Question

Suppose that the travel time from your home to your office is a normal random variable with mean 40 minutes and standard deviation 7 minutes on Mondays-Thursdays, and a normal random variable with mean 35 minutes and standard deviation 5 minutes on Fridays.

(a) If you want to be 95 percent certain that you will not be late for an office meeting at 1PM on a Monday, what is the latest time at which you should leave home?

(b) Suppose that you leave home every morning at 7:10AM to start your shift at 8AM. In a working year with 50 weeks and 250 working days, how many times on average will you be late to work with that schedule?

	1.	a) 12:02 b) 14
	2.	a) 12:08    b) 5
	3.	a) 12:08 b) 17
	4.	a) 12:30 b) 5
	5.	a) 12:30 b) 17

explain your answer

Answer 1

Solution:
Given in the question
travel mean time from your home on monday to thursday = 40 minutes
Standard deviation = 7 minutes
travel mean time from your home on friday = 35 minutes
Standard deviation = 5 minutes
Solution(a)
Office meeting time on Monday = 1PM
He want to 95 percent certain that you will not be late for an office meeting at 1PM on monday
So p-value = 0.95 and Z-score from Z table is 1.645
So the latest time at which you should leave home can be calculated as
X = Mean + Z-score *Standard deviation = 40 + 1.645*7 = 40+11.515 = 51.515 or 52 minutes before the time
So latest time at which you should leave home = 52 minutes before 1 PM i.e. 12:08 PM
Solution(b)
Shift time = 8AM
Leaving time = 7:10 AM
So Leaving before time = 50 minutes
Total working days = 250
So Total Monday to thursday = 200
Total friday = 50
Probability of getting late on Monday to Thursday can be calculated using standard normal curve as follows:
Z-score = (X - Mean)/SD = (50-40)/7 = 1.43
From Z table, we found probability of getting late = 0.0764
So Total number of late days in Monday to Thursday = 200*0.0764 = 15.3 or 16 days
Probability of getting late on Friday can be calculated using standard normal curve as follows:
Z-score = (X - Mean)/SD = (50-35)/5 = 3
From Z table, we found probability of getting late on friday= 0.00135
So Total number of late days in Friday = 50*0.00135 = 0.0675 or 1 day
So total late days = 16 +1 =17 days
So Its correct answer is C. i.e. 12.08 and 17