# What is the standard deviation of the sample average (sample size = 16) from a Laplace(0,1) population?

I've been running some simulations, and it seems clear to me that the Sample Mean $\overline{X}$ from a Laplace$(0,1)$ population is distributed normal with mean $0$. But I need to come up with the standard deviation and I'm stumped.

Specifically, I am looking at the sample mean (from a sample of size 16) from the Laplace(0,1) population. How could I find the standard deviation?

Suppose that $X_1,X_2,\dots,X_n$ are independent random variables, each with Laplace distribution, parameters $0$ and $1$. Let $Y=X_1+\cdots +X_n$. It looks as if you want the standard deviation of $\frac{Y}{n}$. We calculate the variance.
The random variable $Y$ has $\chi^2$ distribution. However, we do not need to know that to calculate the variance. By independence, the variance of $Y$ is the sum of the variances of the $X_i$. It follows that the variance of $\frac{Y}{n}$ is $\frac{1}{n}$ times the variance of (any) $X_i$.
For the variance of $X_1$, since the mean is $0$, we need to calculate $$\int_{-\infty}^\infty \frac{x^2}{2}e^{-|x|}\,dx.$$ By symmetry this is $$\int_0^\infty x^2e^{-x}\,dx.$$ The integral can be calculated using two integrations by parts. Alternately, it is closely related to the variance of the exponential with parameter $1$.
You should end up concluding that $\frac{Y}{n}$ has variance $\frac{2}{n}$.