Q:

Example of the Central Limit Theorem

Consider the sum of $n=14$ iid random variables $X_1,\dots,X_{14}$ as $S_{14}=X_1+\dots+X_{14}$ and that each of them has PDF 
$$f_{X_i}(x)=\frac{1}{\theta}e^{-\frac{x-\delta}{\theta}},\qquad x>\delta$$ and 0 otherwise. So the mean of this PDF is $\mu=E[X_i]=\theta+\delta$ and the variance $\sigma^2=\mathrm{Var}(X_i)=\theta^2$.

Now let $\theta=2$, $\delta=0$, $$Y=\frac{\sqrt{28}Z}{\sqrt{S_{14}}}$$ for $Z\sim\mathcal{N}(0,1)$. I want to find the probability $P(Y<2.467)$. 

I tried to bring the LHS to the form 
$$W=\frac{\frac{S_{n}}{n}-\mu}{\frac{\sigma}{\sqrt{n}}}$$
and apply the Central Limit Theorem since it is known that $W\sim\mathcal{N}(0,1)$ but $Z$ and $S_{14}$ both at the LHS prevent me to do so. 

A:

By the law of large numbers (not the CLT), $S_n/n\to E(X)=2$ hence one can expect that $S_{14}/28\approx2$, in which case $Y=Z\sqrt{28/S_{14}}$ is approximately $Z$ and $$P(Y<2.467)\approx P(Z<2.467)$$ which I will let you compute.
To refine slightly the result, and check the size of the error made in approximating $S_{14}/28$ by $2$, one could use the central limit theorem on the sequence $(X_i)$, to the effect that $$V_n=\frac{S_n-2n}{2\sqrt{n}}$$ is approximately standard normal hence $$Y=\frac{Z}{\sqrt{1+V_{28}/(2\sqrt{14})}}\approx Z\cdot\left(1-\frac1{4\sqrt{14}}V\right)$$ where $V$ is standard normal and independent of $Z$, in which case $P(Y<2.467)$ might be better approximated by $$P(Y<2.467)\approx P\left(Z\cdot\left(1-\frac1{4\sqrt{14}}V\right)<2.467\right)$$ This refinement should be tested numerically but my gut feeling is that it does not change significantly the formula you were supposed to produce using simply the LLN, namely, $$P(Y<2.467)\approx P\left(Z<2.467\right)$$

