I'd like to use the central limit theorem that (for large enough n), the random variable $$ Z = \sqrt{n} \left({\bar X \over \mu} -1\right) \sim \mathcal{N}(0,1) $$ and derive an approximate 100(1- $\alpha$)% confidence interval for $\mu$ when $n$ is large.
I see the relationship between the CLT, the normal distribution and the 100(1- $\alpha$)% interval: the CLT tells us that for the random variable above if there's a large enough $n$, we can get a p.d.f out of it. Also if we were to find the area under this curve, it'd give us some $1 -\alpha$
I'm not sure where I'd start in "showing" this is true and then deriving this confidence interval for $\mu$
$$ \lim_{n \to \infty} \mathbb{P}\left( Z_{a/2} \le \frac{\sqrt{n}(\bar{X} - \mu)}{\sigma} \le Z_{1-a/2} \right) = 1-a. $$
Hence, an approximate CI for finite $n$ is $$ \left( \bar{X} - Z_{1-a/2}\frac{\sigma}{n}\le \mu \le \bar{X} + Z_{1-a/2}\frac{\sigma}{n}\right) $$