The times for running a 5K for a local charity event are Normally distributed with a mean time of 28 minutes and standard deviation of 5.4 minutes. Which time represents the 60th percentile?
Find the z-table here.

A. 19.6 minutes
B. 26.6 minutes
C. 29.4 minutes
D. 36.4 minutes

29.4 minutes represents the 60th percentile

Step-by-step explanation:

Given :  The times for running a 5K for a local charity event are Normally distributed, with a mean time of 28 minutes and standard deviation of 5.4 minutes.

To Find : Which time represents the 60th percentile?

 19.6 minutes

26.6 minutes

29.4 minutes

36.4 minutes

Solution:

time represents the 60th percentile

Means  60 % data = 0.6 below this

from z table ,  z score for this =  0.253

z score  = ( Value - Mean ) / SD

=> 0.253  = ( Value - 28 ) /5.4

=> 1.3662  =  Value - 28

=>  Value = 29.3662

=>  Value = 29.4

29.4 minutes represents the 60th percentile