Using the Central Limit Theorem to calculate a mean from Poisson distributed random variables

Question

Firstly, I am studying the basic concepts of statistics and so any explanations, advice and suggestions are more than appreciated. Onto the problem- I am given the central limit theorem and understand its intuition (that the distribution of the means of any distribution converge to the normal distribution with increasing number of samples), but I do not know how to apply it to this scenario:

The Central Limit Theorem

Let $X_1$, $X_2$, . . . , $X_n$ be independent, identically distributed random variables with mean μ and variance $σ^2$. Then:

􏰁 a. $\sum_{i=1}^n$$X_i$ ∼ N(nμ, n$σ^2$), approximately.

b. $\bar{X}$ ∼ N(μ,$\frac{σ^2}{n}$), approximately.

The approximation improves as n → ∞.

Question 1) The number of typing errors made on a page follows a Poisson distribution with mean 2. Use the central limit theorem to calculate (approximately) the probability that there are more than 950 typing errors in a 450 page book.

Answer 1

Let $X_i$ be the number of errors on a single page in a sample of $n=450$ pages. We want to calculate the probability that $\sum X_i>950$

According to CLT, with $n=450$, $\sigma=\sqrt{2}$ and $\mu =2$, $\sum X_i$~$N(900,900)$. And $\bar{X}$ ~ $(2,\frac{2}{450})$

So, $Pr(\bar{X}>\frac{950}{450})=Pr(\frac{\bar{X}-2}{\sqrt{2}/\sqrt{450}}>\frac{\frac{950}{450}-2}{\sqrt{2}/\sqrt{450}})=Pr(Z>\frac{\frac{950}{450}-2}{\sqrt{2}/\sqrt{450}})=0.04779035$

So the probability would be $0.04779035$.