Is $\frac{S_n-nE(X_1)}{\sqrt{n Var(X_1)}}$ a standardized random variable?

Question

I want to know if this statement is true or false.

The random variables $X_1,X_2,X_3,...$ fulfil the requirements of the central limit theorem. Let $S_n = \sum_{i=1}^n X_i$.

Then

$$\frac{S_n-nE(X_1)}{\sqrt{n Var(X_1)}}$$

is a standardised random variable.

I looked up the central limit theorem, and the expected value has to be $0$ and the variance needs to be $1$ for this to be true. I think the expected value here is in fact $0$, but I don't know whether the variance is $1$.

Answer 1

For the expectation: $E[S_n-nE(X_1)]=E[S_n]-nE[X_1]=nE[X_1]-nE[X_1]=0$.

For the variance: $Var(\frac{S_n-nE(X_1)}{\sqrt{n Var(X_1)}})=\frac{1}{n Var(X_1)}Var(S_n-nE(X_1))=\frac{1}{n Var(X_1)}Var(S_n)=1$