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Let $U_i, i=1,...,300$ be iid r.v's from the uniform distribution on $[-\frac{1}{2},\frac{1}{2}]$. Calculate, using the CLT, $$P(\sum_{i=1}^{300}U_i \le 3).$$

My solution:

By the CLT, for large enough $n$ the random variable $$Z=\frac{\sum_{i=n}^{300}U_i - n \mu}{\sigma \sqrt{n}}$$

is distributed $\sim N(0,1)$, where $\mu=E[U_i], \sigma=\sqrt{V[U_i]}$, for any $1\le i \le300$.

So, $$P(\sum_{i=1}^{300}U_i \le 3)=P({\sum_{i=1}^{300}U_i -300\mu \over \sqrt{300} \sigma} \le {3 - 300\mu \over \sqrt{300} \sigma})=...$$

calculating the moments $\mu=0, \sigma=\sqrt{\frac{1}{12}}$, we obtain

$$...=P(Z\le 0.6)=0.7257$$

I used this table.

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up vote 1 down vote accepted

What you have written seems sensible. When I tried the following code in R

sampsize <- 300
cases <- 100000
matdat <- matrix(runif(sampsize*cases, min=-1/2, max=1/2), ncol=sampsize)
mean(rowSums(matdat) <= 3)

I got the result [1] 0.72506 which is close enough to your result to suggest your answer is probably correct.

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