How to use common density function and central limit theorem to compute P(S > 70)? Say I have a common density function as the following:
$f(x) =  \begin{cases}
       {2(1-x),} &{0 < x < 1}\\
       {0} &{otherwise}\\
     \end{cases} $
Let's say $S = X_1 + ... + X_{200}$. I want to find $P(S>70)$. Now this is what I thought on how to solve it, I figured that the 'easiest' way to solve this is by using the central limit theorem. But how to find $P(S>70)$ exactly using the central limit theorem wasn't straightforward. My intuition right away was to compute the mean and the variance of $X_1,...,X_{200}$ (note these values are independent continuous random variables)because I thought it will be needed to solve $P(S>70)$, however I am quite unsure on how to tackle this one after computing those values, also kind of unsure if I am on the right track in general. Any ideas?  
 A: We will need to assume the $X_i$ are independent.
First we compute the mean and variance of the $X_i$. We have
$$E(X_i)=\int_0^1 (x)(2)(1-x)\,dx=\frac{1}{3}.$$
Also
$$E(X_i^2)=\int_0^1 (x^2)(2)(1-x)\,dx=\frac{1}{6},$$
and therefore $\text{Var}(X_i)=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}$.
Thus $S$ has mean $\frac{200}{3}$. The variance of $S$ is $\frac{200}{18}$. This is because the variance of a sum of independent random variables is the sum of the variances. So the standard deviation of $S$ is $\frac{10}{3}$.
Informally, $S$ has a reasonably close to normal distribution, because it is the sum of a fairly large number of independent identically distributed random variables with a "nice enough" distribution.
So our probability is approximately the probability that a normal with mean $\frac{200}{3}$ and standard deviation $\frac{10}{3}$ is greater than $70$. Some software will let you find this probability directly. 
A more old-fashioned (but still useful) approach is to note that we want
$$\Pr\left(Z\gt \frac{70-\frac{200}{3}}{\frac{10}{3}}\right),$$
where $Z$ is standard normal. Now you can use a table of the standard normal, or software, to finish. We end up wanting $\Pr(Z\gt 1)$. 
