
Central Limit Theorem
From my understanding, according to the central limit, $\displaystyle T = X_1 + X_2 + ... + X_n$ should behave (roughly) like an N(0,1) distribution for a large enough n.
I'm trying to show this by simulation. I created 1000 $\displaystyle X_i$ iid ~U[0,1]. So according to CLT, T~N(0,1). But how would I show this?

According to Central Limit Theorem, $\displaystyle \frac{Tmean(T)}{sd(T)}$ follows a $\displaystyle N(0,1)$ distribution for large n. Also, if $\displaystyle X_i$'s are iid $\displaystyle U(0,1)$ random variables, then their sum do not follow $\displaystyle N(0,1)$ distribution.

the average of uniforms, when standardized should be approximately normal.
same goes for the sum

so $\displaystyle \frac{T  \frac{n}{2}}{n*\sqrt{\frac{1}{12}}}$ is approximetaly N(0,1)?

since these are iid with a second moment we have
$\displaystyle {\bar X{1\over 2}\over \sqrt{1\over 12n}}\to N(0,1)$
we have $\displaystyle {n\bar X{n\over 2}\over n\sqrt{1\over 12n}}\to N(0,1)$

So if I were to plot that, I would get a rouch bell curve, correct? But I try to do this on R
for (i in 1:1000){
+ T=(sum(runif(1000))(1000/2))/1000*sqrt(1/12*1000)
+ V[i]=T
+ }
I create runif(1000) creates 1000 U[0,1] and T is me trying to standardize it using CLT. I do this 1000 times, but I dont get a bell curve when I plot V.

Did you try.......$\displaystyle {T_n{n\over 2}\over \sqrt{n\over 12}}$
because yours is wrong, that n in the denominator is incorrect.