I've been given the following problem to solve (hastily translated to English) in an undergraduate course in probability theory: A small library has 46 running meters of books. The thickness of each book can be seen as independent stochastic variables which all have the expected value of 1.8cm and standard deviation 0.7cm. Approximate the probability of the library containing more than 2500 books.
I'm certain it's an application of the central limit theorem but can't really get my head around it. I've started with some notation: given that $X_i$ is the thickness of book $i$, we have that $\sum_{i=1}^nX_i=4600$, $\mu=1.8$ and $\sigma=0.7$.
Via the central limit theorem one could approximate the probability of the sum of $n$ books thickness' being between $a$ and $b$ cm as $$P(a<\sum_{i=1}^nX_i\leq b)\approx\Phi\left(\frac{b-n\mu}{\sigma\sqrt{n}}\right)-\Phi\left(\frac{a-n\mu}{\sigma\sqrt{n}}\right)=\Phi\left(\frac{b-1.8n}{0.7\sqrt{n}}\right)-\Phi\left(\frac{a-1.8n}{0.7\sqrt{n}}\right)$$ but that's not really the probability I'm looking for since the total length is known (and $n$ not).
Another different approach could be to let $Y_k$ denote the no. of books in meter $k$ and calculate $$P\left(\sum_{k=1}^{46}Y_k >2500\right)$$ given that $\sum_{i=1}^nX_i=4600$ but to do so with the given information one must formulate how $Y$ depends on $X$ and that I am unsure of. Do anyone have a solution?
The problem is equivalent to asking for the probability that $2500$ books fit inside $46$ m or $4600$ cm. The combined thickness of $2500$ books is approximately normally distributed (by CLT) with mean $2500×1.8=4500$ cm and standard deviation $\sqrt{2500}×0.7=35$ cm. The desired probability then becomes $$P(\mathcal N(4500,35^2)<4600)=P(Z<100/35=20/7)=0.99786$$