Answer :
Given:
Given the two sample data:
A: 4.39, 4.47, 4.21, 3.86, 3.72, 3.64, 3.77, 4.60, 3.82, 4.87, 3.94, 2.91, 4.88, 4.21
B: 4.8.3.9.2.8. 3.8. 4.7. 1.7.3.6, 2.2.4.8.3.2. 1.9
Required: Coefficient of variation of data A and B
Explanation:
First, find the CV of sample data A.
Find the mean.
[tex]\begin{gathered} \bar{x}=\frac{\sum x_i}{n} \\ =\frac{4.39+4.47+4.21+3.86+3.72+3.64+3.77+4.60+3.82+4.87+3.94+2.91+4.88+4.21}{14} \\ =\frac{57.29}{14} \\ =4.092143 \end{gathered}[/tex]Find the sum of squares.
[tex]\begin{gathered} \text{ Sum of squares}=\sum(x_i-\bar{x})^2 \\ =3.738236 \end{gathered}[/tex]Calculate the standard deviation.
[tex]\begin{gathered} s=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n-1}} \\ =\sqrt{\frac{3.738236}{14-1}} \\ =0.536243 \end{gathered}[/tex]Coefficient of variation of sample data A is
[tex]\begin{gathered} CV=\frac{s}{\bar{x}} \\ =\frac{0.536243}{4.092143} \\ =0.131 \\ =13.1\% \end{gathered}[/tex]Now, find the coefficient of variation of sample data B.
Find the mean.
[tex]\begin{gathered} \bar{x}=\frac{\sum x_i}{n} \\ =\frac{4.8+3.9+2.8+3.8+4.7+1.7+3.6+2.2+4.8+3.2+1.9}{11} \\ =\frac{37.4}{11} \\ =3.4 \end{gathered}[/tex]Find the sum of squares.
[tex]\begin{gathered} \text{ Sum of squares =}\sum(x_i-\bar{x})^2 \\ =13.04 \end{gathered}[/tex]Calculate the standard deviation.
[tex]\begin{gathered} s=\sqrt{\frac{\sum(x_i-\bar{x})^2}{n-1}} \\ =\sqrt{\frac{13.04}{11-1}} \\ =1.141928 \end{gathered}[/tex]Calculate the coefficient of variation.
[tex]\begin{gathered} CV=\frac{s}{\bar{x}} \\ =\frac{1.141928}{3.4} \\ =0.336 \\ =33.6\% \end{gathered}[/tex]Final Answer: CV of Data Set A = 13.1%
CV of Data Set B = 33.6%