Question

Two brands of Tyres are tested with the following results:

Life ( '000 miles)	:	20-25	25-30	30-35	35-40	45-50	50-55
Brand X		8	15	12	18	13	9
Brand Y		6	20	32	30	12	0

Which brand has a greater variation?

Answer 1

For Brand 'X'

Life	Mid value m	f	fm	Deviation from mean $x=m-\overline{\mathit{X}}$ $\overline{\mathit{X}}=35.17$	x2	fx2
20−25	22.5	8	180	−12.67	160.53	1284.24
25−30	27.5	15	412.5	−7.67	58.83	882.45
30−35	32.5	12	390	−2.67	7.13	85.56
35−40	37.5	18	675	2.33	5.43	97.74
40−45	42.5	13	552.5	7.33	53.73	698.49
45−50	47.5	9	427.5	12.33	152.03	1368.27
		Σf =75	Σfm =2637.5			Σfx2= 4416.75

$$\begin{gathered} \overline{\mathit{X}}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{2637.5}{75}=35.17 \\ S{D}_X=\sqrt{\frac{\mathrm{\Sigma }f{x}^2}{\mathrm{\Sigma }f}}=\sqrt{\frac{4416.75}{75}}=\sqrt{58.89}=7.67 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{variation}=\frac{S{D}_X}{\overline{\mathit{X}}}\times 100 \\ =\frac{7.67}{35.17}\times 100 \\ =21.81\% \end{gathered}$$

For Brand 'Y'

Life	Mid value m	f	fm	Deviation from mean $x=m\mathit{-}\overline{\mathit{X}}$ $\overline{\mathit{X}}=33.60$	x2	fx
20−25	22.5	6	135	−11.1	123.21	739.26
25−30	27.5	20	550	−6.1	37.21	744.2
30−35	32.5	32	1040	−1.1	1.21	38.72
35−40	37.5	30	1125	3.9	15.21	456.3
40−45	42.5	12	510	8.9	79.21	950.52
45−50	47.5	0	0	13.9	193.21	0
		Σf=100	Σfm=3360			$\mathrm{\Sigma }f{x}_{}^2=2929$

$$\begin{gathered} \overline{X}=\frac{\mathrm{\Sigma }fm}{\mathrm{\Sigma }f}=\frac{3360}{100}=33.60 \\ S{D}_Y=\sqrt{\frac{\mathrm{\Sigma }f{x}_{\mathrm{B}}^2}{\mathrm{\Sigma }f}}=\sqrt{\frac{2929}{100}}=\sqrt{29.29}=5.41 \\ \mathrm{Coefficient}\mathrm{of}\mathrm{variation}=\frac{S{D}_Y}{\overline{X}}\times 100 \\ =\frac{5.41}{33.6}\times 100 \\ =16.101 \end{gathered}$$

As the coefficient of variation is more for brand X, it is said to be more variable than brand Y.