Question

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 20.05% of the 60 days have a travel time that is at least ?x.

You may find this standard normal distribution table or this list of software manuals useful. Give your answer rounded to two decimal places.

Answer 1

Solution :

Given that ,

mean = = 35.6

standard deviation = = 10.3

n = 60

=   = 35.6 and

= / n = 10.3/ 60 = 1.33

P(Z z ) = 20.05%

1 - P(Z < z ) = 0.2005

P(Z < z ) = 1 - 0.2005

P(Z < z ) = 0.7995

P(Z < 0.84) = 0.7995

z = 0.84

Using z-score formula

X = z*+

= 0.84*1.33+35.6

= 36.72