Question

Answer the following :

Compute coefficient of variations for the following data to show whether the variation is greater in the yield or in the area of the field:

Year	Area (in acres)	Yield (in lakhs)
2011 - 12	156	62
2012 - 13	135	70
2013 - 14	128	68
2014 - 15	117	76
2015 - 16	141	65
2016 - 17	154	69
2017 - 18	142	71

Answer 1

xi	`x_"i" - bar(x)`	`(x_"i" - barx)^2`
156	17	289
135	– 4	16
128	– 11	121
117	– 22	484
141	2	4
154	15	225
142	3	9
`sumx_"i"` = 973		`sum(x_"i" - barx)^2` = 1148

Here, n = 7, `bar(x) = (sumx_"i")/"n" = 973/7` = 139

Var (X) = `sigma_x^2 = 1/"n" sum(x_"i" - barx)^2 = 1148/7` = 164

∴ S.D. = `sigma_x = sqrt("Var(X)")`

= `sqrt(164)`

= 12.8062

Now, C.V. = `100 xx sigma_x/bar(x)`

= `100 xx 12.8062/139`

= 9.21%

yi	`y_"i" - bar(y)`	`(y_"i" - bary)^2`
62	– 6.71	45.02
70	1.29	1.66
68	– 0.71	0.50
76	7.29	53.14
65	– 3.71	13.76
69	0.29	0.0841
71	2.29	5.24
`sumy_"i"` = 481		`sum(y_"i" - bary)^2` = 119.4

Here, n = 7, `bar(y) = (sumy_"i")/"n" = 481/7` = 68.71

Var (Y) = `sigma_y^2 = 1/"n" sum(y_"i" - bary)^2 = 119.4/7` = 17.0571

∴ S.D. = `sigma_y = sqrt("Var(Y)")`

= `sqrt(17.0571)`

= 4.13

Now, C.V. = `100 xx sigma_y/bar(y)`

= `100 xx 4.13/68.71`

= 100 × 0.0601

= 6.01%

∴ the variation is greater in the area of the field.