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.

  • 1
    $\begingroup$ Why do you understand that the distribution of the averages is normal? For me, the size is too small for the sample to converge. Note that if $X$ is an exponential, then $X \sim \Gamma(1,\lambda)$ and $ \sum_{i=1}^{n} X_i \sim \Gamma(n,\lambda)$. Maybe that helps you! :) $\endgroup$ Commented Oct 19, 2018 at 2:57
  • $\begingroup$ I don't know that Greek letter, or how to search for the equation you have given. $\endgroup$ Commented Oct 19, 2018 at 16:14
  • $\begingroup$ Addendum, I'm tutoring a student in a beginner's statistic class at a community college, and I'm actually learning statistics with her, so that's why I'm uncertain what to do. However, given how this course is not supposed to be difficult, and this problem comes from the chapter on the central limit theorem, I do think it's probably an application of the CLT. $\endgroup$ Commented Oct 19, 2018 at 18:13

2 Answers 2


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:

$$ f(x|p) = \frac{(np)^n}{\Gamma(n)} x^{n-1} e^{-np x}$$


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 :)


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