Question

8.7. The servicing of a machine requires two separate steps, with the time needed for the first step being an exponential random variable with mean $.2$hour and the time for the second step being an independent exponential random variable with mean $.3$hour. If a repair person has $20$machines to service, approximate the probability that all the work can be completed in $8$ hours.

Short answer

The probability is $.1075$.

Step-by-step solution

Given information

An exponential random variable with mean $.2$hour and an independent exponential random variable with mean $.3$ hour.

Explanation

Let $X_{1,j}$represents the time (in hours) needed for first step of servicing $j$th machine and $X_{2,k}$represents the time (also in hours) needed for second step of servicing $k$th machine. The two random variables are independent, the variable $X_{1,j}$is an exponential random variable with parameter${\lambda }_1$with mean.
${\mu }_1=E\left[X_{1,j}\right]=\frac{1}{{\lambda }_1}=.2$
Then, the variable $X_{2,k}$is an exponential random variable with parameter${\lambda }_2$ with mean:
${\mu }_2=E\left[X_{2,k}\right]=\frac{1}{{\lambda }_2}=.3$
Finally,
${\lambda }_1=\frac{1}{.2},{\lambda }_2=\frac{1}{.3}$
Hence, the appropriate variances are:
$$\begin{gathered} {\sigma }_1^2=Var\left(X_{1,j}\right)=\frac{1}{{\lambda }_1^2}=.04 \\ {\sigma }_2^2=Var\left(X_{2,k}\right)=\frac{1}{{\lambda }_2^2}=.09 \end{gathered}$$

Explanation

A repair person has $20$machines to service and let $Y$denote the total time of servicing $i$th machine,$i=1,2,\dots ,20$. Then,$Y_i=X_{1,i}+X_{2,i}$
and the total time of finishing all the work is,
$Y=\sum _{i=1}^{20}{Y}_i=\sum _{i=1}^{20}\left(X_{1,i}+X_{2,i}\right)$
The probability that all the work can be completed in 8 hours is:
$P\{Y\le 8\}$
Since,$\left(X_{1,1}+X_{2,1}\right),\left(X_{2,1}+X_{2,2}\right),\dots ,\left(X_{1,20}+X_{2,20}\right)$is a sequence of independent and identically distributed random variables, each with mean
$\mu =E\left[X_{1,i}+X_{2,i}\right]={\mu }_1+{\mu }_2=.5$
The variance is,
${\sigma }^2=Var\left(X_{1,i}+X_{2,i}\right)={\sigma }_1^2+{\sigma }_2^2=.13$

Probability calculation

Using The central limit theorem:
$$\begin{gathered} P\{Y\le 8\}=P\left\{\frac{Y-20\mu }{\sigma \sqrt{2}}\le \frac{8-20\mu }{\sigma \sqrt{2}}\right\}\approx \mathrm{\Phi }\left(\frac{8-20\mu }{\sigma \sqrt{20}}\right) \\ =\mathrm{\Phi }\left(\frac{8-20\left(.5\right)}{\sqrt{20\left(.13\right)}}\right)=\mathrm{\Phi }(-1.24)=1-\mathrm{\Phi }\left(1.24\right) \\ =1-.8925=.1075 \end{gathered}$$

Hence, the probability is $.1075$.