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# The sum and sum of squares corresponding to length X (in cms) and weight y (in gms) of 50 plant products are given below.

∑^{50}_{i = 1}(x_{i}) = 212, ∑^{50}_{i = 1}(x_{i})² = 902.8, ∑^{50}_{i = 1}(y_{i}) = 261, ∑^{50}_{i = 1}(y_{i})² = 1457.6

Which is more varying, the length or the weight?

**Solution:**

∑^{50}_{i = 1}(x_{i}) = 212, ∑^{50}_{i = 1}(x_{i})² = 902.8

Here, N = 50

Mean,

x = 1/N ∑^{50}_{i = 1}x_{i}

= 212/50

= 4.24

(σ_{1}²) = 1/N ∑^{50}_{i = 1}(x_{i} - x)²]

= 1/50 ∑^{50}_{i = 1}(x_{i} - 4.24)²]

= 1/50 ∑^{50}_{i = 1}(x_{i})² - 8.48x + 17.97]

= 1/50 ∑^{50}_{i = 1}(x_{i})² - 8.48 ∑^{50}_{i = 1}x_{i} + 17.97 x 50]

= 1/50 [902.8 - 8.48 × (212) + 898.5]

= 1/50 [1801.3 - 1797.76]

= 1/50 × 3.54

= 0.07

Standard variation σ_{2} (length)

= √0.07 = 0.26

C.V(length)

= standard deviation/mean x 100

= 0.26/4.24 × 100

= 6.13

∑^{50}_{i = 1}(y_{i}) = 261, ∑^{50}_{i = 1}(y_{i})² = 1457.6

Here, N = 50

Mean,

y = 1/N ∑^{50}_{i = 1}y_{i}

= 261/50

= 5.22

Variance,

(σ_{2}²) = 1/N ∑^{50}_{i = 1}(y_{i} - y)²]

= 1/50 ∑^{50}_{i = 1}(y_{i} - 5.22)²]

= 1/50 ∑^{50}_{i = 1}(y_{i})² - 10.44y_{i} + 27.24]

= 1/50 ∑^{50}_{i = 1}(y_{i})² - 10.44 ∑^{50}_{i = 1}y_{i} + 27.24 × 50]

= 1/50 [1457.6 - 10.44 × (261) + 1362]

= 1/50 [2819.6 - 2724.84]

= 1/50 × 94.76

= 1.89

Standard variation σ_{2} (length)

= √1.89

= 1.37

C.V(length)

= standard deviation/mean × 100

= 1.37/5.22 × 100

= 26.24

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

Therefore, weights vary more than the lengths

NCERT Solutions Class 11 Maths Chapter 15 Exercise 15.3 Question 5

## The sum and sum of squares corresponding to length X (in cms) and weight y (in gms) of 50 plant products are given below.

∑^{50}_{i = 1}(x_{i}) = 212, ∑^{50}_{i = 1}(x_{i})² = 902.8, ∑^{50}_{i = 1}(y_{i}) = 261, ∑^{50}_{i = 1}(y_{i})² = 1457.6

Which is more varying, the length or the weight?

**Summary:**

Thus, we have observed from the above explanation that weights vary more than the lengths

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