91Ó°ÊÓ

Given that it takes a random time, uniformly distributed over$(0,.5)$, to replace a failed bulb.

Assume that the bulbs are used one at a time, whereby if the $i$th bulb has failed, $R_i$is the time (in hours) to replace the bulb.

Let the lifetimes $X_i$be independent exponential random variables with parameterslocalid="1649759533550" $\mathrm{λ}$. Then,

${\mathrm{μ}}_X-E\left[X_i\right]-\frac{1}{\mathrm{λ}}\text{and}{\mathrm{σ}}_X^2-Var\left(X_i\right)-\frac{1}{{\mathrm{λ}}^2},$

and since it is given that the lifetimes are variables with the mean of$5$hours, we have that$\mathrm{λ}-1/5$. So,

${\mathrm{μ}}_X-5\text{and}{\mathrm{σ}}_X^2-25\text{.}$

On the other hand, the replacement times $R_i$are independent random variables. These variables are uniformly distributed$(0,.5)$. So,

${\mathrm{μ}}_n-E\left[R_i\right]-\frac{.5+0}{2}-.25$and${\mathrm{σ}}_n^2-Var\left(R_i\right)-\frac{(.5-0{)}^2}{12}-.0208$.

Let's $X$denote the total lifetime of all $100$light bulbs:

$X-\underset{1}{\overset{100}{∑}}{X}_i$

and let $R$denote the total replacement time of $99$light bulbs:

$R-\underset{1}{\overset{99}{∑}}{R}_i$

At first, notice that the probability that all bulbs have failed by time $550$is equal to the following probabilities:

$P\left\{\underset{1}{\overset{100}{∑}}{X}_i+\underset{1}{\overset{99}{∑}}{R}_i≤550\right\}-P\{X+R≤550\}.(*)$

Because of the independence of$X_i$, independence of $R_i$and also obviously between $X$and$R$, we have:

$E[X+R]-E\left[X\right]+E\left[R\right]-100{\mathrm{μ}}_X+99{\mathrm{μ}}_R-524.75$

and

$Var(X+R)-Var\left(X\right)+Var\left(R\right)-100{\mathrm{σ}}_X^2+99{\mathrm{σ}}_R^2-2502.0592$.

Finally, to approximate the desired probability $(*)$we use the central limit theorem:

localid="1649759682045" $$\begin{gathered} P\{X+R≤550\}-P\left\{\frac{(X+R)-524.75}{\sqrt{2502.0592}}≤\frac{550-524.75}{\sqrt{2502.0592}}\right\} \\ ≈\mathrm{Φ}(.5{)}^{\text{Table}5.1\text{(textbook, Chapter 5)}} \\ =.6915 \end{gathered}$$