =1-e^{-1.5+1.2}\\ =1-e^{-0.3}\\ =1-0.7408\\ =0.2592\end{array}$

$P(X<1500|X\ge 1200)=0.259$

Hence, the probability that the time between component failures is less than 1500 hours is 0.259.