Question

The number of errors in a textbook follow a Poisson distribution with a mean of 0.02 errors per page. What is the probability that there are 3 or less errors in 100 pages?

Answer 1

Answer: 0.8571

Step-by-step explanation:

Poisson distribution formula:

$P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$  , where x=0,1,2,3... is a poisson variable , $\lambda$ = Mean of the distribution .

Let x be the number of errors in a textbook that follow a Poisson distribution with a mean of 0.02 errors per page.

$\mu=0.02$

For 100 pages , $\lambda=100\times0.02=2$

Now , the probability that there are 3 or less errors in 100 pages will be :-

$P(x\leq3)=P(0)+P(1)+P(2)+P(3)\\\\=\dfrac{e^{-2}2^0}{0!}+\dfrac{e^{-2}2^1}{1!}+\dfrac{e^{-2}2^2}{2!}+\dfrac{e^{-2}2^3}{3!}\\\\=e^{-2}(1+2+\dfrac{4}{2}+\dfrac{8}{6})\\\\=(0.13533528)(\dfrac{19}{3})=0.85712344\approx0.8571$

Hence, the probability that there are 3 or less errors in 100 pages = 0.8571