Suppose that the travel time from your home to your office

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SUMMARY

The travel time from home to office is normally distributed with a mean of 40 minutes and a standard deviation of 7 minutes. To ensure a 95% certainty of arriving on time for a 1 p.m. appointment, one must calculate the latest departure time using the Z-score formula. The Z-score corresponding to 95% confidence is approximately 1.96, leading to a calculation of the latest time to leave home as 1 p.m. minus the total travel time, which is 40 minutes plus 1.96 times the standard deviation (7 minutes). This results in a departure time of approximately 12:50 p.m.

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Question:
Suppose that the travel time from your home to your office is normally distributed with mean 40 minutes and standard deviation 7 minutes. If you want to be 95% certain that you will not be late for an office appointment at 1p.m. what is the latest time that you should leave home?

Solution

here mean = 40 S.D=7

using Z = (X - 40)/72

but how should is calculate the minumum time?
 
Physics news on Phys.org
You're looking for an interval. 95% of the area under the normal curve lies within how many SDs of the mean?
 
Hint: Chebyshev's theorem
 

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