Let $X_1,...,X_{200}$ independent random variables which have a distribution of $U(0,1)$. Find:
$$P(\prod_{i=1}^{200}X_i\le 0.34^{200})$$
First, I tried to deal with the product, since I can't deal with it, and I can only deal with sums using CLT, after alot of thought I've done this step:
$P(\ln(\prod_{i=1}^{200}X_i)\le 200\ln(0.34))=P(\frac{1}{200}\sum_{i=0}^{200}\ln(X_i)\le \ln(0.34))$
Now I have a sum of i.i.d, so let $S_{200}=\frac{1}{200}\sum_{i=0}^{200}\ln(X_i)$.
next, in order to use CLT, I must know $E(\ln(X_i))$ and $Var(\ln(X_i))$.
I can see that $X_i\sim U(0,1) \Longrightarrow$ The possible results of $\ln(X_i)$ are in $(-\infty,0]$, but I still cannot reach conclusions about how $\ln(X_i)$ is distributed.
So, I've tried to use the definition to calculate:
$E(\ln(X_i))=\int_{-\infty}^{\infty}\ln(x)f_X(x)dx=\int_0^1\ln(x)1dx=x\ln(x)-x|^1_0=-1$.
$E([\ln(X_i)]^2)=\int_0^1\ln(x)^2dx=\ln(x)^2x-\int_0^12\ln(x)dx=\ln(x)^2x-2(x\ln(x)-x)|^1_0=2.$
Applying CLT:
$P(\frac{S_{200}-200*(-1)}{\sqrt{200*3}}\le\frac{\ln(0.34)+200}{\sqrt{600}})=\phi(8.12)$ --- Which I can't calculate, so I surely have some mistakes, but I can't really find them.
I would appreciate any help in pointing out where did I mess up & would be happy to see more ways of how to solve this question (tbh I hated the integrals, they needed integration by parts etc..).