A line up for tickets to a local concert had an average mean

A line up for tickets to a local concert had an average (mean) waiting time of 20 minutes with a standard deviation of 4 minutes.

a) What percentage of the people in the line waited for more than 28 minutes?

b) If 2000 ticket buyers were in line, how many of them would expect to wait for less than 16 minutes?

Waiting time for tickets for a concert ~ N(20,4)

a) P(X > 28) = P( Z > (28 -20)/4) = P(Z > 2) = 1 - P( Z < 2) = 1 - 0.9772 = 0.0228

Therefore, 2.28% people in the line will wait for more than 28 minutes

b ) P(X < 16) = P(Z < (16-20)/4) = P(Z < -1) = 0.1587

As there are 2000 ticket buyers 2000* 0.1587 = 317.4 = 317 ticket buyers in the line will wait for less than 16 minutes