Word Problem: Probability of Y books Fitting in Book Case

Question

Problem:

You have $4600$ cm of book case. The thickness of the books are independently distributed with $X \sim N(1.8$ cm$,0.7^2)$.

Approximately determine what the probability of having more than 2500 books is.

My solution so far:

My first thought is to define a new random variable for the amount of books $Y$~$N(\mu,\sigma^2)$, and then determine $\mathbb{P}(Y>2500)$. But how do I relate $X$'s known distribution to $Y$'s unknown one?

Answer 1

You have that $X_k\sim\mathcal N(1.8\textrm{cm},0.7^2\textrm{cm}^2)$ independently for all $\{X_k\}_{k\in\Bbb N^+}$ being the widths of books to be placed on a shelf of length $4600$cm.

Now, let $Y_N=\sum_{k=1}^N X_k$ , and so $Y_N \sim \mathcal N(1.8\textrm{cm}\,N, 0.7^2\textrm{cm}^2\,N)$

$Y_N$ is thus the combined width of $N$ books.

You want to find the probability that at least $2500$ books can fit on the shelf of the given length. $$\mathsf P(Y_{2500}\leq 4600)$$