Question

1. 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?  

Mean ______________             Standard deviation ____________________

Predicted percentage ______________________________

Actual percentage _____________________________

Comparison ___________________________________________________

______________________________________________________________

Drive (miles):

Drive (miles)

4

6

20

20

25

25

25

28

29

33

36

36

36

36

36

40

42

54

55

63

63

71

73

73

76

76

76

78

80

80

80

88

88

94

94

Answer 1

Mean and standard deviation using the AVERAGE and STDEV functions are

Mean = 52.54

Standard deviation = 26.573

Predicted percentage = NORM.DIST(40,mean,standard deviation, TRUE)

= NORM.DIST(40,52.54, 26.573, TRUE)

= 0.3185 or 31.85%

There are 35 total data values and only 15 are below 40

So, actual % = (15/35)*100 = 42.86%

Actual percentage of data values below 40 is higher as compared to the predictive percentage because 42.86% is higher than 31.85%. So, we can say that the actual percentage provides higher percentage as compared to predictive percentage.