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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$X_1,鈥�,X_{40}$.

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

The total time needed to process these books is $\underset{i}{鈭�}{X}_i$

localid="1650281811298" $$\begin{gathered} P\left(\underset{i}{鈭�}{X}_i>420\right)=P(\stackrel{炉}{X}>10.5) \\ =P\left(\sqrt{40}\frac{\stackrel{炉}{X}-10}{3}>\sqrt{40}\frac{10.5-10}{3}\right) \\ 鈮�1-桅\left(\sqrt{40}\frac{10.5-10}{3}\right) \\ =1-桅\left(\frac{\sqrt{10}}{3}\right) \\ =0.1459 \end{gathered}$$

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

$$\begin{gathered} X_1+鈰�+X_{25}鈮�240 \\ P\left(\underset{i=1}{\overset{25}{鈭�}}{X}_i鈮�240\right)=P\left({\stackrel{炉}{X}}_{25}鈮�9.6\right) \\ =P\left(\sqrt{25}\frac{{\stackrel{炉}{X}}_{25}-10}{3}鈮�\sqrt{25}\frac{9.6-10}{3}\right) \\ 鈮�桅\left(5路\frac{9.6-10}{3}\right) \\ =桅\left(-\frac{2}{3}\right) \\ =0.2525 \end{gathered}$$