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

answer: procent 54 wynosi 33,33%

but there’s no any figure there!