From the prices of shares X and Y given 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 
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Solution
Let A_{x} = 51
\[x_i\]

\[d_i = x_i  51\]

\[{d_i}^2\]

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 
\[\sum d_i = 0\]

\[\sum d_i^2 = 350\]

Here, we have \[n = 10, \bar{X} = 51\]
\[ \sigma^2 = \frac{\sum {d_i}^2}{n}  \left( \frac{\sum d_i}{n} \right)^2 \]
\[ = \frac{350}{10}  \left( \frac{0}{10} \right)^2 \]
\[ = 35  0\]
\[ = 35\]
\[\sigma = \sqrt{35} = 5 . 91\]
\[{CV}_x = \frac{5 . 91}{51} \times 100\]
\[ = 11 . 58\]
\[ = 11 . 58\]
Let A_{y} =105
\[x_i\]

\[d_i = x_i  105\]

\[{d_i}^2\]

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 
\[\sum d_i = 0\]

\[\sum d_i^2 = 40\]

\[n = 10, \bar{Y} = 105\]
\[ \sigma^2 = \frac{\sum {d_i}^2}{n}  \left( \frac{\sum d_i}{n} \right)^2 \]
\[ = \frac{40}{10}  \left( \frac{0}{10} \right)^2 \]
\[ = 4  0\]
\[ = 4\]
\[\sigma = \sqrt{4} = 2\]
\[{CV}_y = \frac{2}{105} \times 100\]
\[ = 1 . 90\]
Since CV of prices of share Y is lesser than that of X, prices of shares Y are more stable.
Concept: Statistics  Statistics Concept
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