Originally Posted by

**heathrowjohnny** Let $\displaystyle X_{1}, X_{2}, \ldots, X_{n} $ be a random sample from a distribution with mean $\displaystyle \mu $ and variance $\displaystyle \sigma^{2} $. Then, in the limit as $\displaystyle n \to \infty $, the standardized versions of $\displaystyle \bar{X} $ and $\displaystyle T_0 $ have the standard normal distribution. That is $\displaystyle \lim_{n \to \infty} P \left (\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \leq z \right) = P(Z \leq z) = \Phi(z) $ and $\displaystyle \lim_{n \to \infty} P \left (\frac{T_0 - n\mu}{\sqrt{n}\sigma} \leq z \right) = P(Z \leq z) = \Phi(z) $.

My question is, why should this be the case? Why shouldn't $\displaystyle \bar{X} $ and $\displaystyle T_{0} $ have a lognormal distribution, an exponential distribution, etc..?