The time needed to complete a in a particular college course is normally distrib
ID: 3301664 • Letter: T
Question
The time needed to complete a 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 gym in one hour or less (to 4 decimals)?
What is the probability that a student will complete the gym in more than 60 minutes but less than 75 minutes (to 4 decimals)? Assume that the class has 60 students and that the gym period is 90 minutes in length.
How many students do you expect will be unable to complete the gym in the allotted time (to the next whole number)?
Explanation / Answer
Answer:
The time needed to complete a 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 gym in one hour or less (to 4 decimals)?
Z value for 60 minutes, z =(60-83)/13 = -1.77
P( x < 60) = P( z < -1.77)
= 0.0384
What is the probability that a student will complete the gym in more than 60 minutes but less than 75 minutes (to 4 decimals)? Assume that the class has 60 students and that the gym period is 90 minutes in length.
Z value for 60 minutes, z =(60-83)/13 = -1.77
Z value for 75 minutes, z =(75-83)/13 = -0.62
P( 60<x<75)m = P( -1.77<z<-0.62) =P( z<-0.62) – P( z < -1.77)
=0.2676-0.0384
=0.2292
How many students do you expect will be unable to complete the gym in the allotted time (to the next whole number)?
Z value for 90 minutes, z =(90-83)/13 = 0.54
P( x >90) = P( z > 0.54) = 0.2946
0.2946*60 =17.676
We expect 18 students will be unable to complete the gym in the allotted time.
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