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From the Prices of Shares X And Y Given Below: Find Out Which is More Stable in Value: X:35545253565852505149y:108107105105106107104103104101 - Mathematics

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 Ax = 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 \[n = 10, \bar{X} = 51\]
\[ \sigma^2 = \frac{\sum {d_i}^2}{n} - \left( \frac{\sum d_i}{n} \right)^2 \]
\[ = \frac{350}{10} - \left( \frac{0}{10} \right)^2 \]
\[ = 35 - 0\]
\[ = 35\]

\[\sigma = \sqrt{35} = 5 . 91\]

\[{CV}_x = \frac{5 . 91}{51} \times 100\]
\[ = 11 . 58\]
Let Ay =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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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 32 Statistics
Exercise 32.7 | Q 10 | Page 48
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