For length,

Mean $ \bar{x}= \large\frac{\sum x_i}{n}$

$\qquad= \large\frac{212}{50}$

$\qquad= 4.24$

Standard deviation $\sigma= \sqrt {\large\frac{\sum x_i^2}{\sum n} - \bigg( \large\frac{\sum x_i}{n} \bigg)^2 }$

$\qquad= \sqrt { \large\frac{902.8}{50} -\bigg(\large\frac{212}{50}\bigg)^2}$

$\qquad= \sqrt { \large\frac{902.8 \times 50-(212)^2}{50 \times 50}}$

$\qquad= \large\frac{1}{50} $$ \sqrt { 45140-44944}$

$\qquad= \large\frac{14}{50} =\frac{7}{25} $$=0.28$

Step 3:

Coefficient of variation : $ =\large\frac{\sigma}{\bar {x}} $$\times 100$

$\qquad= \large\frac{0.28}{4.24} $$ \times 100$

$\qquad= \large\frac{28}{4.24}$

$\qquad= 6.604$

For weight,

Step 1:

Mean $ \bar{x}= \large\frac{\sum x_i}{n}$

$\qquad= \large\frac{261}{50}$

$\qquad= 5.22$

Step 2:

Standard deviation $\sigma= \sqrt {\large\frac{\sum x_i^2}{\sum n} - \bigg( \large\frac{\sum x_i}{n} \bigg)^2 }$

$\qquad= \sqrt{\large\frac{1457.6}{50} \bigg( \large\frac{261}{50}\bigg)^2 }$

$\qquad= \large\frac{1457.6 \times 50 -(261 \times 261)}{50 \times 50}$

$\qquad= \large\frac{1}{50} $$\sqrt {72880 -68121}$

$\qquad= \large\frac{\sqrt {4759}}{50}$$=1.37$

Step 3:

Coefficient of variation : $ =\large\frac{\sigma}{\bar {x}} $$\times 100$

$\qquad= \large\frac{1.37}{5.22} $$ \times 100$

$\qquad= \large\frac{137}{5.22}$

$\qquad= 26.24$

COMPARISON:

Variability depends upon coefficient of variation.

Higher the coefficient of variation, higher is the variability.

$\therefore $ coefficient of variation of weight is higher as compared to coefficient of variation of length.

$\therefore $ weight is more varying than length.