Question

The time needed to complete a final examination in a particular college course is normally distributed with a mean of 100 minutes and a standard deviation of 20 minutes. Answer the following questions.

(a)What is the probability of completing the exam in one hour or less?

(b) What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes?

(c) Assume that the class has 90 students and that the examination period is 130 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time

Answer 1

Solution :

Given that ,

mean = = 100 minutes

standard deviation = = 20 minutes

a) P(x 60)

= P[(x - ) / (60 - 100) / 20]

= P(z -2)

Using z table,

= 0.0228

b) P(60 < x < 75) = P[(60 - 100) / 20) < (x - ) /  < (75 - 100) / 20) ]

= P(-2 < z < -1.25)

= P(z < -1.25) - P(z < -2)

Using z table,

= 0.1056 - 0.0228

= 0.0828

c) n = 90

P(x > 130) = 1 - p( x< 130)

=1- p P[(x - ) / < (130 - 100) / 20]

=1- P(z < 1.5)

Using z table,

= 1 - 0.9332

= 0.0668

= 0.0668 * 90 = 6 students