Question

Calculate quartile deviation and its relative measure from the following data:

X	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
f	5	10	13	18	14	8

Answer 1

X	f	cf
0 - 10	5	5
10 - 20	10	15
20 - 30	13	28
30 - 40	18	46
40 - 50	14	60
50 - 60	8	68
	N = 68

Q₁ = Size of `("N"/4)^"th"` value

= Size of `(68/4)^"th"` value

= Size of 17^th value

Thus Q₁ lies in the class 20 - 30 and corresponding values are L = 20, `"N"/4` = 17, pcf = 15, f = 13, C = 10.

Q₁ = `"L" + (("N"/4 - "pcf"))/"f" xx "C"`

= `20 + ((17 - 15)/13) xx 10`

= `20 + 2/13 xx 10`

= `20 + 20/13`

= 20 + 1.53846

= 20 + 1.5385

= 21.5385

Q₃ = Size of `("3N"/4)^"th"` value

= Size of `(3 xx 68/4)^"th"` value

= Size of (3 × 17)^th value

= Size of 51^th value

Thus Q₃ lies in the class 40 - 50 and corresponding values are L = 40, `"3N"/4` = 51, pcf = 46, f = 14, C = 10

Q₃ = `"L" + (("3N"/4 - "pcf"))/"f" xx "C"`

= `40 + ((51 - 46)/14) xx 10`

= `40 + 5/14 xx 10`

= `40 + 50/14`

= 40 + 3.5714

= 43.5714

QD = `1/2 ("Q"_3 - "Q"_1)`

= `1/2 (43.5714 - 21.5385)`

= `1/2 (22.0329)`

= 11.01645

QD = 11.02

Relative measure, coefficient of QD = `("Q"_3 - "Q"_1)/("Q"_3 + "Q"_1)`

= `(43.5714 - 21.5385)/(43.5714 + 21.5385)`

= `22.0329/65.1099`

= 0.33839

= 0.3384