The sum and sum of squares corresponding to length x (in cm) and weight y

Solution:

$\sum_{i=1}^{50} x_{i}=212, \sum_{i=1}^{50} x_{i}^{2}=902.8$

Here, $N=50$

$\therefore$ Mean, $\bar{x}=\frac{\sum_{i=1}^{50} y_{i}}{N}=\frac{212}{50}=4.24$

Variance $\left(\sigma_{1}^{2}\right)=\frac{1}{N} \sum_{i=1}^{30}\left(x_{i}-\bar{x}\right)^{2}$

$=\frac{1}{50} \sum_{i=1}^{50}\left(\mathrm{x}_{i}-4.24\right)^{2}$

$=\frac{1}{50} \sum_{i=1}^{50}\left[x_{i}{ }^{2}-8.48 x_{i}+17.97\right]$

$=\frac{1}{50}\left[\sum_{i=1}^{40} x_{i}{ }^{2}-8.48 \sum_{i=1}^{50} x_{i}+17.97 \times 50\right]$

$=\frac{1}{50}[902.8-8.48 \times(212)+898.5]$

$=\frac{1}{50}[1801.3-1797.76]$

$=\frac{1}{50} \times 3.54$

$=0.07$

$\therefore$ Stan dard deviation, $\sigma_{1}$ (Length) $=\sqrt{0.07}=0.26$

$\therefore$ C.V. $($ Length $)=\frac{\text { Stan dard deviation }}{\text { Mean }} \times 100=\frac{0.26}{4.24} \times 100=6.13$

$\sum_{i=1}^{50} y_{i}=261, \sum_{i=1}^{5} y_{i}^{2}=1457.6$

Mean, $\overline{\mathrm{y}}=\frac{1}{\mathrm{~N}} \sum_{\mathrm{i}=1}^{50} \mathrm{y}_{\mathrm{i}}=\frac{1}{50} \times 261=5.22$

Variance $\left(\sigma_{2}^{2}\right)=\frac{1}{N} \sum_{i=1}^{30}\left(y_{i}-\bar{y}\right)^{2}$

$=\frac{1}{50} \sum_{i=1}^{50}\left(y_{i}-5.22\right)^{2}$

$=\frac{1}{50} \sum_{i=1}^{50}\left[y_{i}^{2}-10.44 y_{i}+27.24\right]$

$=\frac{1}{50}\left[\sum_{i=1}^{50} y_{i}{ }^{2}-10.44 \sum_{i=1}^{50} y_{i}+27.24 \times 50\right]$

$=\frac{1}{50}[1457.6-10.44 \times(261)+1362]$

$=\frac{1}{50}[2819.6-2724.84]$

$=\frac{1}{50} \times 94.76$

$=1.89$

$\therefore$ Stan dard deviation, $\sigma_{2}$ (Weight) $=\sqrt{1.89}=1.37$

$\therefore$ C.V. $($ Weight $)=\frac{\text { Stan dard deviation }}{\text { Mean }} \times 100=\frac{1.37}{5.22} \times 100=26.24$

Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.