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?

Question

Anabelle

Mathematics

5 November, 23:57

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?

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Answers (1)

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Trevin Austin · Answer 1

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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