Central Limit Theorem on Independent random variables The Prompt asks me to use Central limit theorem to compute $$P(\sum_{n=1}^{300} X_n > -21)$$
Where F(t) is the distribution function for random variables $X_n,$ for $n = 1, 2, \dots, 300$.
$$F(t) = \begin{cases}
  0, &\text{for }t\lt -2\\    
  \frac{1}{4}t + \frac{1}{2}  &\text{for } t \in [-2, 2]\\
  1, &\text{for } t \geq 2
\end{cases}$$ 

The solution as I tried to solve it,
$$X_1(x) = \begin{cases}
\frac{1}{4} & x \in [-2, 2)\\
0 & x\notin [-2, 2)
\end{cases}$$
$$EX_1 = \int_{-2}^{2}\frac{1}{4}x \cdot dx = 0$$
$$EX^2_1 = \int_{-2}^{2}\frac{1}{4}x^2 \cdot dx = \frac{4}{3}$$
Hence, $Var(X) = \frac{4}{3}$ with which we can compute the probability.
$$P(\sum_{n = 1}^{300} X_n > -21) = 1 - P(\sum_{n = 1}^{300} X_n < -21)$$
$$\approx 1 - P\left(\frac{S_{300} - EX_{1}S_{300}}{\sqrt{Var \cdot S_{300}}}\right)$$ 
$$1 -\Phi\left(\frac{-21 - 0}{\sqrt{300 \cdot \frac{4}{3}}}\right) = 1 -(1 - \Phi(1.05)) = 0.146859 $$
where  $X_n$ $\sim N(0, 1)$
 A: Further Comments: Your (independent) random variables $X_i \sim \mathsf{Unif}(-2,2).$ Thus, $E(X_i) = 0$ and $Var(X_i) = 4/3.$ Then $E(S) = 0$ and 
$Var(S) = 300(4/3) = 400,$ so that $SD(S) = 20.$ Finally,
$P(S > -21) \approx P(Z > -1.05) = 1 - \Phi(-1.05) = 0.8531,,$ where $Z$ is
standard normal. [When using normal tables, it is a good idea to make a rough
sketch of the normal curve and the desired area; that can't prevent small
mistakes, but can help prevent gross ones.]
# Computation in R
1-pnorm(-1.05)
## 0.8531409

Note: A pretty good approximation can be found using simulation, which also
illustrates how well the Central Limit Theorem works for the sum of 300
independent uniform random variables.
The figure below shows a histogram of a million realizations of $S$ along
with the density curve of $\mathsf{Norm}(\mu=0,\, \sigma=20).$ The probability
you seek is the area (in the histogram or under the curve) to the right
of the vertical broken line.

s = replicate(10^6,  sum(runif(300, -2, 2)))
mean(s > -21)
## 0.853196  # aprx P(S > -21), correct to about 2 or 3 places

hist(s, prob=T, br=50, col="skyblue2", main="Simulated Distribution")
  curve(dnorm(x, 0, 20), add=T, lwd=2)
  abline(v = -21, col="red", lwd=2, lty="dashed")

