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

Which is more varying, the length or weight?

Asked by Abhisek | 11 months ago |  86

Solution :-

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

Answered by Abhisek | 11 months ago

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