As everybody, when sampling a distribution, I was told that the population average can be estimated to be within the range of:

$$\mu \in \bar{x} \pm Z\_{\text{conf}}\frac{s}{\sqrt{n}}$$

where $Z\_{\text{conf}}=1.96$ for a confidence of 95%.

What I do not remember is whether this assumes the population data is Normally distributed, or just the sampling errors.

Below is a simulation (in R) of sampling of an Exponential distribution (and I have tried others), and this formula seems to work very well for non-Normal distribution. But I just want to be sure.

```
correct <- 0
pop_mean <- 800
nsim <- 100000
for(i in 1:nsim) {
    n <- 50
    s <- rexp(n, 1/pop_mean)
    mu <- mean(s)
    stderr <- sd(s) / sqrt(n)
    correct <- correct + (mu-1.96*stderr <= pop_mean & pop_mean <= mu+1.96*stderr)
}
print(correct/nsim)  # should give 0.95

```