The question at hand is as follows... The time intervals (measured in hours) between arrivals of emails to your mailbox can be modelled as i.i.d. exponential random variables with parameter 5, namely $$f_T(t) = 5e^{-5t}, \hspace{5mm}t \geq 0$$ Use the Central Limit Theorem to estimate the probability that more than 20 emails arrive to your mailbox in an hour.
I am sort of confused on how to represent the i.i.d. random variables as a sum of r.vs because it is not determined how many emails are sent. Is $n = 20$ since we are aked to find $\mathbb{P}(20 \text{ emails arrive withing an hour})$? I guess I am more so just confused on how to set this problem up. Thanks for any advice.
Let $T_1,T_2,\ldots$ denote the i.i.d. time intervals between the arrivals of your emails, and let $S_N:=T_1+\cdots+T_N$. The idea is to say that, thanks to the CLT, $$X_N:=\frac{S_N-\mathbb E[S_N]}{\sqrt{\text{Var}(S_N)}}$$ is approximately distributed like a standard normal variable $Y$ (at least for $N$ sufficiently large). Hence, for any given $t\in\mathbb R$, we have $\mathbb P(X_{21}\le t)\approx\mathbb P(Y\le t)$, which we can estimate by looking at the table of the normal distribution. Now, you are asked to estimate the probability $\mathbb P(S_{21}\le1)$. Can you relate it to $\mathbb P(X_{21}\le t)$ for some value of $t$?