Using the Central Limit Theory to solve $\lim_{n\rightarrow \infty} \mathbb{P}(n-\sqrt n \lt X_1+X_2+\cdots+X_n\lt n+\sqrt n)$

Question

$X_1,\ldots,X_n$ are independent random variables that are uniformly distributed between 0 and 2. What is:

$$\lim_{n\rightarrow \infty} \mathbb{P}(n-\sqrt n \lt X_1+X_2+\cdots+X_n\lt n+\sqrt n)$$

Attempt:

$\sigma^2=\frac{1}{3}$ $\mu=1$

So is the idea to get what we have into the form of the following so we can use z-score values?

$$\mathbb{P}\left(a\lt\frac{X_1+X_2+\cdots+X_n -\mu_x}{\displaystyle\frac{\sigma_x}{\sqrt n}}\lt b\right)$$ and then n$\rightarrow \infty$ we get $\Phi(b)-\Phi(a)$. Is my thought process correct?

Edit: my attempt:

Set $X_1+X_2+\cdots+X_n =\bar X$

Subtract $n$ from both sides and we get: $\mathbb{P}\left(\left|\bar X -n\right|\lt \sqrt n\right)$

$$\mathbb{P}\left(\frac{\left|\bar X -n\right|}{\sqrt{n}}\lt 1\right)$$

And now I believe I am stuck.

Answer 1

We know $$\lim\limits_{n\to \infty}{{\sum_{i=1}^nX_i\over n}-\mu\over \sigma/\sqrt{n}}\sim N(0,1)$$$$\implies \lim\limits_{n\to \infty}{\sum X_i-n\over \sqrt{n\over 3}}\sim N(0,1)$$$$\implies \lim\limits_{n\to\infty}P(n-\sqrt{n}<X_1+\cdots+X_n<n+\sqrt{n})=\Phi(\sqrt{3})-\Phi(-\sqrt3)=2\Phi(\sqrt{3})-1$$