Consider $(X_n)_{n \in \mathbb{N}}$ of i.i.d. random variables with $X_n \sim \text{Uniform}(0,1)$. We set $$\overline{X}_n = \frac{1}{N}\sum_{i=1}^N X_i$$ and we calculate $\mathbb{E}(\overline{X}_n) = \frac{1}{2}$ and $Var(\overline{X}_n) = \frac{1}{12N}$. Now, we further set $$I = \frac{1}{\sqrt{2\pi}}\int_0^1 e^{-\frac{1}{2}x^2} \: dx$$
I want to show by applying the central limit theorem that $$\lim_{N \rightarrow \infty} \mathbb{P}(0 \leq \sqrt{12N}(\overline{X}_n -\frac{1}{2}) \leq 1) = I$$ So my guess is that the solution is something like $$\lim_{N \rightarrow \infty} \mathbb{P}(0 \leq \sqrt{12N}(\overline{X}_n -\frac{1}{2}) \leq 1) = \mathbb{P}(0 \leq Y \leq 1)$$ where $Y \sim \text{Normal}(0,1)$. And this is then simply $$\mathbb{P}(0 \leq Y \leq 1) = \int_0^1 f_Y(x) \: dx = \frac{1}{\sqrt{2\pi}}\int_0^1 e^{-\frac{1}{2}x^2} \: dx = I$$ But my question is, how do I show with the CLT that $\sqrt{12N}(\overline{X}_n -\frac{1}{2})$ is $Y \sim \text{Normal}(0,1)$?
Keep in mind $Var X = \frac{1}{12}$, so $\sigma = \frac{1}{\sqrt{12}}$. Now, by CLT $$ P(\frac{S_n - n \mu}{\sigma \sqrt{n}}) = P(\frac{\sqrt{n}(\frac{S_n}{n} - \mu)}{\sigma} \leq 1) = \Phi(1) $$ Since $\sigma = \frac{1}{\sqrt{12}}, \mu=0.5$ it is exactly you expression, just plug in the values.