# 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

 X($x_i$) Y($y_i$) $x_i^2$ $y_i^2$ 35 108 1225 11664 54 107 2916 11449 52 105 2704 11025 53 105 2809 11025 56 106 8136 11236 58 107 3364 11449 52 104 2704 10816 50 103 2500 10609 51 104 2601 10816 49 101 2401 10201 =510 = 1050 =26360 =110290

For X,

Mean , $\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{510}{10} = 51$

Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum x_i^2 - (\sum x_i)^2 \right ]$

$\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(26360) - (510)^2 \right ] \\ = \frac{1}{100}.\left [263600 - 260100 \right ] \\ \\ = 35$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{35} = 5.91$

C.V.(X) = $\frac{\sigma}{\overline x}\times100 = \frac{5.91}{51}\times100 = 11.58$

Similarly, For Y,

Mean , $\overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i = \frac{1050}{10} = 105$

Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum y_i^2 - (\sum y_i)^2 \right ]$

$\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(110290) - (1050)^2 \right ] \\ = \frac{1}{100}.\left [1102900 - 1102500 \right ] \\ \\ = 4$

We know,  Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{4} = 2$

C.V.(Y) = $\frac{\sigma}{\overline y}\times100 = \frac{2}{105}\times100 = 1.904$

Since C.V.(Y) < C.V.(X)

Hence Y is more stable.

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