On the wikipedia page in my native language it is stated that according to the central limit theorem for $X_i$ iid with finite mean and variance we have that $\bar{X}_n$ is approximately normal. Isn’t this wrong? Take for example $X_i$ bernoulies then $0<\bar{X}_n<1$. So it can’t be normally distributed right?
I always thought that it only holds for the rescaled average. Can someone help me out here?
It is true that if $X_i \sim Bernoulli(p)$ that $0 \leq \bar{X} \leq 1$. But also, $\sigma^2 = Var(\bar{X}) = \frac{1}{n}p(1-p)$, so as $n$ grows larger the variance gets smaller. The Central Limit Theorem says that $\bar{X} \xrightarrow{dist} N(\mu, \sigma^2)$, and so when $n$ gets larger the part of the normal distribution that sits outside of $[0, 1]$ will flatten out so that practically none of the probability mass remains.
If you look at it in terms of the distribution of $Z = \frac{X - \mu}{\sigma}$, you'll see that the range of $Z$ is something like $\left[-\sqrt{\frac{np}{1-p}}, \sqrt{\frac{n(1-p)}{p}} \right]$, which tends to $(-\infty, \infty)$ as $n \rightarrow \infty$.