34 of consumers read the ingredients listed on a products la

34% of consumers read the ingredients listed on a product\'s label.

if 290 consumers are randomly selected, what is the probability that between 81 and 110 of them read the ingredients listed on a product\'s label?

Solution

mean=n*p=290*0.34=98.6

standard deviation =sqrt(n*p*(1-p))=sqrt(290*0.34*(1-0.34))=8.06697

So the probability is

P(81<X<110) = P((81-98.6)/8.06697 <(X-mean)/s <(110-98.6)/8.06697)

=P(-2.18<Z<1.41) =0.9061 (from standard normal table)