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

 

Answers (1)

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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