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

First we have to calculate Mean for Length x,

\(Mean=\overline{X}=\dfrac{\displaystyle\sum x_i}{n}\)

\( Variance= \dfrac{1}{N^2}[N\displaystyle\sum{f_i}x_i^2-(\displaystyle\sum{f_i}x_i)^2] \)

= \( (\dfrac{1}{50^2})[(50\times 902.8)-212^2]\)

= \( \dfrac{1}{2500}(45140-44944)\)

= \( \dfrac{196}{2500}\)

= 0.0784

Standard deviation σ = \( \sqrt{Variance}\)

= \( \sqrt{0.0784}\)  = 0.28

\(\) \( C.V._X=\dfrac{σ}{X}\times 100\)

= \( \dfrac{0.28}{4.24}\times 100\)

= 6.603

Now we have to calculate  mean of weight y

= \( \overline{Y}=\dfrac{\displaystyle\sum y_i}{n}\)

= \(\dfrac{261}{50}=5.22\)

\(Variance= \dfrac{1}{N^2}[N\displaystyle\sum{f_i}y_i^2-(\displaystyle\sum{f_i}y_i)^2] \)

= \( (\dfrac{1}{50^2})[(50\times 1457.6)-261^2]\)

= \( \dfrac{1}{2500}(72880-68121)\)

= \( \dfrac{4759}{2500}\)

= 1.9036

Standard deviation σ = \( \sqrt{Variance}\)

= \( \sqrt{1.9036}\)  = 1.37

So, co-efficient of variation of team B,

\( C.V._Y=\dfrac{σ}{X}\times 100\)

= \(\dfrac{1.37}{5.22}\times 100\)

= 26.24

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

Therefore, weights vary more than the lengths