The time at which the mailman delivers the mail to ace bike shop follows a normal distribution with mean 2:00 pm and standard deviation of 15 minutes.

There are three questions related to this problem.

First, the probability of the mail will arrive after 2:30 PM

Find the z-score of 2:30 which is 30 minutes after 2:00.

z (2:30) = (2:30 - 2:00) / 15 = - 30/15 = - 2

P (x < 2:30) = P (z<-2) = 0.0228

Second, the probability of the mail will arrive at 1:36 PM

Find the z-score of 1:36 which is 24 minutes before 2:00.

z (1:36) = (1:36 - 2:00) / 15 = - 24/15 = - 1.6

P (x < 1:36) = P (z<-1.6) = 0.0548

Lastly, the probability of the mail will arrive between 1:48 PM and 2:09 PM

Find the z-score of 1:46 and 2:09 PM which will result to a z value of 0.034599

P (1:48 < x < 2:09) = P (z<0.034599) = 0.5138