NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.15.3 Q.2
From the prices of shares X and Y below, find out which is more stable in value:
|
X |
35 |
54 |
52 |
53 |
56 |
58 |
52 |
50 |
51 |
49 |
|
Y |
108 |
107 |
105 |
105 |
106 |
107 |
104 |
103 |
104 |
101 |
The prices of the shares X are:
35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Here, the number of observations, N = 10
Mean = Σi=110 xi ÷ N
= 510 ÷ 10
= 51
The following table is obtained corresponding to shares X.
Now, varience σ12 = (1 ÷ N) × Σi=110 (xi – )2 = 350 ÷ 10 = 35
So, Standard deviation (σ1) = √35 = 5.91
CV (Shares X) = (σ1 ÷ ) × 100 = (5.91 ÷ 51) × 100 = 11.58
The prices of share Y are:
108, 107, 105, 105, 106, 107, 104, 103, 104, 101
Mean = Σi=110 yi ÷ N
= 1050 ÷ 10
= 105
The following table is obtained corresponding to shares Y.
Now, varience σ22 = (1 ÷ N) × Σi=110 (yi – )2 = 40 ÷ 10 = 4
So, Standard deviation (σ2) = √4 = 2
CV (Shares X) = (σ2 ÷ ) × 100 = (2 ÷ 105) × 100 = 1.9 < 11.58
C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Thus, the prices of shares Y are more stable than the prices of shares X.