Question

Calculate Standard Deviation of the following two series . Which series has more variability:

A:	58	59	60	65	66	52	75	31	46	48
B:	56	87	89	46	93	65	44	54	78	68

Answer 1

For A

S. No.	(X)	Deviation $$\begin{gathered} \left(x=\mathrm{X}-\overline{\mathrm{X}}\right) \\ \overline{\mathrm{X}}=56 \end{gathered}$$	x2
1 2 3 4 5 6 7 8 9 10	58 59 60 65 66 52 75 31 46 48	2 3 4 9 10 −4 19 −25 −10 −8	4 9 16 81 100 16 361 625 100 64
	ΣX = 560		Σx2 = 1376

$$\begin{gathered} \overline{\mathit{X}}=\frac{\Sigma X}{N}=\frac{560}{10}=56 \\ SD\left(\sigma \right)=\sqrt{\frac{\mathrm{\Sigma }{x}^2}{N}}=\sqrt{\frac{1376}{10}}=\sqrt{137.6}=11.7 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{variation}=\frac{\sigma }{\overline{X}}\times 100 \\ =\frac{11.7}{56}\times 100 \\ =20.89\% \end{gathered}$$

For B

S. No.	Y	Deviation $$\begin{gathered} \left(y=Y-\overline{y}\right) \\ \left(\overline{y}=68\right) \end{gathered}$$	y2
1 2 3 4 5 6 7 8 9 10	56 87 89 46 93 65 44 54 78 68	−12 19 21 −22 25 −3 −24 −14 10 0	144 361 441 484 625 9 576 196 100 0
	ΣY = 680		Σy2 = 2936

$$\begin{gathered} \overline{Y}=\frac{\mathrm{\Sigma }Y}{N}=\frac{680}{10}=68 \\ SD\mathit{}\left(\sigma \right)=\sqrt{\frac{\mathrm{\Sigma }{y}^2}{N}}=\sqrt{\frac{2936}{10}}=\sqrt{293.6}=17.1 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{variation}=\frac{\mathrm{\sigma }}{\overline{\mathit{Y}}}\times 100 \\ =\frac{17.1}{68}\times 100=\frac{1710}{68} \\ =25.14\% \end{gathered}$$
Since the coefficient of variation of series B is more, it is said to have more variability.