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NCERT Class XI Mathematics - Statistics - Solutions

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Question : 27 of 34
Marks: +1, -0
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below :
∑i=150xi\sum\limits_{i=1}^{50} x_i = 212 , ∑i=150xi2\sum\limits_{i=1}^{50} x_i^2 = 902.8 ; ∑i=150yi\sum\limits_{i=1}^{50} y_i = 261 , ∑i=150yj2\sum\limits_{i=1}^{50} y_j^2 = 1457.6
Which is more varying, the length or weight?
Solution:  
We have , ∑i=150xi\sum\limits_{i=1}^{50} x_i = 212 , ∑i=150xi2\sum\limits_{i=1}^{50} x_i^2 = 902.8
Now, xˉ\bar{x} = 21250\frac{212}{50} = 4.24
σx2\sigma_x^2 = 150\frac{1}{50} × 902.8 - (21250)2\left(\frac{212}{50}\right)^2
= 18.056 – 17.978 = 0.078
∴ σx\sigma_x = 0.078\sqrt{0.078} = 0.28
Also, ∑i=150yi\sum\limits_{i=1}^{50} y_i = 261 , ∑i=150yj2\sum\limits_{i=1}^{50} y_j^2 = 1457.6
yˉ\bar{y} = 26150\frac{261}{50} = 5.22
σy2\sigma_y^2 = 150\frac{1}{50} × 1457.6 - (26150)2\left(\frac{261}{50}\right)^2 = 29.152 - 27.248 = 1.904
∴ σy\sigma_y = 1.904\sqrt{1.904} = 1.38
C.V. of length (x) = 0.284.24\frac{0.28}{4.24} × 100 = 6.6
C.V. of weight (y) = 1.385.22\frac{1.38}{5.22} × 100 = 26.4
Since C.V. of weight > C.V. of length
Thus weight have more variability than length.
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