# Simplify probability with central limit theorem?

Assume $X_1, X_2, X_3,\ldots$ are independent and identically distributed exponential random variables with mean of $2$. If $\bar X_n$ the sample mean is normally distributed $\sim N(0,1)$, how large is $n$ so that $P[|\bar X_n -2|\le 0.001] \ge .95$? Use the approximation $P[\bar X_n \ge 2] = .05$

Since it's normal, can we rewrite the central limit theorem as? $$P\left[\frac{\bar X_n - \mu}{\frac{\sigma}{\sqrt n}} \le \frac{Z}{n}\right]$$

I'm not sure how to proceed

• $\sqrt{n} \frac{\overline{X}_n-2}{2}$ is approximately N(0,1) distributed under this circumstance. – Ian Oct 14 '16 at 0:58
• so is the $\frac{Z}{n}$ justified on the right hand side? is it equal to $.001/n$ or is the whole fraction .001? – MoronicHero Oct 14 '16 at 1:13
• I don't know what you meant by $Z/n$. The point is that $\frac{\overline{X}_n-\mu}{\frac{\sigma}{\sqrt{n}}}$ is approximately N(0,1) and you just work from there. – Ian Oct 14 '16 at 1:28
• ok i see. Z is $\bar X_n$ so I didn't need to divide by n. – MoronicHero Oct 14 '16 at 1:31