Question

According to a​ study, nearly 80​% of workers in law libraries are satisfied with their job. Assume the true proportion of law librarians are satisfied with their job is 0.8. In a random sample of 20 law​ librarians, what is the probability that at most 4 are unsatisfied with their​ job?

Answer 1

0.618

Explanation:

This is a binomial distribution. We are to determine the probability of people unsatisfied. If we denote the probability of being satisfied with
`q`, then
`q=0.8`. Let
`p` be the probability of being unsatisfied. Then
`p+q=1`

`p=1-q =1-0.8=0.2`

Since the random sample is 20 and
`p` is small enough, we can use the Poisson probability function

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

where
`\lambda` is the mean and is given by
`\lambda=np=20*0.2=4`.

Denoting the probability of at most 4 unsatisfied with
`P(x\le4)`,

`P(x\le4)=P(x=0)+P(x=1)+P(x=2)+P</p><p>(x=3)+P(x=4)`

`P(x=0)= (e^(-4)4^0)/(0!) = e^(-4)`

`P(x=1)= (e^(-4)4^1)/(1!) = 4e^(-4)`

`P(x=2)= (e^(-4)4^2)/(2!) = 8e^(-4)`

`P(x=3)= (e^(-4)4^3)/(3!) = e^(-4)(32)/(3)`

`P(x=4)= (e^(-4)4^4)/(4!) = e^(-4)(32)/(3)`

`P(x\le4)=e^(-4)(1+4+8+(32)/(3)+(32)/(3))=0.018*34.33 = 0.618`