91Ó°ÊÓ

The time between failures of a component was approximated by an exponential distribution with mean $\mathrm{θ}$.

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

Therefore, X follows an exponential distribution with mean$\mathrm{θ}$

The p.d.f of $X$is

$f\left(x\right)=\frac{1}{\mathrm{θ}}{e}^{−\frac{x}{\mathrm{θ}}};xâ‰\yen 0$

a.

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

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

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 $P(Xâ‰\yen 1200)$.

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 .

$\begin{array}{c}P(X<1500|Xâ‰\yen 1200)=\frac{P(1200≤X<1500)}{P(Xâ‰\yen 1200)}\\ =\frac{{\displaystyle \underset{1200}{\overset{1500}{∫}}f\left(x\right)dx}}{{\displaystyle \underset{1200}{\overset{\mathrm{∞}}{∫}}f\left(x\right)dx}}\\ =\frac{{\displaystyle \underset{1200}{\overset{1500}{∫}}\frac{1}{100}{e}^{\frac{−x}{1000}}dx}}{{\displaystyle \underset{1200}{\overset{\mathrm{∞}}{∫}}\frac{1}{100}{e}^{\frac{−x}{1000}}dx}}\\ =\frac{−1000{\left[e^{\frac{−x}{1000}}\right]}_{1200}^{1500}}{−1000{\left[e^{\frac{−x}{1000}}\right]}_{1200}^{\mathrm{∞}}}\\ =\frac{e^{\frac{−1500}{1000}}−e^{\frac{−1200}{1000}}}{e^{\frac{−\mathrm{∞}}{1000}}−e^{\frac{−1200}{1000}}}\\ =\frac{e^{−1.5}−e^{−1.2}}{−e^{−1.2}}\\ =\frac{e^{−1.5}}{−e^{−1.2}}+\frac{−e^{−1.2}}{−e^{−1.2}}\\ =1−e^{−1.5+1.2}\\ =1−e^{−0.3}\\ =1−0.7408\\ =0.2592\end{array}$

$P(X<1500|Xâ‰\yen 1200)=0.259$

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