Question 1106626: 3- 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 10 minutes.
a- What percentage of the class is expected to complete the exam within an hour?
b- What percentage of the class is predicted to be unable to complete the exam within 100 minutes?
c- If the time limit is 110 minutes and there are 200 students in the class, how many of them do you expect to be unable to complete the exam within the allocated time?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Using the normal distribution, z-values need to be calculated for a, b, c
:
z-value = (X - mean) / standard deviation
:
a) z-value = (60 - 90) / 10 = -3.0
lookup the probability(P) associated with a z-value = -3.0
P ( X < 60 ) = 0.0013 = 0.13%
:
b) P ( X > 100 ) = 1 - P ( X < 100 )
z-value = ( 100 - 90 ) / 10 = 1.0
P ( X < 100 ) = 0.8413
P ( X > 100 ) = 1 - 0.8413 = 0.1587 = 15.87%
:
c) z-value = (110 - 90 ) / 10 = 2.0
P ( X < 110 ) = 0.9772
P ( X > 110 ) = 1 - 0.9772 = 0.0228
200 * 0.0228 = 4.56 approximately 5 students
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