Question

The time needed to complete a final examination in a particular college course is normally distributed with a mean of 77 minutes and a standard deviation of 12 minutes. Answer the following questions. 

a. What is the probability of completing the exam in one hour or less (to 4 decimals)? 

b. What is the probability that a student will complete the exam in more than 60 minutes but less than 75 minutes (to 4 decimals)? 

c. 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 nearest whole number)?  

Answer 1

Part a)
X ~ N ( µ = 77 , σ = 12 )
P ( X < 60 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 60 - 77 ) / 12
Z = -1.4167
P ( ( X - µ ) / σ ) < ( 60 - 77 ) / 12 )
P ( X < 60 ) = P ( Z < -1.4167 )
P ( X < 60 ) = 0.0783

Part b)
X ~ N ( µ = 77 , σ = 12 )
P ( 60 < X < 75 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 60 - 77 ) / 12
Z = -1.4167
Z = ( 75 - 77 ) / 12
Z = -0.1667
P ( -1.42 < Z < -0.17 )
P ( 60 < X < 75 ) = P ( Z < -0.17 ) - P ( Z < -1.42 )
P ( 60 < X < 75 ) = 0.4338 - 0.0783
P ( 60 < X < 75 ) = 0.3555

Part c)
X ~ N ( µ = 77 , σ = 12 )
P ( X > 90 ) = 1 - P ( X < 90 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 90 - 77 ) / 12
Z = 1.0833
P ( ( X - µ ) / σ ) > ( 90 - 77 ) / 12 )
P ( Z > 1.0833 )
P ( X > 90 ) = 1 - P ( Z < 1.0833 )
P ( X > 90 ) = 1 - 0.8607
P ( X > 90 ) = 0.1393

Number of students would be 60 * 0.1393 = 8.385 ≈ 8 students would not able to complete the exam on time.