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

Variance,

(σ₁²) = 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 σ₂ (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,

(σ₂²) = 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 σ₂ (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

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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