Question

The following table gives weights of the students of two classes. Calculate the coefficient of variation of the two distributions. Which series is more variable?

Weight (in kg)	Class A	Class B
30-40	22	13
40-50	16	10
50-60	12	17

Answer 1

We construct the following table to compute C.V.:

For class A as well as class B, the mid-values of the class-intervals are 35, 45, 55.

Class A
xi	fi	fixi	`x_"i" - barx` `bar(x)` = 43	`f_"i"(x_"i" - barx)^2`
35	22	770	– 8	1408
45	16	720	2	64
55	12	660	12	1728
	`sumf_"i"` = 50	`sumf_"i"x_"i"` = 2150		`sumf_"i"(x_"i" - barx)^2` = 3200

For Class A

`bar(x) = (sumf_"i"x_"i")/(sumf_"i") = 2150/50` = 43

Standard deviation = `sqrt((sumf_"i"(x_"i" - barx)^2)/(sumf_"i")`

= `sqrt(3200/50)`

= 8

∴ C.V. = `"S.D."/"Mean" xx 100`

= `8/43 xx 100`

= 18.6

Class B
xi	fi	fixi	`x_"i" - barx` `bar(x)` = 46	`f_"i"(x_"i" - barx)^2`
35	13	455	– 11	1573
45	10	450	– 1	10
55	17	935	9	1377
	`sumf_"i"` = 40	`sumf_"i"x_"i"` = 1840		`sumf_"i"(x_"i" - barx)^2` = 2960

For Class B

`bar(x) = (sumf_"i"x_"i")/(sumf_"i") = 1840/40` = 46

Standard deviation = `sqrt((sumf_"i"(x_"i" - barx)^2)/(sumf_"i")`

= `sqrt(2960/40)`

= 8.6

∴ C.V. = `"S.D."/"Mean" xx 100`

= `8.6/46 xx 100`

= 18.7

∴ (C.V.)_B < (C.V.)_A, class B is more variable.