Find the mean and standard deviation of the DRIVE variable by using =AVERAGE(A2:

Question

Find the mean and standard deviation of the DRIVE variable by using =AVERAGE(A2:A36) and =STDEV(A2:A36). Assuming that this variable is normally distributed, what percentage of data would you predict would be less than 40 miles? This would be based on the calculated probability. Use the formula =NORM.DIST(40, mean, stdev,TRUE). Now determine the percentage of data points in the dataset that fall within this range. To find the actual percentage in the dataset, sort the DRIVE variable and count how many of the data points are less than 40 out of the total 35 data points. That is the actual percentage. How does this compare with your prediction?   (10 points)

Mean ______________             Standard deviation ____________________

Predicted percentage ______________________________

Actual percentage _____________________________

Comparison ___________________________________________________

______________________________________________________________

What percentage of data would you predict would be between 40 and 70 and what percentage would you predict would be more than 70 miles? Subtract the probabilities found through =NORM.DIST(70, mean, stdev, TRUE) and =NORM.DIST(40, mean, stdev, TRUE) for the “between” probability. To get the probability of over 70, use the same =NORM.DIST(70, mean, stdev, TRUE) and then subtract the result from 1 to get “more than”. Now determine the percentage of data points in the dataset that fall within this range, using same strategy as above for counting data points in the data set. How do each of these compare with your prediction and why is there a difference?   (11 points)

Predicted percentage between 40 and 70 ______________________________

Actual percentage _____________________________________________

Predicted percentage more than 70 miles ________________________________

Actual percentage ___________________________________________

Comparison ____________________________________________________

_______________________________________________________________

Why? __________________________________________________________

________________________________________________________________

Mean ______________             Standard deviation ____________________

Predicted percentage ______________________________

Actual percentage _____________________________

Comparison ___________________________________________________

______________________________________________________________

43315446151362542125651624522424444 46651632165415114256556121221252663 13415142213315533614221543335644653 ts 6423.5 6 3 2 3 6 4 5 3 2 3 2 2 2 3 3 3 4 2 3 5 1 6 6 6 5 1 3 1 12 44456447543643754445443434475524444 FFFFFFMMMFFFFMFFF F M F M F M M M F M F M M M M M M M 8747568778775784476776677878747678 10 C 5 5 9 8 6 8 11 10 9 8 12 9 9 7 8 8 7 8 8 9 8 8 11 12 9 9 11 8 12 11 9 1 1 1 1 B tat-OR MI PA FL MI MI SC MI OR IL FL NY PA TX PA SC NY OH IL MI GA CA TX NV TX FL NY CA CA PA CA MI MI KY OR A-fr-3 2 8 4 8 7 3 3 3 4 3 76 8 6 6 6 28 55 40 4 2 2 3 8 9 5 4 8 6 76 78 76 7 94 6

Explanation / Answer

Using the functions of Excel, we got the following results:

Mean = 51.2

Standard deviation = 25.49948

Predicted percentage ( less than 40 miles) = 0.33

Actual percentage (less than 40 miles ) = 14/35 = 0.40

Comparison = Predicted probability is less than actual percentage

Predicted percentage between 40 and 70 = 0.44

Actual percentage = 8/35 = 0.23

Comparison = Predicted probability is greater than actual percentage

Predicted percentage more than 70 miles = 0.23

Actual percentage = 13/35 = 0.37

Comparison = Predicted probability is less than actual percentage

Reason for difference: The sample is small here, which is causing the difference. It is only 35 here

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.