Question

The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

$\sum_{i=1}^{50}x_i=212,\sum_{i=1}^{50}x_i^2=902.8,\sum_{i=1}^{50}y_i=261,\sum_{i=1}^{50}y_{i=1}^2=1457.6$

Which is more varying, the length or weight?

Answer 1

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

Here, N = 50

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

$Varience(σ^2)=\frac{1}{N}\sum_{i=1}^{50}(x_i-\bar{x})^2$

$=\frac{1}{50}\sum_{i=1}^{50}({x_i}-4.24)^2$

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

$=\frac{1}{50}\sum_{i=1}^{50}[{x_i^2}-8.48\sum_{i=1}^{50}x_i+17.97×50]$

$\frac{1}{5}[902.8-8.48×(212)+898.5]$

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

$=\frac{1}{50}×3.54$

=0.07

∴ standard deviation, σ1 (Length)=√0.07=0.26

∴ C.V (Length)= $\frac{standard \,deviation}{Mean}×100=\frac{0.26}{4.24}×100=6.13$

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

$∴\,Mean,\bar{y}=\sum_{i=1}^{50}y_i=\frac{1}{50}×261=5.22$

$Varience(σ_2^2)=\frac{1}{N}\sum_{i=1}^{50}(y_i-\bar{y})^2$

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

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

$=\frac{1}{50}\sum_{i=1}^{50}[{y_i^2}-10.44\sum_{i=1}^{50}y_i+27.24×50]$

$\frac{1}{5}[1457.6-10.44×(261)+1362]$

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

$=\frac{1}{50}×94.76$

$1.89$

∴ standard deviation, σ2 (Weight)=√1.89=1.37

∴ C.V (Weight) = $\frac{standard \,deviation}{Mean}×100=\frac{1.37}{5.22}×100=26.24$

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