Question

A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was approximately normally distributed with mean 25 and standard deviation 7.

a. What proportion of students studied more than 20 hours?

b. What proportion of students studied less than 16 hours?

c. What proportion of the students studied between 25 and 40 hours?

Answer 1

= 25

= 7

(a) To find P(X>20):

Transforming to Standard Normal Variate:
Z = (X - )/

= (20 - 25)/7 = - 0.7143

Table Of Area Under Standard Normal Curve gives area = 0.2611

So, proportion of students studied more than 20 hours = 0.5 + 0.2611 = 0.7611

(b)

To find P(X < 16):
Z = (16 - 25)/7 = - 1.2857

Table givss area = 0.4015.

So,

proportion of students studied less than 16 hours = 0.5 - 0.4015 = 0.0985

(c) To fins P(25 < X < 40):

Z = (40 - 25)/7 = 2.1429

Table gives area = 0.4838

So,

proportion of the students studied between 25 and 40 hours = 0.4838.