Question

Suppose a200 -page book has, on the average, one misprint every 10 pages. On about how many pages would you expect to find2misprints?

Short answer

The number of pages is .$N_p=1$

Step-by-step solution

Given Information  

A 200 page book has, on the average, one misprint every10 pages.

Definition of the Poisson distribution. 

A discrete frequency distribution that indicates the likelihood of a set of independent occurrences occurring at a specific moment.

Find the number of pages.

The formula for Poisson distribution states that.$P_n=\frac{{\mu }^n{e}^{-\mu }}{n!}$

The average is given below.

$\mu =0.1$

Probabilities are given below.

$\begin{array}{c}{P}_n=\frac{{\left(0.1\right)}^n{e}^{-0.1}}{n!}\\ P_2=\frac{{\left(0.1\right)}^2{e}^{-0.1}}{2!}\\ =0.0045\end{array}$

The number of the pages is given below.

$\begin{array}{c}{N}_p=N\times P_2\\ =0.0045\times 200\\ \approx 1\end{array}$

The number of pages is .$N_p=1$