I have the following question
Suppose $X_1, .. X_n$ are i.i.d with $\mu = 12/5$ and $\sigma = 28/45$. Show using the CLT an approximation to $P(a < \sum_{i=1}^nX_i < b)$ for large $n$ and constants $a,b.$
My work: Because the data is i.i.d then $X = X_1 .. X_n \sim N(12/5, (28/45)^2),$ the CLT is given as the following: $\frac{X-\mu}{\sigma/\sqrt{n}}.$ Therefore, we must find $$a<\frac{X-(12/5)}{(28/45)/\sqrt{n}}<b.$$
However, I'm not sure how to proceed from here, any suggestions?
Formally, the CLT says that as $n$ grows large ($n \to \infty$), $$ \sqrt{n} \Big(\frac{1}{n}\sum_{i=1}^n X_i - \mu\Big) \to N(0, \sigma^2), $$ "in distribution." Roughly speaking, you can interpret this statement as for $n$ large enough, we have $$ \frac{1}{n} \sum_{i=1}^n X_i \approx N(\mu, \sigma^2/n). $$
Returning to your problem, you want to compute \begin{align*} P(a < \sum_i X_i < b) &= P(a/n < \frac{1}{n}\sum_i X_i < b/n) \\ &\approx P(a/n < N(\mu, \sigma^2/n) < b/n) \\ &= P(\sqrt{n}/\sigma (a/n - \mu) < N(0, 1) < \sqrt{n}/\sigma (b/n - \mu))\\ &= \Phi(\sqrt{n}/\sigma (b/n - \mu)) - \Phi(\sqrt{n}/\sigma (a/n - \mu)), \end{align*} where $\Phi(t) = P(N(0, 1) \leq t)$ is the CDF of a standard Normal random variable.