Central Limit Theorm with Exponential Distribution

Question

So the question asks,

"$Y_1, Y_2$... are i.i.d. with distribution Exponential(3). Use the central limit theorem to estimate $P(\sum^{1600}_{i=1}Y_i \le 540)$."

So I understand that we subtract and divide the Mean = 1/3 and root of Variance = $\sqrt{1/9}$ = 1/3 like so:

$P(S \le 540) = P((S-1/3)/1/3 \le (540-1/3)/1/3) = P((S-1/3)/1/3 \le 1619) \approx \phi(1619)$

And then we approximate to phi values, but the answer I've been given is $\phi(0.5)$

Is there something wrong with my method or calculations?

Answer 1

Your mistake is that you're using the mean and variance of $Y_i$, not of their sum.

The mean of a sum is the sum of the means, and the variance of a sum of independent random variables is the sum of the variances.

If each $Y_i$ has mean $1/3$ and variance $1/9$, $S = \sum_{i=1}^{1600} Y_i$ has mean $\mu = 1600/3$ and variance $1600/9$, thus standard deviation $\sigma = \sqrt{1600/9} = 40/3$. By the CLT, it is approximately normal. Thus the given value of $540$ corresponds to a $z$ value of $(540 - \mu)/\sigma = 1/2$. So indeed $\mathbb P(S \le 540) \approx \Phi(1/2)$ where $\Phi$ is the standard normal CDF.