Question

jenny bird who lives in the san francisco bay area, commutes by car from home to work. she knows that it takes her an average of 28 minutes for this commute in the morning. However, due to the variability in the traffic situation every morning, the standard deviation of these commutes is 5 minutes. suppose the population of her morning commute times has a normal distribution with a mean of 28 minutes and a standard deviation of 5 minutes. Jenn has to be at work at 8:30 am every morning. By what time must she leave home in the morning so that she is late for work at most 1% of the time?

Answer 1

Mean time of travelling = 28 minutes

Standard deviation of time = 5 minutes

X be the difference between 8:30 and the time she should leave for work

if she leaves at t or before t she will be late only for work only 1% of times

P[ X < t ] = 1 - 0.01 = 0.99

P[ ( X - 28 )/5 < ( t - 28 )/5 ] = 0.99

P[ Z < ( t - 28 )/5 ] = 0.99

P[ Z < 2.33 ] = 0.99

t - 28 = 2.33*5 = 11.65

t = 28 + 11.65 = 39.65

She should leave 39.65 minutes before 8:30, so that she is late for work only 1% of the times

or, she should leave at 7:50 from home to work so that she is late for work only 1% of the times