Question

Univeristy of Calagary's students average 60 hours of sleep per week. Sleep is a normally distributed variable with a standard deviation of 6 hours per week. Determine the amount of sleep that is exceeded by no more than 2.5% of students.

Answer 1

Using standard normal table,

P(Z < z) = 2.5%

= P(Z < z) = 0.025  

= P(Z <-1.96 ) = 0.025

z =-1.96 Using standard normal z table,

Using z-score formula  

x= z * +

x= -1.96 *6+60

x= 48.24

x=48