For length
$\mathrm{N}=50, \sum_{i=1}^{50} x_i=212, \sum_{i=1}^{50} x_i^2=902.8$
$\bar{x}=\frac{\sum x_i}{N}=\frac{212}{50}=4.24$
Standard deviation $=\sigma$
$=\frac{1}{N}\left[N \sum x_i^2-\left(\sum x_i\right)^2\right]$
$=\frac{1}{50} \sqrt{50 \times 902.8-(212)^2}$
$=\frac{\sqrt{196}}{50}=\frac{14}{50}=0.28$
Coefficients of variation $=\frac{\sigma}{\bar{x}} \times 100$
$=\frac{0.28}{4.24} \times 100=6.6$
For weight, $\mathrm{N}=50, \sum_{i=1}^{50} y_i=261, \sum_{i=1}^{50} y_i^2=1457.6$
Mean $\bar{x}=\frac{\sum y_i}{N}=\frac{261}{50}=5.22 \mathrm{gm}$.
Standard deviation, $\sigma$
$\begin{aligned}
&=\frac{1}{N} \sqrt{N \sum y_i^2-\left(\sum y_i\right)^2} \\
&=\frac{1}{50} \sqrt{50 \times 1457.6-(261)^2}=\frac{\sqrt{4759}}{50}=1.38 \\
&\therefore \text { Coefficients of variation }(\text { C.V })=\frac{\sigma}{\overline{\mathrm{x}}} \times 100 \\
&=\frac{1.38}{5.22} \times 100=26.4
\end{aligned}$
Coefficients of variation of weights is more than that of length
$\therefore$ Weight is more varying than length.