91影视

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

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

And, ${蟽}_2^2=\mathrm{Var}鈦�\left(X_{2,k}\right)=\frac{1}{{位}_2^2}=.09$.

A repair person has $20$machines to service and let $Y$represent the total time of servicing $i$th machine, $i=1,2,鈥�,20$. Then, for each $i$,
$Y_i=X_{1,i}+X_{2,i}$
Then the total time of finishing all the work is:
$Y=\underset{i=1}{\overset{20}{鈭�}}{Y}_i=\underset{i=1}{\overset{20}{鈭�}}\left(X_{1,i}+X_{2,i}\right)$

The sequence with mean is:

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

The variance is:

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

The probability is:
$P\{Y鈮�t\}$
Using The central limit theorem:
$$\begin{gathered} P\{Y鈮�t\}=P\left\{\frac{Y-20渭}{蟽\sqrt{2}}鈮�\frac{t-20渭}{蟽\sqrt{2}}\right\} \\ 鈮�桅\left(\frac{t-20渭}{蟽\sqrt{20}}\right)=桅\left(\frac{t-20(.5)}{\sqrt{20(.13)}}\right) \\ 桅\left(\frac{t-20(.5)}{\sqrt{20(.13)}}\right)鈮�.95 \\ 鈬�\frac{t-20(.5)}{\sqrt{20(.13)}}鈮�桅(.95)=1.64 \\ t鈮�12.64. \end{gathered}$$

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