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.