Central Limit Theorem on Independent random variables

Question

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 $n = 1, 2, \dots, 300$. $$F(t) = \begin{cases} 0, &\text{for }t\lt -2\\ \frac{1}{4}t + \frac{1}{2} &\text{for } t \in [-2, 2]\\ 1, &\text{for } t \geq 2 \end{cases}$$

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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 $X_n$ $\sim N(0, 1)$

Answer 1

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")