In a 2010 survey 58 of Generation Y GenY adults were found t

In a 2010 survey, 58% of Generation Y (GenY) adults were found to pay their monthly bills on time.

Suppose a new sample of 50 people from GenY is chosen. What is the probability that at most 25 of them pay their monthly bills on time?

Solution

mean=n*p=50*0.58 =29

standard deviation =sqrt(n*p*(1-p))=sqrt(50*0.58*(1-0.58)) =3.489986

So the probability is

P(X<25) = P((X-mean)/s <(25-29)/3.489986)

=P(Z<-1.15) =0.1251 (from standard normal table)