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?______________________________________________ Drive (miles) 54 78 71 94 80 36 36 6 28 55 36 6 42 33 36 80 63 40 4 25 80 76 20 73 25 88 76 29 76 63 71 42 54 80 36

Explanation / Answer

mean= 51.2 miles

Standard deviation= 25.50

predicted percentage = Cumulative Probability = NORM.DIST(40, mean, stdev,TRUE) = 0.33

Actual percentage =15/35 =0.43=43%

Comparison : THE Actual percentage is greater than predicted percentage.

Question 2 :

Predicted percentage between 40 and 70 = NORM.DIST(70, mean, stdev, TRUE) - NORM.DIST(40, mean, stdev, TRUE) = 0.77 - 0.33 = 0.44 = 44%

actual percentage = 7/35 = 0.2 = 20%

predicted percentage more than 70 miles = 1 - 0.77 = 23%

actual percentage = 13/35 = 37%

Comparison : We see that actual percentage between 40 to 70 is quite less than predcited but for above 70, it is just opposite.

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