Question

The number of errors in a textbook follow a poisson distribution with a mean of 0.01 errors per page. what is the probability that there are 3 or less errors in 100 pages? round your answer to four decimal places (e.g. 98.7654).

Answer 1

Mean number of errors in each page = 0.01
Mean number of errors in 100 pages = 0.01*100=1

It is possible to use the cumulative distribution function (CMF), but the math is a little more complex, involving the gamma-function. Tables and software are available for that purpose.

Thus it is easier to evaluate with a calculator for the individual cases of k=0,1,2 and 3.

The Poisson distribution has a PMF (probability mass function)


`P(k):=(\lambda^ke^(-\lambda))/(k!)`
with λ = 1
=>

`P(0):=(1^0e^(-1))/(0!)=0.3678794`

`P(1):=(1^1e^(-1))/(1!)=0.3678794`

`P(2):=(1^2e^(-1))/(2!)=0.1839397`

`P(3):=(1^3e^(-1))/(3!)=0.0613132`
=>

`P(k<=3)=P(0)+P(1)+P(2)+P(3)=0.9810118`
or
P(k<=3)=0.9810 (to four decimal places)