Question

1. 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?  

Predicted percentage between 40 and 70 ______________________________ Actual percentage _____________________________________________ Predicted percentage more than 70 miles ________________________________ Actual percentage ___________________________________________ Comparison ____________________________________________________ _______________________________________________________________ Why? __________________________________________________________ ________________________________________________________________
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
52.54285714
26.57325375
0.318459615
0.318459615
0.318459615

Answer 1

Using excel function AVERAGE and STDEV to find the mean and standard deviation of the data

we get

Mean =52.54

standard deviation = 26.57

Predicted percentage between 40 and 70 = NORM.DIST(70, mean, stdev, TRUE) - NORM.DIST(40, mean, stdev, TRUE)

setting the values, we get

= NORM.DIST(70, 52.54, 26.57, TRUE) - NORM.DIST(40, 52.54, 26.57, TRUE)

= 0.7445 - 0.3185

= 0.4260 or 42.60%

There are only 6 data values between 40 and 70

So, actual % = (6/35)*100 = 14.14%

SECOND PART

Using excel function AVERAGE and STDEV to find the mean and standard deviation of the data

we get

Mean =52.54

standard deviation = 26.57

Predicted percentage above 70 = 1- NORM.DIST(70, mean, stdev, TRUE)

setting the values, we get

= 1- NORM.DIST(70, 52.54, 26.57, TRUE)

=1- 0.7445

= 0.2555 or 25.55%

There are only 14 data values above 70

So, actual % = (14/35)*100 = 40.00%

Actual percentage for data values between 40 and 70 is lower as compared to the predicted percentage for data values between 40 and 70 because less data values are present between the 40 and 70 in the given data sample. Actual percentage for data values 70 is higher as compared to the predicted percentage for data values above 70 because more data values are present above 70 in the given sample.