Question

The time needed to complete a final examination in a particular college course is normally distributed with a mean of 83 minutes and a standard deviation of 13 minutes. Answer the following questions. Round the intermediate calculations for z value to 2 decimal places. Use Table 1 in Appendix B. What is the probability of completing the exam in one hour or less (to 4 decimals)? What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes (to 4 decimals)? Assume that the class has 60 students and that the examination period is 90 minutes in length. How many students do you expect will be unable to complete the exam in the allotted time (to the next whole number)?

Answer 1

Part a)

P ( X <= 60 )
Standardizing the value

Z = ( 60 - 83 ) / 13
Z = -1.77

P ( X <= 60 ) = P ( Z < -1.77 )
P ( X <= 60 ) = 0.0384

Part b)

P ( 60 < X < 75 )
Standardizing the value

Z = ( 60 - 83 ) / 13
Z = -1.77
Z = ( 75 - 83 ) / 13
Z = -0.62
P ( -1.77 < Z < -0.62 )
P ( 60 < X < 75 ) = P ( Z < -0.62 ) - P ( Z < -1.77 )
P ( 60 < X < 75 ) = 0.2692 - 0.0384
P ( 60 < X < 75 ) = 0.2307

Part c)

P ( X > 90 ) = 1 - P ( X < 90 )
Standardizing the value

Z = ( 90 - 83 ) / 13
Z = 0.54

P ( Z > 0.54 )
P ( X > 90 ) = 1 - P ( Z < 0.54 )
P ( X > 90 ) = 1 - 0.7054
P ( X > 90 ) = 0.2946

N = 60 , hence 60 * 0.2946 = 17.676   18 students.