A paint company claims their paint will be completely dry wi

A paint company claims their paint will be completely dry within 45 minutes after application. Recently, customers have complained drying times are longer than the claimed 45 minutes. A consumer advocate group takes a random sample of 25 paint specimens and records their drying times. The average drying time x is 50. Consider dryng time, for all test specimens, to be normally distributed with ? = 6.

Suppose the claimed drying time is true, that is ? = 45 minutes, what is the probability of observing a sample mean of x = 50 or greater from a sample size of 25? (Round your answer to four decimal places.)
And can somebody link me to the table they use to get the P value because the table I have only goes up to 3.4 and I have Z = 4.1667 for this problem.

Solution

So the probability of observing a sample mean of x = 50 or greater from a sample size of 25 is

P(xbar>50) = P((xbar-mean)/(s/vn) >(50-45)/(6/sqrt(25)))

=P(Z>4.17)

=0 (from standard normal table)