Question

Assume that the finishing times in a New York City 10-kilometer road race are
normally distributed with a mean of 61 minutes and a standard deviation of 9
minutes. Let X be a randomly selected finishing time.

a. What is the probability that a that a person running a 10-kilometer road race
in New York City, will end up with finishing times average between 52 and 70
minutes for a road race?

b. Find the value of the 95% percentile, P 95 , for this random variable.

c. Use the information from part (b) to circle the correct choice and to fill in the
blanks in the following sentence:
% of finishing time average (at least/at most) minutes in a

New York City 10-kilometer road race.

Answer 1

Solution:
Given in the question
the finishing times in a New York City 10-kilometer road race are normally distributed with
Mean() = 61
Standard deviation() = 9
Solution(a)
We need to calculate probability that a that a person running a 10-kilometer road race in New York City, will end up with finishing times average between 52 and 70 minutes for a road race i.e. P(52<X<70) =? which can be calculated as
P(52<X<70) = P(X<70) - P(X<52)
Here we will use standard normal distribution, First we will calculate Z-score which can be calculated as
Z-score = (X-)/ = (70-61)/9 = 1
Z = (52-61)/9 = -1
From Z table we found p-value
P(52<X<70) = P(X<70) - P(X<52) = 0.8413 - 0.1587 = 0.6826
So there is 68.26% probability that a that a person running a 10-kilometer road race in New York City, will end up with finishing times average between 52 and 70 minutes for a road race.
Solution(b)
We need to calculate 95th percentile i.e. p-value = 0.95
From Z table we found Z-score = 1.645
So finishing time average can be calculated as
X = + Z-score * = 61 + 1.645*9 = 61 + 14.81 = 75.81
95th percentile is 75.81 minutes.
Solution(c)
95% of finishing time is atleast 75.81 minutes in a new york city 10-kilometer road race.