Consider the following for the two (02) items that follow: The sum...

Mean of $X$, $\bar{X} = \dfrac{\Sigma X}{n} = \dfrac{200}{50} = 4$

Mean of $Y$, $\bar{Y} = \dfrac{\Sigma Y}{n} = \dfrac{250}{50} = 5$

Variance of $X$, $\sigma_X^2 = \dfrac{\Sigma X^2}{n} - (\bar{X})^2 = \dfrac{900}{50} - 4^2 = 18 - 16 = 2$

Standard deviation of $X$, $\sigma_X = \sqrt{2}$

Variance of $Y$, $\sigma_Y^2 = \dfrac{\Sigma Y^2}{n} - (\bar{Y})^2 = \dfrac{1400}{50} - 5^2 = 28 - 25 = 3$

Standard deviation of $Y$, $\sigma_Y = \sqrt{3}$

Coefficient of variation of $X$, $CV_X = \dfrac{\sigma_X}{\bar{X}} \times 100 = \dfrac{\sqrt{2}}{4} \times 100 = 25\sqrt{2} \approx 35.35\%$

Coefficient of variation of $Y$, $CV_Y = \dfrac{\sigma_Y}{\bar{Y}} \times 100 = \dfrac{\sqrt{3}}{5} \times 100 = 20\sqrt{3} \approx 34.64\%$

Since $35.35\% > 34.64\%$, the coefficient of variation of $X$ is strictly more than the coefficient of variation of $Y$.

Answer: Coefficient of variation of $X$ is strictly more than coefficient of variation of $Y$.