6. Find the mean and standard deviation of the DRIVE variable by using =AVERAGE(

Question

6. 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 ___.51.7 Standard deviation ____25.80 Predicted percentage ____42.42% Actual percentage ________40% Comparison _The predicted percentage is 12.42 % larger. The results are quite close though. They are only different by 2 % so that’s actually pretty accurate.

Explanation / Answer

Using the command ,=AVERAGE(A2:A36) in excel and after press the enter key we have,

Mean = 51.7

Using the command ,=STDEV(A2:A36) in excel and after press the enter key we have,

Standard deviation = 25.80

Assuming that this variable is normally distributed, what percentage of data predicted would be less than 40 miles is given by,

Using the command ,=NORM.DIST(40, mean, stdev,TRUE) in excel and after press the enter key we have,

Predicted percentage = 0.33 i.e 33%

To find the actual percentage we sort the DRIVE variable and there are 14 data points are less than 40 out of the total 35 data points. That is the actual percentage is 14/35 = 0.40 i.e. 40%

Comparision -

The predicted percentage is 33% which is smaller than the actual percentage.

DRIVE SORT 36 4 20 6 88 6 6 20 71 25 42 25 76 28 63 29 36 33 63 36 38 36 28 36 55 38 33 39 40 40 80 42 86 42 83 54 4 54 39 55 25 63 25 63 54 71 54 71 81 73 73 76 29 76 76 77 78 78 77 80 42 81 36 83 71 86 94 88 6 94

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