Using Central Limit Theorem to approximate. $X_1,X_2 \dots X_n$ are independent R.V(random vars) that are uniform $\in$ [0, 1]
and let $S_n = X_1 + \dots + X_n$.
Now, I am trying to use the Central Limit Theorem to give an approximation of 
P$(S_{100} \in [40, 60]).$
This seems like a fairly straightforward question. But I'm not sure how to approach it.
This is my attempt so far:
Since $X_i$ are uniformly distributed, they are uniform RV.
then E[X] = $\frac{40 + 60}{2}$ = 50 and my Variance = $\frac{(60-40)^2}{12} = 33.333$
I think I'm lost. Would appreciate any help and guidance! thanks
 A: While $X$ has a uniform distribution, $S$ does not. Therefore
you cannot use directly the mean and variance formulas that apply for
a uniform distribution to find $E(S)$ and $V(S),$ as you have tried to do.
Here is an outline of what you need to do.
By the formulas for a uniform distribution, you have $E(X_i) = 1/2$
and $V(X_i) = 1/12.$ To find $E(S)$ you proceed as follows.
using a rule that says "expectation of sum is sum of expectations":
$$E(S_{100}) = E(X_1 + X_2 + \cdots + X_{100}) 
= \sum_{i=1}^{100} E(X_i) = 100(1/2) = 50.$$
Because the $X_i$ are independent, you can use a similar method
to add the 100 variances to get $V(S_{100}) = 100(1/12) = 100/12 =  8.3333.$ Then $SD(S_{100}) = \sqrt{100/12} = 2.887.$ 
Now, by the Central Limit Theorem, $S_{100} \approx W,$ where
 $W \sim \mathrm{Norm}(\mu = 50, \sigma = 2.887).$ So your problem becomes
the evaluation of $P(40 \le W \le 60) \approx P(40 \le S_{100} \le 60).$  I assume you know how to standardize $W$ and use standard normal tables to evaluate $P(40 \le W \le 60)$.
Using software I get an answer that is very nearly 1. This makes
sense because a normal distribution has almost all of its area
within three standard deviations of the mean. This would be
the interval $50 \pm 3(2.887)$ or roughly $(41.3, 58.7),$ which
is contained in $(40, 60).$
A: Let's start by looking at the approximate distribution of $S_n$ when $n$ is large, by applying the CLT.  Since you have underlying uniform values $X_1,X_2,X_3,... \sim \text{IID U}(0,1)$, you have $\mathbb{E}(X_i) = \tfrac{1}{2}$ and $\mathbb{V}(X_i) = \tfrac{1}{12}$.  The sum of the first $n$ values has the corresponding moments:
$$\mathbb{E}(S_n) = \frac{n}{2}
\quad \quad \quad
\mathbb{V}(S_n) = \frac{n}{12}.$$
Thus, for large $n$ you can apply the CLT to get the approximate distribution $S_n \overset{\text{Approx}}{\sim} \text{N}( n/2, n/12 )$, or if you prefer to standardise you get:
$$\frac{2S_n - n}{\sqrt{n/3}} \overset{\text{Approx}}{\sim} \text{N} (0,1).$$
Thus, you have the general rule:
$$\begin{aligned}
\mathbb{P}(s_* \leqslant S_n \leqslant s^*)
&= \mathbb{P} \Big( \frac{2s_* - n}{\sqrt{n/3}} \leqslant \frac{2S_n - n}{\sqrt{n/3}} \leqslant \frac{2s^* - n}{\sqrt{n/3}} \Big) \\[6pt]
&\approx \Phi \Big( \frac{2s^* - n}{\sqrt{n/3}} \Big) - \Phi \Big( \frac{2s_* - n}{\sqrt{n/3}} \Big). \\[6pt]
\end{aligned}$$
For your particular values you have:
$$\begin{aligned}
\mathbb{P}(40 \leqslant S_{100} \leqslant 60)
&\approx \Phi \Big( \frac{2 \cdot 60 - 100}{\sqrt{100/3}} \Big) - \Phi \Big( \frac{2 \cdot 40 - 100}{\sqrt{100/3}} \Big) \\[6pt]
&= \Phi \Big( \frac{20}{\sqrt{100/3}} \Big) - \Phi \Big( - \frac{20}{\sqrt{100/3}} \Big) \\[12pt]
&= \Phi ( 2 \sqrt{3} ) - \Phi ( -2 \sqrt{3} ) \\[12pt]
&= \Phi ( 3.464102 ) - \Phi ( -3.464102 ) \\[12pt]
&= 0.999468. \\[12pt]
\end{aligned}$$
