Question

Can you show me how to do it in excel terms??

The time needed to complete a final examination in a particular college course is normally distributed with a mean of 79 minutes and a standard deviation of 8 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 test in one hour or less (to 4 decimals)?
  

What is the probability that a student will complete the test 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 test in the allotted time (to the next whole number)?

Answer 1

The time needed to complete a final examination in a college course is normally distributed with a mean of 79 minutes and a standard deviation of 8 minutes. Thus, X - N(u,o?). The Z-score is then written as follows: Z-X-11 --9579 - N(0,1) The probability of completing the test in one hour or less (in 60 minutes or less) is derived as follows: P(X360)=P(4=4560-4) < 60-79 . 8 =P(:5–2.375) Using excel this probability is calculated as follows: X fx =NORMSDIST(-2.375) D E F C 0.008774 Thus, P(X 560)=P(: 5-2.375) =0.008774 -0.0088 Hence, the probability of completing the test in one hour or less is 0.0088.

b. The probability that the student will complete the test in more than 60 minutes but less than 75 is derived as follows: 60-u X-u 75 - u P(60 c. It is assumed that the class has 60 students and the examination period is 90 minutes in length. The probability that students will not be able to complete the test in the allotted time, that is, the test is not completed in 90 minutes is obtained as follows: P(X >90)=1-P(X <90) 1-PX-4 90- - - =1-P(:51.375) The probability is calculated in Excel as follows: X fc =NORMSDIST(1.375) B C 0.9154343 D E F Thus, P(X > 90)=1-P(=51.375) =1-0.9154343 = 0.08457 Thus, the expected number of students that are not able to complete the exam in 90 minutes is derived as follows: Mean = np = 60 x 0.08457 = 5.0742 Hence, the expected number of students who will be unable to complete the test in the allotted time is 5.

c. It is assumed that the class has 60 students and the examination period is 90 minutes in length. The probability that students will not be able to complete the test in the allotted time, that is, the test is not completed in 90 minutes is obtained as follows: P(X >90)=1-P(X <90) 1-PX-4 90- - - =1-P(:51.375)

The probability is calculated in Excel as follows: X fc =NORMSDIST(1.375) B C 0.9154343 D E F Thus, P(X > 90)=1-P(=51.375) =1-0.9154343 = 0.08457 Thus, the expected number of students that are not able to complete the exam in 90 minutes is derived as follows: Mean = np = 60 x 0.08457 = 5.0742 Hence, the expected number of students who will be unable to complete the test in the allotted time is 5.

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