Assume that the average price for a movie is 1093 Assume the

Assume that the average price for a movie is 10.93. Assume the population standard deviation is 0.59 and that a sample of 32 theaters was randomly selected.

What is the probability that the sample mean will be less than $11.11 ?

Solution

So the probability that the sample mean will be less than $11.11 is

P(xbar<11.11) = P((xbar-mean)/(s/vn) <(11.11-10.93)/(0.59/sqrt(32)))

=P(Z<1.73) =0.9582 (from standard normal table)