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  • 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?

5. 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?

Answers (1)

best_answer

For length x,

Mean, \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i =\frac{212}{50}= 4.24

We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(902.8) - (212)^2 \right ] \\ \\ = \frac{196}{2500} =0.0784

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{0.0784} = 0.28

C.V.(x) = \frac{\sigma}{\overline x}\times100 = \frac{0.28}{4.24}\times100 = 6.603

For weight y,

Mean,

Mean, \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i =\frac{261}{50}= 5.22

We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]

\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(1457.6) - (261)^2 \right ] \\ \\ = \frac{4759}{2500} =1.9036

We know,  Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{1.9036} = 1.37

C.V.(y) = \frac{\sigma}{\overline y}\times100 = \frac{1.37}{5.22}\times100 = 26.24

Since C.V.(y) > C.V.(x)

Therefore, weight is more varying.

Posted by

HARSH KANKARIA

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