Can we use chi-square distribution and central limit theorem to find the approximate normal distribution?

Question

If $X_1,\ldots,X_i,\ldots,X_n$ are same normal distribution, $X_i \sim \operatorname{Normal}(0,σ^2)$, and they are independent. $$ Z = \frac{\sum_{i=1}^n X_i^2} n. $$

What is the distribution of the square of the normal distribution?like $X_i^2$,and,what is it mean and variance?

I am trying to turn this Z into a normal distribution

can we use chi-square distribution and central limit theorem to find the approximate normal distribution ?

How to do it？

I do not quite understand the chi-square distribution and central limit theorem, could you answer this question in detail?

Any help would be much appreciated!

re-edit：

I do this works: $$ Z = \frac{\sum_{i=1}^n X_i^2} n= σ^2\sum_{i=1}^n \left(\frac{X_i}{σ}\right)^2. $$ this is a chi-square distribution,and mean $= nσ^2$, var${}=2nσ^2$.

is this right?

and how to use CLT to find the approximate normal distribution?

Answer 1

There is no positive normal random variable!