How to logically interpret this question on normal distribution (travel time)?

Question

I was trying to comprehend the following question:

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

There are a few things which are confusing me. I know the formula for normal distribution but I can't logically understand the question yet.

It is said that the mean travel time is $40$ minutes. The question is basically asking me what minimum travel time I need to give to myself so that with $95%$ probability I will reach office at $1$ p.m. Right? I'm not being able to convert this into the math. Any hints will be appreciated.

Perhaps, let's take a simpler question: What would be probability that I will reach on time if I begin my journey exactly $40$ minutes (which is the standard deviation) before $1$ p.m?

Answer 1

Maybe try a case and think about it this way:

Suppose you left with $40$ minutes to spare. So long as it takes you $40$ minutes or less to arrive, you are on time. What is the probability that if you choose a time randomly from this distribution, you will get a time of $40$ minutes or less? HINT: You may want to think about a $z$-score for $40$.

With that idea in mind, you should see why you want $95$% of the values of the travel time to be less than or equal to your travel time to arrive on time. Given that probability or proportion of values, could you perhaps find a $z$-score corresponding to that proportion? What travel time would that $z$-score correspond to?

Answer 2

Hint: The probability that you will need less than $ x $ minutes to arrive at work is $$P (X <x)=P (\frac {X-40}{7}<\frac {x-40}{7})$$ You know that $\frac {X-40}{7}=Z $ follows a Standard Normal distribution and you want the above probability to be equal to $95\% $. Look up a table of CDF values for $ Z $ to see what $\frac {x-40}{7}$ should be, and solve for $ x $.

Answer 3

A normal distribution is symmetric. So, if the mean is 40 and the s.d. is 7, the odds of it being plus or minus 1-sigma are about 67% (68.5 or so, but easier to think two-thirds). In other words, 67% of the time it will be between 33 and 47 minutes, and of those, half the time (33% of the time in total) in the upper half and an equal amount (33% or so) in the lower. In other words, the remaining roughly 33% it will be split between less than 33 or more than 47.

That's one-sigma. Two-sigma covers the range from 26-33 and 47-54. 95% of the odds split this range - about 2.5% in each.

This question seems engineered to force you to do a little lookup. 95% of he time in total you will be there between 26-40 minutes or 40-54. Were the question "if you want to be 97.5% certain to be there in 40 minutes" you should leave 14 minutes early, at the 26 minute mark - because delays that make the trip take 14 minutes longer occur only 2.5% of the time ( you don't care about the 2.5% - or really the 97.6% that make it shorter than 54 minutes; this is the 'one-tailed/two-tailed' problem.

Now, look up in a table the actual value that has a one-sided probability of 5%.