For a sample $X_1, \ldots, X_n$ from an exponential distribution with unknown (rate) parameter $\lambda$, the sum $S = \sum_{i=1}^n X_i$ is a sufficient statistic.
Moreover, this has a Gamma distribution with parameters $n$ and $\lambda$.
For $n = 5000$, the normal approximation should be quite good.
That Gamma distribution has mean $\mu = n/\lambda$ and standard deviation
$\sigma = \sqrt{n}/\lambda$. So we have $ \mu - c \sigma \le S \le \mu + c \sigma$ with probability $0.95$, where $c \approx 1.96$. Since
$\mu \pm c \sigma = (n \pm c \sqrt{n})/\lambda$, this translates to a 95\% confidence interval for $1/\lambda$ of $[S/(n + 1.96 \sqrt{n}), S/(n - 1.96 \sqrt{n}]$.
Note that the median of the exponential distribution with parameter $\lambda$ is
$\sqrt{2}/\lambda$. So the confidence interval for the median is
$[\sqrt{2} S/(n + 1.96 \sqrt{n}), \sqrt{2} S/(n - 1.96 \sqrt{n})]$$
Let's label the $N$ drives with numbers $i \in \{1, 2, \ldots, N\}$ and denote the random lifetimes of each drive as $T_1, T_2, \ldots, T_N$. Each one is independent and identically distributed as an exponential random variable with mean $\tau$. Let $$F_{T_i}(t) = \Pr[T_i \le t]$$ be the cumulative distribution function that gives the probability that drive $i$ has failed by time $t$.
Now let $T_{(1)}$ represent the failure time of the first drive to fail when all $N$ drives are operated simultaneously. Then we have $$T_{(1)} = \min_i (T_1, T_2, \ldots, T_N);$$ that is to say, it is the minimum of the set of random failure times of all the drives. So for instance, if $N = 5$ and we ran each drive until failure and observed $(T_1, T_2, T_3, T_4, T_5) = (10, 25, 34, 15, 9)$, then we have $T_{(1)} = 9$, the smallest observed failure time.
The question you are interested in is, what is $$F_{T_{(1)}}(t) = \Pr[T_{(1)} \le t]?$$ Well, it is easier to work with the complementary probability--the survival function $$S_{T_{(1)}}(t) = \Pr[T_{(1)} > t] = 1 - \Pr[T_{(1)} \le t] = 1 - F_{T_{(1)}}(t).$$ The survival function of the first/minimum failure time is $$\Pr[T_{(1)} > t] = \Pr[\min(T_1, T_2, \ldots, T_N) > t] = \Pr[(T_1 > t) \cap (T_2 > t) \cap \cdots \cap (T_N > t)],$$ because if the smallest of the $T_i$ exceeds $t$, we know that all of the $T_i$ exceed $t$; and vice versa--if all of the $T_i$ exceed $t$, then the smallest also exceeds $t$. This is why we switched from the CDF to the survival, because this logic doesn't work properly if we used the CDF, since $T_{(1)} \le t$ does not guarantee that the other $T_i$ are also $t$ or less; they can be greater.
Now because each $T_i$ is independent, the probability of the intersection of the events $(T_1 > t) \cap (T_2 > t) \cap \cdots \cap (T_N > t)$ is simply the product of the probability of each event; i.e., $$\Pr[T_{(1)} > t] = \Pr[T_1 > t] \Pr[T_2 > t] \cdots \Pr[T_N > t].$$ And because each of the $T_i$ are identically distributed, the LHS is simply the $N^{\rm th}$ power of any single probability of a single drive surviving past time $t$: $$\Pr[T_{(1)} > t] = (\Pr[T_1 > t])^N.$$ Written in terms of the CDF, we then have $$F_{T_{(1)}}(t) = 1 - S_{T_{(1)}}(t) = 1 - (\Pr[T_1 > t])^N = 1 - (1 - F_{T_1}(t))^N. \tag{1}$$
Note that our derivation does not use the fact that the $T_i$ are exponentially distributed, so formula $(1)$ is distribution-free. The only requirement is that the $T_i$ are independent and identically distributed.
Best Answer
The expected value, or average of a variable with exponential distribution is $\frac{1}{\lambda}$.