Probability unif

Question

By the Central Limit Theorem, I am asked estimate the probability that P$\leq\ $$\mathsf E(P)$. The answer of 50% is what I am getting but this seems incorrect and can't seem to convince myself otherwise. If anyone could shed some light on this as to why this incorrect or even how to think about this problem properly, it would be greatly appreciated.

Answer 1

If you want to use the CLT on a product, take logs. $\log(P)=\sum_{i=1}^5\log(X_i).$ So by the CLT (if $5$ were a lot larger), $P$ would be a log-normal.

So the question is what is the probability for a log-normal to be less than its mean, which is not $50\%$. Looking up the relevant stuff for the standard parametrization on wikipedia, $$ P(X\le E(X)) = \int_0^{e^{\mu+\sigma^2/2}} \frac{1}{x\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2\sigma^2}(\log(x)-\mu)^2} = \int_{-\infty}^{\frac{\sigma}{2}} \frac{1}{\sqrt{2\pi}}e^{-u^2/2}du = \Phi(\sigma/2).$$ So it's going to depend on the variance of $\log(X)$.