Question

Determine $t$ so that the probability that the repair person in Self-Test Problem 8.7 finishes the $20$ jobs within time t is approximately equal to $.95$.

Short answer

Determined the time $t$as $12.64$ hours.

Step-by-step solution

Given information

The $20$ jobs within time $t$is approximately equal to $.95$.

Explanation

Let $X_{1,j}$denotes the time needed for first step of servicing $j$th machine and $X_{2,k}$denotes the time needed for second step of servicing $k$th machine.

localid="1650025718171" $$\begin{gathered} {\mu }_1=E\left[X_{1,j}\right]=\frac{1}{{\lambda }_1}=.2 \\ {\mu }_2=E\left[X_{2,k}\right]=\frac{1}{{\lambda }_2}=.3 \end{gathered}$$
Finally,
${\lambda }_1=\frac{1}{.2},{\lambda }_2=\frac{1}{.3}$
Hence the appropriate variances are:
localid="1649745208470" ${\sigma }_1^2=\mathrm{Var}\left(X_{1,j}\right)=\frac{1}{{\lambda }_1^2}=.04,$

And, ${\sigma }_2^2=\mathrm{Var}\left(X_{2,k}\right)=\frac{1}{{\lambda }_2^2}=.09$.

Explanation

A repair person has $20$machines to service and let $Y$represent the total time of servicing $i$th machine, $i=1,2,\dots ,20$. Then, for each $i$,
$Y_i=X_{1,i}+X_{2,i}$
Then 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 sequence with mean is:

localid="1649744892768" $\mu =E\left[X_{1,i}+X_{2,i}\right]={\mu }_1+{\mu }_2=.5$

The variance is:

$\sigma =Var\left(X_{1,i}+X_{2,i}\right)={\sigma }_1^2+{\sigma }_2^2=.13$

Probability calculation

The probability is:
$P\{Y\le t\}$
Using The central limit theorem:
$$\begin{gathered} P\{Y\le t\}=P\left\{\frac{Y-20\mu }{\sigma \sqrt{2}}\le \frac{t-20\mu }{\sigma \sqrt{2}}\right\} \\ \approx \Phi \left(\frac{t-20\mu }{\sigma \sqrt{20}}\right)=\Phi \left(\frac{t-20(.5)}{\sqrt{20(.13)}}\right) \\ \Phi \left(\frac{t-20(.5)}{\sqrt{20(.13)}}\right)\approx .95 \\ \Rightarrow \frac{t-20(.5)}{\sqrt{20(.13)}}\approx \Phi (.95)=1.64 \\ t\approx 12.64. \end{gathered}$$

Hence, the $t$is $12.64$hours.