Question

The number of typing errors made by a typist has a Poisson distribution with an average of two errors per page. If more than two errors appear on a given page, the typist must retype the whole page. What is the probability that a randomly selected page does not need to be retyped? (Round your answer to three decimal places.)

Answer 1

Answer: 0.6767

Explanation:

Given : Mean =
`\lambda=2` errors per page

Let X be the number of errors in a particular page.

The formula to calculate the Poisson distribution is given by :_

`P(X=x)=(e^(-\lambda)\lambda^x)/(x!)`

Now, the probability that a randomly selected page does not need to be retyped is given by :-

`P(X\leq2)=P(0)+P(1)+P(2)\\\\=((e^(-2)2^0)/(0!)+(e^(-2)2^1)/(1!)+(e^(-2)2^2)/(2!))\\\\=0.135335283237+0.270670566473+0.270670566473\\\\=0.676676416183\approx0.6767`

Hence, the required probability :- 0.6767