Answer:
See explaination
Step-by-step explanation:
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
(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
(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
