Question

Assume that the time needed to complete a midterm exam for a particular course is normally distributed with a mean of 90 minutes and a standard deviation of 15 minutes.

5- What percentage of the class is expected to complete the exam within 60 minutes?

6- What percentage of the class is predicted to NOT complete the exam within 115 minutes?

7- If the time limit is 100 minutes and there are 200 students in the class, how many of them do you predict to not complete the exam within the allocated time?

Answer 1

Here we assume that the time ( X ) needed to complete a midterm exam for a particular course is normally distributed with a mean of 90 minutes and a standard deviation of 15 minutes.

That is X~ N( = 90,   =15)

5- What percentage of the class is expected to complete the exam within 60 minutes?

Here we want to find P( X < 60 )

Let's use excel:

P( X < 60 ) = "=NORMDIST(60,90,15,1)" = 0.0228 = 2.28%

6- What percentage of the class is predicted to NOT complete the exam within 115 minutes?

Here we want to find P(X > 115) = 1 - P( X < 115) ........( 1 )

P( X < 115) = "=NORMDIST(115,90,15,1)" = 0.9522

Plug this value in equation ( 1 ), we get

P(X > 115) = 1 - 0.9522 = 0.0478 = 4.78%

7- If the time limit is 100 minutes and there are 200 students in the class, how many of them do you predict to not complete the exam within the allocated time?

First we need to find P(X > 100) = 1 - P( X < 100) ........( 2 )

P( X < 100) = "=NORMDIST(100,90,15,1)" = 0.7475

Plug this value in equation ( 2 ), we get

P(X > 100) = 1 - 0.7475 = 0.2525

Total number of students = 200

Therefore, total number of students we predict to not complete the exam within the allocated time

= 200*0.2525 = 50.4985 = 50 ( after rounding to the closest integer).