91影视

The data is collected from parking meters by Brinks and by other contractors and is listed in two samples containing 10 observations each.

The sample coefficient of variation is the quotient of the sample standard deviation to the sample mean. The formula of coefficient of variation is as follows:

$C.V.=\frac{s}{\stackrel{炉}{x}}脳100$

, where

$s$is the sample standard deviation;

$\stackrel{炉}{x}$is the sample mean.

The first sample is defined as collections made when the contractor was Bricks.

The mean of sample 1 is given as follows:

$$\begin{gathered} {\stackrel{炉}{x}}_1=\frac{\left(1.3+1.5+1.3+1.5+1.4+1.7+1.8+1.7+1.7+1.6\right)}{10} \\ =\frac{15.5}{10} \\ =1.55 \end{gathered}$$

The standard deviation of sample 1 is given as follows:

$$\begin{gathered} s_1=\sqrt{\frac{{鈭�}_{i=1}^n{\left(x_i-{\stackrel{炉}{x}}_1\right)}^2}{n_1-1}} \\ =\sqrt{\frac{{\left(1.3-1.55\right)}^2+{\left(1.5-1.55\right)}^2+....+{\left(1.6-1.55\right)}^2}{10-1}} \\ =0.18 \end{gathered}$$

Thus, the mean for sample 1 is equal to 1.55 million dollars, and the standard deviation of sample 1 is equal to 0.18 million dollars.

The coefficient of variation for sample 1 is computed as follows:

$$\begin{gathered} C{V}_1=\frac{s_1}{{\stackrel{炉}{x}}_1}脳100 \\ =\frac{0.18}{1.55}脳100 \\ =11.5\% \end{gathered}$$

Therefore, the coefficient of variation for sample 1 is equal to 11.5%.

The second sample is defined as collections made when the contractor was not Bricks.

The mean of sample 2 is given as follows:

$$\begin{gathered} {\stackrel{炉}{x}}_2=\frac{\left(2.2+1.9+1.5+1.6+1.5+1.7+1.9+1.6+1.6+1.8\right)}{10} \\ =\frac{17.3}{10} \\ =1.73 \end{gathered}$$

The standard deviation of sample 2 is given as follows:

$$\begin{gathered} s_2=\sqrt{\frac{{鈭�}_{i=1}^n{\left(x_i-{\stackrel{炉}{x}}_2\right)}^2}{n_2-1}} \\ =\sqrt{\frac{{\left(2.2-1.73\right)}^2+{\left(1.9-1.73\right)}^2+....+{\left(1.8-1.73\right)}^2}{10-1}} \\ =0.22 \end{gathered}$$

Thus, the mean for sample 2 is equal to 1.73 million dollars, and the standard deviation of sample 2 is equal to 0.22 million dollars.

The coefficient of variation for sample 2 is computed as follows:

$$\begin{gathered} C{V}_2=\frac{s_2}{{\stackrel{炉}{x}}_2}脳100 \\ =\frac{0.22}{1.73}脳100 \\ =12.8\% \end{gathered}$$

Therefore, the coefficient of variation for sample 2 is equal to 12.8%.

For the given two samples, the coefficients of variation for the two samples are comparable in magnitude.

Therefore, the two samples appear to have the sameamount of variation.