91Ó°ÊÓ

The dot plot is as follows:

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

We then note that all ten dots in the dot plot lie to the right to zero which implies the most of the difference (Bottom minus Top) are positive and thus the mean zinc concentration at the bottom appears to be greater than the mean zinc concentration at the top.

It is given that:

$0.015,0.028,0.177,0.121,0.102,0.107,0.019,0.066,0.058,0.111$

The mean is:

$$\begin{gathered} \stackrel{¯}{x}=\frac{{\displaystyle \underset{i-1}{\overset{n}{∑}}}{x}_i}{n} \\ =\frac{0.015+0.028+0.177+0.121+0.102+0.107+0.019+0.066+0.058+0.111}{10} \\ =\frac{0.804}{10} \\ =0.0804 \end{gathered}$$

The sample variance is then as:

$$\begin{gathered} s^2=\frac{\underset{}{∑}{(x-\stackrel{¯}{x})}^2}{n-1} \\ =\frac{{(0.015-0.0804)}^2+(0.028-0.0804{)}^2+(0.177-0.0804{)}^2+(0.121-0.0804{)}^2+(0.102-0.0804{)}^2+(0.107-0.0804{)}^2+(0.019-0.0804{)}^2+(0.066-0.0804{)}^2+(0.058-0.0804{)}^2+(0.111-0.0804{)}^2}{10-1} \\ =0.0027 \end{gathered}$$

The sample standard deviation is then,

$$\begin{gathered} s=\sqrt{s^2} \\ =\sqrt{0.0027} \\ =0.0523 \end{gathered}$$

The difference are on average $0.0804$milligrams which varies by$0.0523$ milligrams on average.