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 such that 17.36% of the 60 days have a travel time that is at least . You may find this standard normal distribution table or this list of software manuals useful. Give your answer rounded to two decimal places.

Answer:the answer is 45.28 minutesStep-by-step explanation:Let X be the travel time in minutesX is normally distributed with μ= 36.6 and σ=10.3 i.e. X~N (35.6, 10.3)The travel time x such that 17.36% of the 60 days have a travel time that is at least x is given by,P(X>x)= 0.1736P(X<x)= 0.8264P(Z<x−35.6/10.3)= 0.8264P(Z<0.940)=0.8264x−35.6/10.3= 0.940x=0.940* 10.3+35.6=45.28Therefore, x is equal to 45.28 minutes