E(X) and Standard Deviation of Independent Exponential Distributions Regarding a question, where the lifespan of a machine part is exponentially distributed with $\lambda=1/10$. When asked about the average lifespan of $100$ independent components, why is $\mathbb{E}(X)=10$ and $SD(X)=\frac{10}{\sqrt{100}}$? Can someone explain how one arrives at the standard deviation, and why its not equal to $\mathbb{E}(X)$, or why we don't use the Gamma Distribution (where $\mathbb{E}(X)=\frac{r}{\lambda}$ and $SD(X) =\frac{\sqrt{r}}{\lambda}$ respectively in this case? Am I misunderstanding the problem conceptually?
 A: So we are considering independent $X_1, \ldots, X_{100}$, where each $X_i$ is exponentially distributed and their mean $X := \frac 1{100}\sum_{i=1}^{100} X_i$. We want to compute its exceptation and its standard derivation. First we note, that by linearity of exceptation:
\[
\mathbb E[X] = \frac 1{100} \sum_{i=1}^{100} \mathbb E[X_i] = \frac 1{100} \cdot 100 \cdot 10 = 10 \]
and by independence we have addivity of the variance, 
\[
\sigma^2(X) = \frac 1{100^2} \sum_{i=1}^{100} \sigma^2(X_i) = \frac 1{100^2} \cdot 100 \cdot 10^2 = 1
\]
Hence $\mathrm{SD}(X) = \sqrt{\sigma^2(X)} = 1$.
Another way is using the Gamma-distribution, as you suggested. We have $X_i \sim \Gamma(1, 10)$ for each $i$, hence, by independence $\sum_i X_i \sim \Gamma(100, 10)$ and therefore $X = \frac 1{100}\sum_i X_i \sim \Gamma(100, 1/10)$. As a $\Gamma(k,\theta)$ distributed variable has expectation $k\theta$ and standard derivation $\sqrt k\theta$, we have $\mathbb E[X] = 100 \cdot \frac 1{10} = 10$ and $\mathrm{SD}(X) = \sqrt{100} \cdot \frac 1{10} = 1$.
