The sum and sum of squares corresponding to length $x$ (in $cm$) and weight $y$ (in $gm$) of $50$ plant products are given below:
$\sum\limits_{i = 1}^{50} {{x_i} = 212, \sum\limits_{i = 1}^{50} {x_i^2} = 902.8, \sum\limits_{i = 1}^{50} {{y_i} = 261, \sum\limits_{i = 1}^{50} {y_i^2 = 1457.6} } }$
Which is more varying,the length or weight?

(B) For length $x$:
Mean $\bar{x} = \frac{212}{50} = 4.24$
Variance $\sigma_x^2 = \frac{1}{N} \sum x_i^2 - (\bar{x})^2 = \frac{902.8}{50} - (4.24)^2 = 18.056 - 17.9776 = 0.0784$
Standard deviation $\sigma_x = \sqrt{0.0784} = 0.28$
Coefficient of Variation $C.V._x = \frac{\sigma_x}{\bar{x}} \times 100 = \frac{0.28}{4.24} \times 100 \approx 6.60$
For weight $y$:
Mean $\bar{y} = \frac{261}{50} = 5.22$
Variance $\sigma_y^2 = \frac{1}{N} \sum y_i^2 - (\bar{y})^2 = \frac{1457.6}{50} - (5.22)^2 = 29.152 - 27.2484 = 1.9036$
Standard deviation $\sigma_y = \sqrt{1.9036} \approx 1.38$
Coefficient of Variation $C.V._y = \frac{\sigma_y}{\bar{y}} \times 100 = \frac{1.38}{5.22} \times 100 \approx 26.44$
Since $C.V._y > C.V._x$,the weight varies more than the length.