Question

Place Engine Size output (starting with label) in E2 -> EngncSIzC Place Hwy Fuel Eff output (starting with labe) in H2-> Hwy Fuci Effcicncy .155 0.0855 3.1558 .0857 0.7151 Mean -> Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Mcan -> Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Vlaximium Sum 34.2714 .0863 34.5000 7.0000 9.0888 0.6958 0.0122 1.3757 0.1782 Range Minimum Maximum 8233 1.7687 um Count -> Count Count -> ount Q1 Q3 2.5246 3.7869 Q1 Q3 25.0000 12.0000

Choose one answer for the purple blanks, answer number for orange blanks

Thanks!

Answer 1

Engine size data has approximately same values of Mean, Median and Mode, So Engine size data follows approximately normal distribution.

The Empirical rule for normal distribution state that :

Approximately 68% of the observations fall within 1 standard deviation of the mean or area between z= -1 and z=1 is 0.68
Approximately 95% of the observations fall within 2 standard deviations of the mean or area between z= -2 and z= 2 is 0.95
Approximately 99.7% of the observations fall within 3 standard deviations of the mean or area between z= -3 and z= 3 is 0.997

We are given two limits 1.7256 and 4.5859

Let us first find the z scores for both numbers.

z = $\frac{x-\mu }{\sigma }$ ;  µ is mean and σ is standard deviation.

For x = 1.7256

z = $\frac{1.7256-3.1558 }{0.7151 }$   

z = -2

For x = 4.5859

z = $\frac{4.5859-3.1558 }{0.7151 }$

z = 2

Therefore according to Empirical rule

Approximately 95% of the observations fall within 2 standard deviations of the mean or area between z= -2 and z= 2 is 0.95

#7 ) The Empirical rule predicts that about 95 percent of engine size data should lies between 1.7256 and 4.5859 liters , which are +/- 2 standard deviation away from the mean.

#8) The coefficient of variation (CV) is the best measure of variability between two groups.

CV for engine size = $\frac{\sigma }{\mu }*100$ =  $\frac{0.7151 }{3.1558 }*100$ = 22.66%

CV for Hwy fuel efficiency =  $\frac{\sigma }{\mu }*100$ =  $\frac{9.0888 }{34.2714}*100$ = 26.52%

As CV for Hwy fuel efficiency is greater than CV for engine size,

Hwy fuel efficiency varies most.