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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?

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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 $\mathrm{SD}(X) = \sqrt{100} \cdot \frac 1{10} = 1$.

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Thanks for the clear solution, but regarding the solution using Gamma-distribution, I was under the assumption that $\lambda = \frac{1}{10}$, and that for Gamma, $\mathbb{E}[X] = \frac{r}{\lambda}$, why is $\lambda = 10$ in the case of $X = \Gamma(1,10)$ When you apply linearity of expectations, does the second argument of Gamme multiply as well? –  zhuyxn Nov 1 '12 at 10:37
There are different common ways to denote the parameters for a Gamma-distribution. I know $\theta$ and $k$, which should correspond to your $r$, $\lambda$ via $r = k$ and $\lambda = \theta^{-1}$, so in $r, \lambda$-notation $X_i \sim \Gamma(1, 1/10)$, $\sum X_i \sim \Gamma(100, 1/10)$, $X \sim \Gamma(100, 10)$. Therefore $\mathbb E[X] = 100/10 = 10$ and $\mathrm{SD}(X) = \frac{\sqrt r}{\lambda} = 10/10 = 1$. –  martini Nov 1 '12 at 10:41
Sorry, I'm a bit a confused (possibly over notation), assuming that $\mathbb{E}[X] = \frac{r}{\lambda}$ and $\lambda = \frac{1}{10}$, wouldn't $\mathbb{E}[X] = \frac{100}{\frac{1}{10}} = 1000$? When you find $X$, why does the $\lambda$ get multiplied by 100? –  zhuyxn Nov 1 '12 at 10:53
In your last step in the main solution, when you take $X = \Gamma(100,1/10)$, what does that conceptually mean in terms of dividing the survival time ($\theta$) by 100 from 10 to $\frac{1}{10}$? Why not divide $k$ by 100? –  zhuyxn Nov 1 '12 at 11:49
Well... it's known, that if $Y \sim \Gamma(k, \theta)$, then $\alpha Y \sim \Gamma(k, \alpha\theta)$. –  martini Nov 1 '12 at 12:37
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