Question

Price of a particular item in 10 years in two cities are given below, which city has more stable prices?

City A	55	54	52	53	56	58	52	50	51	49
City B	108	107	105	105	106	107	104	103	104	101

Answer 1

City A			City B
XA	${\mathit{x}}_{\mathit{A}}\mathbf{=}{\mathit{X}}_{\mathit{A}}\mathbf{-}{\overline{\mathit{X}}}_{\mathit{A}}$	${\mathit{x}}_{\mathbf{A}}^{\mathbf{2}}$	XB	${\mathit{x}}_{\mathbf{B}}\mathbf{=}{\mathit{X}}_{\mathbf{B}}\mathbf{-}{\overline{\mathit{X}}}_{\mathit{B}}$	${\mathit{x}}_{\mathbf{B}}^{\mathbf{2}}$
55	2	4	108	3	9
54	1	1	107	2	4
52	−1	1	105	0	0
53	0	0	105	0	0
56	3	9	106	1	1
58	5	25	107	2	4
52	−1	1	104	−1	1
50	−3	9	103	−2	4
51	−2	4	104	−1	1
49	−4	16	101	−4	16
ΣXA=530		$\mathrm{\Sigma }{x}_{\mathrm{A}}^2=70$	ΣXB=1050		$\mathrm{\Sigma }{x}_{\mathrm{B}}^2=40$

$$\begin{gathered} {\overline{X}}_A=\frac{\Sigma X_A}{N}=\frac{530}{10}=53{ \\ }_A^{\sigma }=\sqrt{\frac{\Sigma x_A^2}{N}}=\sqrt{\frac{70}{10}}=\sqrt{7}=2.64 \\ \mathrm{C}.\mathrm{V}(\mathrm{City}\mathrm{A})=\frac{{\sigma }_A}{\overline{X_A}}\times 100=\frac{2.64}{53}\times 100=4.99\% \end{gathered}$$

$$\begin{gathered} {\overline{X}}_B=\frac{\Sigma X_B}{N}=\frac{1050}{10}=105 \\ {\sigma }_B=\sqrt{\frac{\Sigma x_B^2}{N}}=\sqrt{\frac{40}{10}}=2 \\ \mathrm{C}.\mathrm{V}(\mathrm{City}\mathrm{B})=\frac{{\sigma }_B}{\overline{X_B}}\times 100=\frac{2\times 100}{105}=\frac{200}{105}=1.90\% \end{gathered}$$

Since C.V. of city B is less. Therefore, prices are more stable in city B.