NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.15.3 Q.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:
Σi=150 xi = 212, Σi=150 xi2 = 902.8, Σi=150 yi = 261, Σi=150 yi2 = 1457.6
Which is more varying, the length or weight?
Given, Σi=150 xi = 212, Σi=150 xi2 = 902.8
Here, N = 50
So, Mean, x = Σi=150 xi ÷ N = 212 ÷ 50 = 4.24
Variance σ12 = (1 ÷ N) × Σi=150 (xi – x)2
= (1 ÷ 50) × Σi=150 (xi – 4.24)2
= (1 ÷ 50) × Σi=150 [xi 2 – 8.48 xi + 17.97]
= (1 ÷ 50) × [Σi=150 xi 2 – 8.48 Σi=150 xi + 17.97 × 50]
= (1 ÷ 50) × [902.8 – 8.48 × 212 + 898.5]
= (1 ÷ 50) × [1801.3 – 1797.76]
= 3.54 ÷ 50
= 0.07
So, standard deviation (σ1) = √0.07 = 0.26
CV (Length) = (standard deviation ÷ Mean) × 100
= (0.26 ÷ 4.24) × 100
= 6.13
Again, given Σi=150 yi = 261, Σi=150 yi2 = 1457.6
Here, N = 50
So, Mean, y = Σi=150 yi ÷ N = 261 ÷ 50 = 5.22
Variance σ22 = (1 ÷ N) × Σi=150 (yi – y)2
= (1 ÷ 50) × Σi=150 (yi – 5.22)2
= (1 ÷ 50) × Σi=150 [yi 2 – 10.44 yi + 27.24]
= (1 ÷ 50) × [Σi=150 yi 2 – 10.44 Σi=150 yi + 27.24 × 50]
= (1 ÷ 50) × [1457.6 – 10.44 × 261 + 1362]
= (1 ÷ 50) × [2819.6 – 2724.84]
= 94.76 ÷ 50
= 1.89
So, standard deviation (σ2) = √1.89 = 1.37
CV (Length) = (standard deviation ÷ Mean) × 100
= (1.37 ÷ 5.22) × 100
= 26.24
Thus, C.V. of weights is greater than the C.V. of lengths.
Therefore, weights vary more than the lengths.