A person has $100$ light bulbs whose lifetimes are independent exponentials with mean $5$ hours. The bulbs are used one at a time, with a failed bulb being replaced immediately by a new one.
(a) Approximate the probability that there is still a working bulb after $525$ hours.
(b) Suppose that it takes a random time, uniformly distributed over $(0,0.5)$, to replace a failed bulb. Approximate the probability that all bulbs have failed by time $550$.
Here is my work:
Here is what I know:
For a sequence of random variable $X_1, X_2, \ldots, X_n$ where they all have the same distribution we have:
\begin{equation} \mu = E(X_i) \end{equation} \begin{equation} \sigma^2 = \operatorname{Var}(X_i) \end{equation} \begin{equation} S_n = X_1 + X_2 + \cdots + X_n \end{equation} \begin{equation} \overline{X}_n = \frac{S_n}{n}=\frac{X_1 + X_2 + \cdots + X_n}{n} \end{equation} \begin{equation} P\left(a \leq \frac{S_n - n\mu}{\sigma\sqrt{n}} \leq b\right) \approx \Phi(b) - \Phi(a) \end{equation}
$\overline{X}_n$ and $S_n$ approximately follows a normal distribution
so for
$a)$ here is what I did:
Let $X = $ lifetime of each lightbulb, $X_i\sim\exp(\lambda)$, and $\lambda = \frac{1}{5}$
Then $S_n = \sum_iX_i$, $\mu = 5$, $\sigma^2=25$
So I'm looking for $P(S_n>525)$
This turns into \begin{equation} P\left(\frac{S_n - n\mu}{\sigma\sqrt{n}} > \frac{525-n\mu}{\sigma\sqrt{n}}\right) \end{equation}
\begin{equation} P(Z>0.5) = 1 - P(Z<0.5) =0.308 \end{equation}
$b)$ Here is what I did:
Let $U_i\sim\mathrm{uniform}(0, 0.5)$.
Therefor $U_n = \sum_iU_i$
and I said let $L_n = U_n + S_n$ and $\mu_L = E(U) + E(X)$
and since $U_i$ and $X_i$ are independent I said $\sigma_L^2 = \operatorname{Var}(U_i) + \operatorname{Var}(X_i)$
Thus following same logic as $a)$ gives me
\begin{equation} P\left(\frac{L_n - n\mu_L}{\sigma_L\sqrt{n}} < \frac{550-n\mu_L}{\sigma_L\sqrt{n}}\right) \end{equation}
\begin{equation} P(Z<0.5) = 0.691 \end{equation}
Is this correct? I made some assumptions on $b)$ so I'm not sure.
