Question

This week, a very large running race (5K) occured in Denver. The times were normally distributed, with a mean of 21.02 minutes and a standard deviation of 2.24 minutes.
Report your answers accurate to 2 decimals


a. What percent of runners took 23.79 minutes or less to complete the race?  %


b. What time in minutes is the cutoff for the fastest 11.38 %?  Minutes


c. What percent of runners took more than 15.4 minutes to complete the race?

Answer 1

Given for normal distribution,

= 21.02

= 2.24

a) P(X < 23.79)

=> P(Z < (23.79 - 21.02)/2.24 ) )

P(Z < 1.2366) = 0.89

b) Given P(X > x) = 0.1138

we have to find x.

For P(Z > z) = 0.1138, z = 1.207

Hence (X - )/ = 1.207

=> X = 1.207*2.24 + 21.02

=> X = 23.72

c) P(X > 15.4) = P(Z > (15.4 - 21.02)/2.24 ) = P(Z > -2.508) = 0.9939

=> 99.39%