Question

A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average time spent studying for the sample was between 29.0 and 30 hours studying? Round to 4 decimal places.

Answer 1

Solution :

Given that ,

mean = = 25

standard deviation = = 15

n = 36

= 25

=  / n= 15 / 36 =2.5

P(29<    <30 ) = P[(29-25) /2.5 < ( - ) / < (30-25) /2.5 )]

= P(1.6 < Z < 2)

= P(Z <2 ) - P(Z < 1.6)

Using z table

=0.9772 - 0.9452

=0.0320

probability= 0.0320