Can we use chi-square distribution and central limit theorem to find the approximate normal distribution? 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?
 A: There is no positive normal random variable!
A: The binomial distribution can not reach negative value but can still be approximated by normal approximation, so I don't think Murthy's answer is persuasive enough.
  And actually, $$(\frac{X_{i}}{\sigma })^{2}\sim \chi ^{2}(1), with\:finite\:  mean\: 1\: and\: finite\:variance\: 2; since\:Xi\sim i.i.d,\\ (\frac{X_{i}}{\sigma })^{2}\sim i.i.d,\\by\:Central\:Limit\:Theorem,\\ \sqrt{n}(\frac{\sum_{i=1}^{n}((\frac{X_{i}}{\sigma })^{2}-1)}{n})=\sqrt{n}(\frac{\sum_{i=1}^{n}(\frac{X_{i}}{\sigma })^{2}}{n}-1) \overset{d}{\rightarrow} N(0,2)\:as\: n \rightarrow +\infty \\$$ 
A: If $X_i/\sigma\overset{iid}{\sim}N(0,1)$, then $W_i = (X_i/\sigma)^2\overset{iid}{\sim}\chi^2_1$  and $\mathbb{E}(W_i) = 1$, $\mathbb{Var}(W_i) = 2$. Therefore, if $\bar{W}_n$ is the sample mean, we have by the classic CLT
$$
\sqrt{n}(\bar{W}_n-1)\overset{d}{\rightarrow} N(0,2).
$$
Based on the above we obtain that for large $n$,
$$
\bar{W}_n \overset{approx}{\sim}N(1,\dfrac{2}{n}).
$$ 
