91Ó°ÊÓ

The dot plot is as follows:

From the above graph in part (a), we have,

We note that$15ofthe16$ dots in the dot plot lie to the right to zero which implies that most of the differences (Weight minus Groove) are positive and thus mean total distance traveled with the weight loss method appears to exceed the mean total distance traveled with the groove method.

It is given that:

$10.2,2.7,6.4,5.3,7.0,1.8,5,8.6,7.3,3.6,-0.5,7.3,3.6,-0.5,8.4,1,3.7,2.2,0.2$

The mean is :

$$\begin{gathered} \stackrel{¯}{x}=\frac{{\displaystyle \underset{i-1}{\overset{n}{∑}}}{x}_i}{n} \\ =\frac{10.2+2.7+6.4+5.3+7.0+1.8+5+8.6+7.3+3.6+-0.3+8.4+1+3.7+2.2+0.2}{16} \\ =\frac{72.9}{16} \\ =4.5563 \end{gathered}$$

The sample variance is then as:

$$\begin{gathered} s^2=\frac{\underset{}{∑}{(x-\stackrel{¯}{x})}^2}{n-1} \\ =\frac{{(10.2-4.5563)}^2+(2.7-4.5563{)}^2+(6.4-4.5563{)}^2+(5.3-4.5563{)}^2+(7-4.5563{)}^2+(1.8-4.5563{)}^2+(5-4.5563{)}^2+(8.6-4.5563{)}^2+(7.3-4.5563{)}^2+(3.6-4.5563{)}^2+(-0.3-4.5563{)}^2+(8.4-4.5563{)}^2+(1-4.5563{)}^2+(3.7-4.5563{)}^2+(2.2-4.5563{)}^2+(0.2-4.5563{)}^2}{16-1} \\ =10.4040 \end{gathered}$$

The sample standard deviation is then,

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{10.4040} \\ =3.2255 \end{gathered}$$

The difference is $4.5563$thousands of miles on average which varies on average by$3.2255$ thousands of miles.