Question

In Problem$8.7$, suppose that it takes a random time, uniformly distributed over$(0,.5)$, to replace a failed bulb. Approximate the probability that all bulbs have failed by time$550$.

Short answer

Approximate the probability that all bulbs have failed by time $550$is$.6915$.

Step-by-step solution

Given Information.

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

Explanation.

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" $\lambda$. Then,

${\mu }_X-E\left[X_i\right]-\frac{1}{\lambda }\text{and}{\sigma }_X^2-Var\left(X_i\right)-\frac{1}{{\lambda }^2},$

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

${\mu }_X-5\text{and}{\sigma }_X^2-25\text{.}$

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

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

Explanation.

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

$X-\sum _1^{100}{X}_i$

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

$R-\sum _1^{99}{R}_i$

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

$P\left\{\sum _1^{100}{X}_i+\sum _1^{99}{R}_i\le 550\right\}-P\{X+R\le 550\}.(*)$

Explanation.

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

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

and

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

Explanation.

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

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