statistical Math?

Question

8. An executive at Westinghouse drives from his home in the
suburbs near Pittsburgh to his office in the center of
the city. The driving times are normally distributed
with a mean of 35 minutes and a standard deviation of 8
minutes.

a. In what percent of the days will it take him less
than 30 minutes to drive to work?
b. In what percent of the days will it take more than
40 minutes to drive to work?
c. Some days there will be accidents, etc., so the
trip will take longer than usual. How long will the
longest 10 percent of the trips take?

Answer 1

In each case, you need to convert to the "standard normal"
z = (x-µ)/σ

where µ is the mean and σ is the standard deviation.

a) z = (30 - 35)/8 = -0.625

From the standard normal table, N_z(-0.625) = 0.266

So it will take less than 30 minutes for 26.6% of the days.

b) The standard normal table gives the probability that z is less than some number. So to find the probability that it takes more than 40 minutes, find the probability that it takes less than 40 minutes (similar to what I did above for 30 minutes); then the probability that it takes more than 40 minutes is
1 - probability that it takes less than 40 minutes. You should get 26.6%

c) This is the same as asking "For what trip length is the probability 0.9 that a trip takes less than that length of time?"

Find the value .9000 (or something close to that) in the body of the normal distribution tables, and note the corresponding t-value on the edge of the table. In my table, 0.8997 corresponds to 1.28 and 0.9015 corresponds to 1.29.
If you interpolate, the t-value would be 1.2817 (1.28 may well be good enough). Then

z = 1.2817
(x-µ)/σ = 1.2817
(x - 35)/8 = 1.2817
x = 1.2817*8 + 35 = 45.254
(or x = 1.28*8 + 35 = 45.24 if you use the uninterpolated value)

So the longest 10% of the trips take at least 45.254 minutes.