Need help understanding central limit theorem

Question

I am very confused about CLT and have searched on the internet but found nothing that solved my confusion. How can I solve a problem like this with CLT?

Let $Y =\operatorname{Pois}(n)$. Using Normal approximation, aka the CLT, give an estimate of the probability

$$p\Big[|Y-n| \geq 2\sqrt{n}\Big].$$

Answer 1

A Poisson distribution with parameter $n$ has mean $n$ and standard deviation $\sqrt{n}$

so the expression $\mathbb P\Big[|Y-n| \geq 2\sqrt{n}\Big]$ is the probability of being two standard deviations or more away from the mean.

For a normal distribution, this probability would be $\Phi(-2)+1-\Phi(2) \approx 0.0455$, suggesting this might be a reasonable estimate for a Poisson distribution.

How reasonable? It tends to get better as $n$ increases, but for $n=100$ the actual Poisson probability is about $0.0509$ while for $n=101$ the actual Poisson probability is about $0.04112$; this jump happens near values of $n$ which are near squares or halfway between squares. As an illustration with R

n <- 1:101
plot(n, ppois(floor(n-2*sqrt(n)), n) + 1 - ppois(ceiling(n+2*sqrt(n)) -1, n))
abline(h=pnorm(-2) + 1 - pnorm(2), col="red")