Suppose abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. assume the travel times are normally distributed with a standard deviation of 10.3 min. determine the travel time x such that 22.96% of the 60 days have a travel time that is at least x.

Given that:

mean,μ=35.6 min

std deviation,σ=10.3 min

we are required to find the value of x such that 22.96% of the 60 days have a travel time that is at least x.

using z-table, the z-score that will give us 0.2296 is:-1.99

therefore:

z-score is given by:

(x-μ) / σ

hence:

-1.99 = (x-35.6) / 10.3

-20.497=x-35.6

x=35.6-20.497

x=15.103