Question

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

Answer 1

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

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

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

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.