From the prices of shares of 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
Solution:
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,
x = 1/N ∑10i = 1xi
= 1/10 x 510
= 51
The following table is obtained corresponding to shares X.
| xi | (xi - x) | (xi - x)2 |
| 35 | -16 | 256 |
| 54 | 3 | 9 |
| 52 | 1 | 1 |
| 53 | 2 | 4 |
| 56 | 5 | 25 |
| 58 | 7 | 49 |
| 52 | 1 | 1 |
| 50 | - 1 | 1 |
| 51 | 0 | 0 |
| 49 | - 2 | 4 |
| 350 |
(σ1²) = 1/N [N ∑10i = 1(xi - x)
= 1/10 × 350
= 35
(σ1) = √35
= 5.91
C.V.(Shares X) = σ1/X × 100
= 5.91/51 × 100
= 11.58
The prices of the shares Y are
108, 107, 105, 105, 106, 107, 104, 103, 104, 101
Here, the number of observations, N = 10
Mean
y = 1/N ∑10i = 1yi
= 1/10 x 1050
= 105
The following table is obtained corresponding to shares Y.
| yi | (yi - y) | (yi - y)² |
| 108 | 3 | 9 |
| 107 | 2 | 4 |
| 105 | 0 | 0 |
| 105 | 0 | 0 |
| 106 | 1 | 1 |
| 107 | 2 | 4 |
| 104 | - 1 | 1 |
| 103 | - 2 | 4 |
| 104 | - 1 | 1 |
| 101 | - 4 | 16 |
| 40 |
Variance,
(σ2²) = 1/N [N ∑10i = 1(yi - y)
= 1/10 × 40
= 4
Standard deviation,
(σ2) = √4
= 2
C.V.(Shares X) = σ2/X × 100
= 2/105 × 100
= 1.9
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
NCERT Solutions Class 11 Maths Chapter 15 Exercise 15.3 Question 2
From the prices of shares of X and Y below, find out which is more stable in value:
Summary:
From the given data, the prices of shares Y are more stable than the prices of shares X
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