Question

From the data given below state which group is more variable, A or B?

X	35	54	52	53	56	58	52	50	51	49
Y	108	107	105	105	106	107	104	103	104	101

Answer 1

The prices of the shares X are 

35, 54, 52, 53, 56, 58, 52, 50, 51, 49 

Here, the number of observations, N = 10

∴ \(Mean,\bar{x}=\frac{1}{N}\sum_{i=1}^{10}x_i=\frac{1}{10}×510=51\)

The following table is obtained corresponding to shares X

\(x_i\)	\(x_i,-\bar{x}\)	\((x_i,-\bar{x})^2\)
35	-16	256
35	3	9
53	1	1
56	2	4
58	5	25
52	7	49
50	1	1
51	-1	1
49	0	0
-	2	4
-	-	350

Variance(σ²) = \(\frac{1}{N}\sum_{i=1}^{10}(xi-\bar{x})^2=\frac{1}{10}×350=35\)

\(∴\,Standard\,deviation\.(σ_1) =√35=5.91\)

C.V (Shares X)= \(\frac{σ_1}{X}×100=\frac{5.91}{51}×100=11.58\)

The prices of share Y are 

108, 107, 105, 105, 106, 107, 104, 103, 104, 101

∴ Mean, \(\bar{y}=\frac{1}{N}\sum_{i=1}^{10}yi=\frac{1}{10}×1050=105\)

The following table is obtained corresponding to shares Y

\(y_i\)	\(y_i,-\bar{y}\)	\((y_i,-\bar{y})\)
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
-	-	40

Variance(σ²) = \(\frac{1}{N}\sum_{i=1}^{10}(y_i-\bar{y})^2=\frac{1}{10}×40=4\)

\(Standard\,deviation\.(σ_2) =√4=2\)

∴ C.V (Shares )= \(\frac{σ_2}{X}×100=\frac{2}{105}×100=1.9=11.58\)

∴ C.V. of prices of shares X is greater than the C.V. of prices of shares Y

Thus, the prices of shares Y are more stable than the prices of shares X.