How do I compute the distribution of averages of an exponential distribution? I'm given an exponential distribution and told the mean is $0.88$. I believe that means the distribution is notated $Exp(\lambda)$ where $\lambda = \frac{1}{0.88} $; is that correct? Then I'm told that groups of 25 samples of this distribution are averaged and I'm asked to find the distribution of the averages, notated $\bar{X}$.  
So my understanding is that this is an application of the central limit theorem. The distribution of $\bar{X}$ will be normal, and I'm asked to find the mean and standard deviation.
I think the way to look at this is that the exponential distribution is the population with a mean of $0.88$ and also a standard deviation of $0.88$ (I believe that the standard deviation of an exponential distribution is equal to the mean). My sample mean will match the population mean, that is, $0.88$. 
What I'm iffy about is the next step. Is it true that I can apply this?
$$s = \frac{\sigma}{\sqrt{N}}  = \frac{0.88}{5} = 0.176 $$
That is, the sample standard deviation will be the population std deviation, $\sigma = 0.88$ divided by square root of the sample size, $25$?
Does this formula work with any original population for which $\sigma$ can be computed? That's pretty amazing if so.
 A: Yet the rigorous answer is as it follows: 
Note that the exponential distribution is given by: $$ f(x|p) = p e^{-px} $$ where $p$ is your parameter. Note that it is also a gamma $\Gamma(1,p)$distribution. The general gamma $\Gamma(\alpha,\beta)$ is given by: $$ f(x|p) = \frac{1}{\Gamma(\alpha)}\beta^{\alpha} x^{\alpha-1} e^{-\frac{x}{\beta}}$$ so it's not difficult to see that it's a gamma. 
We can evaluate: 
$$ P\left( \frac{X_i}{n} \leq a \right) = P(X_i \leq na)  = \int_{0}^{na} p e^{-px} dx = 1 - e^{-np a}$$ 
Hence, for $Y=\frac{X}{n}$: $$ f(y|p) = np e^{-np y}$$
This happens to be a gamma $\Gamma(1, np)$. So: 
$$ \frac{\sum_{i=1}^{n} X_i}{n} \sim \Gamma(n,np)$$
Finally we evaluate the distribution of that random variable, which is: 
$$ f(x|p) = \frac{(np)^n}{\Gamma(n)} x^{n-1} e^{-np x}$$
A: I am not quite sure you can use CLT. Still, let $X_i$ be an exponential. It's not hard to show that $E(X_i) = \frac{1}{\lambda} $ and $ Var(X_i) = \frac{1}{\lambda^2}$. Then:
$$ P\left( \frac{\sum_{i=1}^{n} X_i}{n} \leq a \right) = P\left( \frac{\sum_{i=1}^{n} X_i - n\mu}{n} \leq a  - \mu \right)  $$
$$P\left( \frac{\sum_{i=1}^{n} X_i - n\mu}{\sigma\sqrt{n}} \leq \frac{(a  - \mu) \sqrt{n}}{\sigma}\right)  = P\left(Z \leq \frac{(a  - \mu) \sqrt{n}}{\sigma} \right)$$  where $Z \sim N(0,1)$. 
Notice that the left side of the inequality is precisely the equation of the central limit theorem. In your case, $n=25$, $\mu =1/\lambda$ and $ \sigma = 1/\lambda^2 $. 
Now you can say it's approximatelly normal. 
$$ P\left( Z \leq \frac{5(a-0.88)}{0.88} \right) $$. 
That doesn't give you an answer about the distribution, though... 
Yes, you can use the equality above. Because since all $X_i$ are iid, you have: 
$$ Var\left(\frac{\sum X_i}{n}\right) = \sum_{i=1}^{n} \frac{1}{n^2} Var(X_i) = \frac{\sigma^2}{n}$$
Now for sd: $$ SD\left(\frac{\sum X_i}{n} \right) = \frac{\sigma}{\sqrt{n}}$$
Note that this is not the standard deviation of your sample, but the standard deviation of your estimator to the populational mean! 
Hope I helped :)
