Each new book donated to a library must be processed. Suppose that the time it takes to process a book has a mean of $10$minutes and a standard deviation of $3$minutes. If a librarian has $40$ books to process,

(a) approximate the probability that it will take more than $420$minutes to process all these books;

(b) approximate the probability that at least $25$ books will be processed in the first $240$ minutes. What assumptions have you made?

Given in the question that, the time taken to process a book has a mean of $10$minutes and a standard deviation of $3$minutes. A librarian has $40$ books to process.

Find the approximate probability that took more than $420$minutes to process and the approximate probability to process at least $25$books to process in the first $240$minutes.

Define the variables at random. The times required to process these books are$X_1,\dots ,X_{40}$.

The assumption is that these times are equally distributed and independent, with a mean of $\mu =10$and a variance of ${\sigma }^2=3^2=9$.Then, using the Central Limit Theorem, find the necessary probability.

The total time needed to process these books is $\sum _i{X}_i$

localid="1650281811298" $$\begin{gathered} P\left(\sum _i{X}_i>420\right)=P(\overline{X}>10.5) \\ =P\left(\sqrt{40}\frac{\overline{X}-10}{3}>\sqrt{40}\frac{10.5-10}{3}\right) \\ \approx 1-\Phi \left(\sqrt{40}\frac{10.5-10}{3}\right) \\ =1-\Phi \left(\frac{\sqrt{10}}{3}\right) \\ =0.1459 \end{gathered}$$

Given in the question that, the time taken to process a book has a mean of $10$minutes and a standard deviation of $3$minutes. A librarian has $40$books to process.

Find the approximate probability that took more than $420$minutes to process and the approximate probability to process at least $25$books to process in the first $240$minutes.

At least 25 books will be processes in the first 240 minutes if and only if

$$\begin{gathered} X_1+\cdots +X_{25}\le 240 \\ P\left(\sum _{i=1}^{25}{X}_i\le 240\right)=P\left({\overline{X}}_{25}\le 9.6\right) \\ =P\left(\sqrt{25}\frac{{\overline{X}}_{25}-10}{3}\le \sqrt{25}\frac{9.6-10}{3}\right) \\ \approx \Phi \left(5·\frac{9.6-10}{3}\right) \\ =\Phi \left(-\frac{2}{3}\right) \\ =0.2525 \end{gathered}$$