Chapter 3.2, Problem 22E

To determine

To find:

The coefficient of variation, CV, for each data set and the larger spread relative to its own mean between the given data sets.

Answer to Problem 22E

Solution:

The coefficient of variation, CV, for data set A and B are 20.66% and 35.28% respectively. The data set corresponding to B has the larger spread relative to its own mean.

Explanation of Solution

Given information:

Given set of data set are,

$\begin{array}{l}\text{A}:31,33,35,39,31,32,30,40,13,41,38,32,37,33\\ \text{B}:2.3,3.5,4.3,2.1,4.8,3.9,2.0,3.3,4.0,1.6,2.2\end{array}$

Formula used:

The formula to calculate the sample standard deviation for given set of data is,

$s=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$

Where, $x_i$ is the $i^{\text{th}}$ data value, $\overline{x}$ is the sample mean, and $N$ is the total number of collection.

The formula to calculate the coefficient of variance of a sample is,

$C.V=\frac{s}{\overline{x}}\times 100\%$

$\overline{x}$ is the sample mean and $s$ is the sample standard deviation.

Calculation:

For data set A, $n_A=14$, Compute ${\overline{x}}_A$ for the given set of data,

$\begin{array}{rll}\overline{x_A}& =& \frac{31+33+35+39+31+32+30+40+13+41+38+32+37+33}{14}\\ & =& \frac{465}{14}\\ & =& 33.2143\end{array}$

Construct the table for sample standard deviation of the given set of data.

$x_{iA}$	$x_{iA}-\overline{x_A}$	${\left(x_{iA}-\overline{x_A}\right)}^2$
31	$31-33.2143=-2.2143$	4.90312449
33	$33-33.2143=-0.2143$	0.04592449
35	$35-33.2143=1.7857$	3.18872449
39	$39-33.2143=5.7857$	33.47432449
31	$31-33.2143=-2.2143$	4.90312449
32	$32-33.2143=-1.2143$	1.47452449
30	$30-33.2143=-3.2143$	10.33172449
40	$40-33.2143=6.7857$	46.04572449
13	$13-33.2143=-20.2143$	408.61792449
41	$41-33.2143=7.7857$	60.61712449
38	$38-33.2143=4.7857$	22.90292449
32	$32-33.2143=-1.2143$	1.47452449
37	$37-33.2143=3.7857$	14.33152449
33	$33-33.2143=-0.2143$	0.04592449
		${\displaystyle \sum {\left(x_{iA}-\overline{x_A}\right)}^2}=612.3571$

Substitute $612.3571$ for ${\left(x_{iA}-\overline{x_A}\right)}^2$ and 14 for $n_A$ in the formula of the standard deviation.

$\begin{array}{rll}{s}_A& =& \sqrt{\frac{{\displaystyle \sum {\left(x_{i_A}-\overline{x_A}\right)}^2}}{n_A-1}}\\ & =& \sqrt{\frac{612.3571}{13}}\\ & =& \sqrt{47.1044}\\ & \approx & 6.8633\end{array}$

The sample standard deviation for the data set A is 6.8633.

Substitute 6.8633 for $S_A$, and 33.2143 for $\overline{x_A}$ in the formula of coefficient of variance.

$\begin{array}{rll}C.V_A& =& \frac{s_A}{\overline{x_A}}\times 100\%\\ & =& \frac{6.8633}{33.2143}\times 100\%\\ & =& 20.66\%\end{array}$

The coefficient of variance for data set A is 20.66%.

For data set B, $n_B=11$, Compute ${\overline{x}}_B$ for the given set of data,

$\begin{array}{rll}\overline{x_B}& =& \frac{2.3+3.5+4.3+2.1+4.8+3.9+2+3.3+4+1.6+2.2}{11}\\ & =& \frac{34}{11}\\ & =& 3.0909\end{array}$

Construct the table for sample standard deviation of the given set of data.

$x_{iB}$	$x_{iB}-\overline{x_B}$	${\left(x_{iB}-\overline{x_B}\right)}^2$
2.3	$2.3-3.0909=-0.7909$	0.62552281
3.3	$3.3-3.0909=0.2091$	0.04372281
4.3	$4.3-3.0909=1.2091$	1.46192281
2.1	$2.1-3.0909=-0.9909$	0.98188281
4.8	$4.8-3.0909=1.7091$	2.92102281
3.9	$3.9-3.0909=0.8091$	0.65464281
2.0	$2.0-3.0909=-1.0909$	1.19006281
3.3	$3.3-3.0909=0.2091$	0.04372281
4.0	$4.0-3.0909=0.9091$	0.82646281
1.6	$1.6-3.0909=-1.4909$	2.22278281
2.2	$2.2-3.0909=-0.8909$	0.79370281
		${\displaystyle \sum {\left(x_{iB}-\overline{x_B}\right)}^2}=11.8891$

Substitute $11.8891$ for ${\left(x_{iB}-\overline{x_B}\right)}^2$ and 11 for $n_B$ in the formula of the standard deviation.

$\begin{array}{rll}{s}_B& =& \sqrt{\frac{{\displaystyle \sum {\left(x_{iB}-\overline{x_B}\right)}^2}}{n_B-1}}\\ & =& \sqrt{\frac{11.8891}{10}}\\ & =& \sqrt{1.18891}\\ & \approx & 1.0904\end{array}$

The sample standard deviation for the data set B is 1.0904.

Substitute 1.0904 for $S_B$, and 3.0909 for $\overline{x_B}$ in the formula of coefficient of variance.

$\begin{array}{rll}C.V_B& =& \frac{s_B}{\overline{x_B}}\times 100\%\\ & =& \frac{1.0904}{3.0909}\times 100\%\\ & \approx & 35.28\%\end{array}$

The coefficient of variance for data set B is 35.28%.

Coefficient of variance for sample B is greater than coefficient of variance for sample A.

$C.V_B>C.V_A$

The data set B has larger spread relative to its own mean.