Question

The following data shows the expected life of two models of T.V.: A and B:

Life (no. of years)	0−2	2−4	4−6	6−8	8−10	10−12
Model A	5	16	13	7	5	4
Model B	2	7	12	19	9	1

Which model has greater uniformity.

Answer 1

For A

Life (No. of years)	Frequency (f)	Mid Values (m)	fm	m2	fm2
0 − 2 2 − 4 4 − 6 6 − 8 8 −10 10 − 12	5 16 13 7 5 4	1 3 5 7 9 11	5 48 65 49 45 44	1 9 25 49 81 121	5 144 325 343 405 484
	Σf = 50		Σfm = 256		Σfm2 = 1706

$$\begin{gathered} {\overline{X}}_A=\frac{\Sigma fm}{\Sigma f}=\frac{256}{50}=5.12 \\ \mathrm{S}.\mathrm{D}.\left(\sigma \right)=\sqrt{\frac{\Sigma f{m}^2}{\Sigma f}-{\left({\overline{X}}_A\right)}^2} \\ =\sqrt{\frac{1706}{50}-{\left(5.12\right)}^2} \\ =\sqrt{34.12-26.21} \\ =\sqrt{7.91}=2.81 \\ \mathrm{C}.\mathrm{V}.\mathrm{of}\mathrm{T}.\mathrm{V}.=\frac{{\sigma }_A}{{\overline{X}}_A}\times 100 \\ =\frac{2.812}{5.12}\times 100 \\ =54.91\% \end{gathered}$$

For B

Life (No. of years)	Frequency (f)	Mid Values (m)	fm	m2	fm2
0 − 2 2 − 4 4 − 6 6 − 8 8 −10 10 − 12	2 7 12 19 9 1	1 3 5 7 9 11	2 21 60 133 81 11	1 9 25 49 81 121	2 63 300 931 729 121
	Σf = 50		Σfm = 308		Σfm2 = 2146

$$\begin{gathered} {\overline{X}}_B=\frac{\Sigma fm}{\Sigma f}=\frac{308}{50}=6.16 \\ \mathrm{S}.\mathrm{D}.\left({\sigma }_B\right)=\sqrt{\frac{\Sigma f{m}^2}{\Sigma f}-{\left({\overline{X}}_B\right)}^2} \\ =\sqrt{\frac{2146}{50}-{\left(6.16\right)}^2} \\ =\sqrt{42.92-37.95} \\ =\sqrt{4.97} \\ =2.23 \\ \mathrm{C}.\mathrm{V}.\mathrm{of}\mathrm{T}.{\mathrm{V}}_{\mathrm{B}}=\frac{{\sigma }_B}{{\overline{X}}_B}\times 100 \\ =\frac{2.23}{6.16}\times 100 \\ =36.21\% \end{gathered}$$

As, C.V of Model B is less than that of C.V of model A therefore model B has greater uniformity.