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.