For the case n=200 use the Central Limit Theorem to approximate the probability. The first part of this question I thought was pretty easy. It was:
The random variables $X_1,X_2,…$ are independent and identically distributed with the Probability Mass Function
$Pr(X=−1)=Pr(X=3)=0.5$
Find the mean:
$-1*0.5+3*0.5=1$
Find the Variance:
$((-1)^2*0.5+3^2*0.5)-E[X]^2=5-1=4$
Now for the second part, I am asked:
a) For the case $n=200$ use the Central Limit Theorem to approximate the probability $Pr(ln(\bar X>0)$, where:
$$\bar X=(1/200)\sum_{i=1}^{200} X_i $$
is the sample mean and ln denotes the natural logarithm (i.e., logarithm with base e).
b) Find the minimum n for which:
$$Pr(\sum_{i=1}^n X_i>190)>0.99$$
I am really struggling when using the central limit theorem as my teacher has not explained it to me very well. I find this question especially hard due to the ln in the probability for part a. And also finding the minimum value in part b as we have not been taught this. Could anyone help by providing useful links to help me solve these or work them through with me?
Any help would be massively appreciated. Thank you!
 A: First of all $\overline X$ is approximately distributed as $\overline X\sim \mathcal N(\mu,\frac{\sigma^2}{n})= \mathcal N(1,\frac{4}{200})$. The expected value of $\overline X$ is the same as for $X_i$. But the variance of $\overline X$ is the variance of  $X_i$ divided by $n$.
Then we have to evaluate how $\ln\left( \overline X\right)$ is distributed with $\ln(X)=Y$
$\approx P\left(\ln\left( \overline X\right)\leq y \right)=P\left( \overline X\leq e^y \right)=\Phi\left(\frac{e^y-1}{\sqrt{\frac{4}{200}}} \right)$
Thus $P \left(\ln(\overline X)>0\right)= P(Y>0)=1-P(Y<0)=1-\Phi\left(\frac{e^0-1}{\sqrt{\frac{4}{200}}} \right)=...$
For b) the sum is approximately distributed as $S_n=\sum_{i=1}^n X_i \sim N(n\cdot \mu, n\cdot \sigma^2)=N(n\cdot 1, 4n)$ Thus the inequality is 
$$P(S_n>190)=1-P(S_n<190)\approx 1-\Phi\left(\frac{190-n}{\sqrt{4n}} \right)>0.99$$
$$0.01>\Phi\left(\frac{190-n}{\sqrt{4n}} \right)$$
$$ \Phi^{-1}(0.01)>\frac{190-n}{\sqrt{4n}}$$
$$-2.326>\frac{190-n}{2\sqrt{n}}$$
$$-4.652\sqrt n>190-n$$
$$n-4.652\sqrt n-190>0$$
With $\sqrt n=x$ we get a quadratic inequality: $x^2-4.652x-190>0$. Solve this inequality and take the positive solution. Then square that value to obtain $n^*$.
