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_(i=1)^(50)x_i=212 ,sum_(i=1)^(50)x_i^2=902. 8 ,sum_(i=1)^(50)y_i=261 ,sum_(i=1)^50 y_ i^2=1457. 6` Which is more varying, the length or weight?

Answer

weight will vary more than the lengths

Explanation

\begin{aligned}
&\sum_{i=1}^{50} \mathrm{X}_{\mathrm{i}}=212, \sum_{\mathrm{i}=1}^{50} \mathrm{X}_{\mathrm{i}}^{2}=902.8\\
&\text { Here, } N=50\\
&\operatorname{Mean} \bar{X}_{i}=\frac{\sum_{i=1}^{50}}{N}=\frac{212}{50}=4.24\\
&\text { Variance }\left(\sigma_{1}^{2}\right)=\frac{1}{N} \sum_{\mathrm{i}=1}^{50}\left(\mathrm{X}_{\mathrm{i}}-\overline{\mathrm{X}}\right)^{2}\\
&=\frac{1}{50} \sum_{i=1}^{50}\left(\mathrm{X}_{\mathrm{i}}-4.24\right)^{2}\\
&=\frac{1}{50} \sum_{\mathrm{i}=1}^{50}\left[\mathrm{X}_{\mathrm{i}}^{2}-8.48 \mathrm{X}_{\mathrm{i}}+17.97\right]\\
&=\frac{1}{50}\left[\sum_{\mathrm{i}=1}^{50} \mathrm{X}_{\mathrm{i}}^{2}-8.48 \sum_{\mathrm{i}=1}^{50} \mathrm{X}_{\mathrm{i}}+17.97 \times 50\right]\\
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