91Ó°ÊÓ

The repair rate for the machines follows exponential distribution with means${\mathrm{μ}}_1=1$,${\mathrm{μ}}_2=2$ ${\mathrm{μ}}_3=0.5$${\mathrm{μ}}_4=0.5$.

a.

Probability density function for exponential distribution is:

$f\left(x\right)=\frac{1}{{\mathrm{μ}}_1}exp\left(\frac{-x}{{\mathrm{μ}}_1}\right),0<x<∞$

Given$\mathrm{μ}=1$

role="math" localid="1660363205292" $f\left(x\right)=\frac{1}{1}\left(exp\left(-\frac{x}{1}\right)\right)$

Find the area to the right of a number exponential distribution:

$\begin{array}{l}A=P\left(x>a\right)\\ =exp\left(-\frac{a}{\mathrm{μ}}\right)\\ =exp\left(-\frac{a}{1}\right)\end{array}$

Now probability for the event that machine repair time exceeds one hour is same as that of probability for the event X>1 is

$$\begin{gathered} P\left(X>1\right)=exp\left(-\frac{1}{1}\right) \\ =exp\left(-1\right) \\ =0.3678 \end{gathered}$$

Hence probability that the repair time for machine 1 exceeds 1 hour is 0.3679

b.

Probability density function for exponential distribution is:

$f\left(x\right)=\frac{1}{{\mathrm{μ}}_2}exp\left(\frac{-x}{{\mathrm{μ}}_2}\right),0<x<∞$

Given

${\mathrm{μ}}_2=2$

$f\left(x\right)=\frac{1}{2}\left(exp\left(-\frac{x}{2}\right)\right)$

The area to the right of a number exponential distribution:

$\begin{array}{l}A=P\left(x>a\right)\\ =exp\left(-\frac{a}{\mathrm{μ}}\right)\\ =exp\left(-\frac{a}{2}\right)\end{array}$

Now probability for the event that machine repair time exceeds one hour is same as that of probability for the event X>1 is

$$\begin{gathered} P\left(X>1\right)=exp\left(-\frac{1}{2}\right) \\ =exp\left(-0.5\right) \\ =0.6065 \end{gathered}$$

Hence probability that the repair time for machine 2 exceeds 1 hour is 0.6065.

c.

Probability density function for exponential distribution is:

$f\left(x\right)=\frac{1}{{\mathrm{μ}}_3}exp\left(\frac{-x}{{\mathrm{μ}}_3}\right),0<x<∞$

Given${\mathrm{μ}}_3=0.5$

$f\left(x\right)=\frac{1}{0.5}\left(exp\left(-\frac{x}{0.5}\right)\right)$

Find the area to the right of a number exponential fistribution:

$\begin{array}{l}A=P\left(x>a\right)\\ =exp\left(-\frac{a}{\mathrm{μ}}\right)\\ =exp\left(-\frac{a}{0.5}\right)\end{array}$

Now probability for the event that machine repair time exceeds one hour is same as that of probability for the event X>1 is

$$\begin{gathered} P\left(X>1\right)=exp\left(-\frac{1}{0.5}\right) \\ =exp\left(-2\right) \\ =0.1353 \end{gathered}$$

Hence probability that the repair time for machine 3 exceeds 1 hour is 0.1353

Probability density function for exponential distribution is:

$f\left(x\right)=\frac{1}{{\mathrm{μ}}_4}exp\left(\frac{-x}{{\mathrm{μ}}_4}\right),0<x<∞$

Given

${\mathrm{μ}}_4=0.5$

$f\left(x\right)=\frac{1}{0.5}\left(exp\left(-\frac{x}{0.5}\right)\right)$

Find the area to the right of a number exponential distribution:

$\begin{array}{l}A=P\left(x>a\right)\\ =exp\left(-\frac{a}{\mathrm{μ}}\right)\\ =exp\left(-\frac{a}{0.5}\right)\end{array}$

Now probability for the event that machine repair time exceeds one hour is same as that of probability for the event X>1 is

$$\begin{gathered} P\left(X>1\right)=exp\left(-\frac{1}{0.5}\right) \\ =exp\left(-2\right) \\ =0.1353 \end{gathered}$$

Hence probability that the repair time for machine 3 exceeds 1 hour is 0.1353

If all four machines fails simultaneously then, the probability that the repair time for the entire system exceed some hour is

$$\begin{gathered} P\left(\begin{array}{c}repairtimeforthe\\ entiresystemexceeds1hour\end{array}\right)=1-\left[\begin{array}{c}\left(1-0.3678\right)×\left(1-0.6065\right)\\ \mathrm{x}\left(1-0.135\right)×\left(1-0.135\right)\end{array}\right] \\ =1-\left[0.632×0.3934×0.8646×0.8646\right] \\ =1-0.1859 \\ P\left(\begin{array}{c}repairtimeforthe\\ entiresystemexceeds1hour\end{array}\right)=0.8140 \end{gathered}$$

Hence the probability that the repair time for the entire system exceeds 1 hour is 0.814