# Question about the CLT as it applies to the exponential distribution

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 $$t$$?
• Thank you for your guidance. I'm afraid I'm still a bit lost. So we are trying to find $\mathbb{P}(S_{21} \leq 1)$ but we found (using CLT) $\mathbb{P}(X_{21} \leq t)$. The inequality we are trying to find is the "probability that at least 20 emails arrive in less than an hour" while the second inequality is the "probability that the approximately normal distribution yields an arrival time less than or equal to some time $t$. I am confused on how to relate the two. Sorry for my lack of understandinf, you did explain it quite well, however the topic is new to me. Apr 18, 2022 at 13:37
• What does the event $S_{21}\le1$ (“more than $20$ emails within one hour”) mean in terms of $X_{21}$? If it helps, write $S_{21}=aX_{21}+b$ where $a:=\sqrt{\text{Var}(S_{21})}$ and $b:=\Bbb E[S_{21}]$. Apr 18, 2022 at 14:05
• Dude, I don't know. I'm really trying but my brain too small. I'm thinking that the event $S_{21} \leq 1$ is equivalent to $X_{21} \geq 21$? But then from here, what step would I even take next? Apr 18, 2022 at 21:45
• It’s a simple degree-one inequation: $S_{21}=aX_{21}+b\le1\iff X_{21}\le\frac{1-b}a=:t$. Apr 19, 2022 at 4:07