91Ó°ÊÓ

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

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${\mathrm{λ}}_1$with mean.
${\mathrm{μ}}_1=E\left[X_{1,j}\right]=\frac{1}{{\mathrm{λ}}_1}=.2$
Then, the variable $X_{2,k}$is an exponential random variable with parameter${\mathrm{λ}}_2$ with mean:
${\mathrm{μ}}_2=E\left[X_{2,k}\right]=\frac{1}{{\mathrm{λ}}_2}=.3$
Finally,
${\mathrm{λ}}_1=\frac{1}{.2},{\mathrm{λ}}_2=\frac{1}{.3}$
Hence, the appropriate variances are:
$$\begin{gathered} {\mathrm{σ}}_1^2=Var\left(X_{1,j}\right)=\frac{1}{{\mathrm{λ}}_1^2}=.04 \\ {\mathrm{σ}}_2^2=Var\left(X_{2,k}\right)=\frac{1}{{\mathrm{λ}}_2^2}=.09 \end{gathered}$$

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

Using The central limit theorem:
$$\begin{gathered} P\{Y≤8\}=P\left\{\frac{Y−20\mathrm{μ}}{\mathrm{σ}\sqrt{2}}≤\frac{8−20\mathrm{μ}}{\mathrm{σ}\sqrt{2}}\right\}≈\mathrm{Φ}\left(\frac{8−20\mathrm{μ}}{\mathrm{σ}\sqrt{20}}\right) \\ =\mathrm{Φ}\left(\frac{8−20\left(.5\right)}{\sqrt{20\left(.13\right)}}\right)=\mathrm{Φ}(−1.24)=1−\mathrm{Φ}\left(1.24\right) \\ =1-.8925=.1075 \end{gathered}$$

Hence, the probability is $.1075$.