While preparing for a test I came across the following problem:
$X_i\sim G_{0.5}$, for $i\geq 1$. Define $Z_n = \frac{1}{\sqrt{n}}\sum_{i=1}^{n}(X_i-2)$. Approximate $P(-1\leq Z_n \leq 2)$ with large enough $n$. Note that $Z_n$ is not "properly" normalized.
Here is my approach:
I started with calculating $\mu = \frac{1}{0.5} = 2$ and $\sigma^2 = \frac{1-0.5}{0.5^2} = 2$.
Next I express $Z_n$ in terms of $\bar{X}_n=\frac{1}{n}\sum_{i=1}^{n} X_i$ which gives $Z_n = \sqrt{n}(\bar{X}_n-2)$.
Now I now from the The Central Limit Theorem that $P(\bar{X}_n\leq \alpha) = P(Z_n\leq \sqrt{n}\frac{\alpha-\mu}{\sigma }) = \Phi(\sqrt{n}\frac{\alpha-\mu}{\sigma})$.
To obtain the $Z_n$ from the CLT I need to divide the $Z_n$ from above by $\sigma$. So I get :
$P(-1/\sigma \leq Z_n/\sigma \leq 2/\sigma) = P(Z_n\leq 2/\sigma)-(1-P(Z_n \leq -1/\sigma)) = \Phi(\sqrt{2}-(1-\Phi(\frac{1}{\sqrt{2}}))=0.6816$.
Is this solution correct now?
You got the right formula for the variance but made a small mistake in computing it.
Also, when you define $\bar X_n$ as the mean of the first $n$ variables, you have $Z_n=\sqrt n\,( \bar X_n -2)$, so you can’t conclude that $Z_n \to 0$. In fact, it converges in distribution (not in probability) to a multiple of the Gaussian.
So since $\sqrt n \to \infty$ and $\bar X_n - 2 \to 0$, the prodcut is indeterminate. But it does converge in distribution to a normal. This is called the Central Limit Theorem.
This exercise probably appeared in the context of CLT, I guess...
Can you take it from here?