Question

Find lower quartile, upper quartile, 7^th decile, 5^th decile and 60^th percentile for the following frequency distribution.

Wages	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency	1	3	11	21	43	32	9

Answer 1

Wages	Frequency	Cumulative Frequency
10 - 20	1	1
20 - 30	3	4
30 - 40	11	15
40 - 50	21	36
50 - 60	43	79
60 - 70	32	111
70 - 80	9	120
	N = 120

Lower quartile, Q₁ = size of `("N"/4)^"th"` value

= size of `(120/4)^"th"` value

= size of 30^th value

Q₁ lies in the class (40 - 50) and its corresponding values are L = 40, `"N"/4` = 30, pcf = 15, f = 21 and C = 10

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

= `40 + ((30 - 15)/21) xx 10`

= `40 + 15/21 xx 10`

= 40 + 7.14

= 47.4

Q₃ = size of `("3N"/4)^"th"` value

= size of `((3 xx 120)/4)^"th"` value

= size of 90^th value

Q₃ lies in the class (60 - 70) and its corresponding values are L = 60, `"3N"/4` = 90, pcf = 79, f = 32 and C = 10.

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

= `60 + ((90 - 79)/32) xx 10`

= `60 + 11/32 xx 10`

= 60 + 3.4375

= 63.4375

= 63.44

7^th decile = D₇ = size of `("7N"/10)^"th"` value

= size of `((7 xx 120)/10)^"th"` value

= size of 84^th value

Thus D₇ lies in the class (60 - 70) and its corresponding values are L = 60, `"7N"/10` = 84, pcf = 79, f = 32 and C = 10.

D₇ = `"L" +[("7N"/10 - "pcf")/"f"] xx "C"`

= `60 + ((84 - 79)/32) xx 10`

= `60 + 5/32 xx 10`

= `60 + 50/32`

= 60 + 1.5625

= 60 + 1.56

= 61.56

5^th decile = D₅ = size of `("5N"/10)^"th"` value

= size of `((5 xx 120)/10)^"th"` value

= size of 60^th value

Thus D₅ lies in the class (50 - 60) and its corresponding values are L = 50, `"5N"/10` = 60, pcf = 36, f = 43 and C = 10.

D₅ = `"L" +[("5N"/10 - "pcf")/"f"] xx "C"`

= `50 + ((60 - 36)/43) xx 10`

= `50 + 24/43 xx 10`

= `50 + 240/43`

= 50 + 5.581

= 55.581

= 55.58

P₆₀ = size of `("60N"/100)^"th"` value

= size of `((60 xx 120)/100)^"th"` value

= size of 72^th value

Thus P₆₀ lies in the class (50 - 60) and its corresponding values are L = 50, `"60N"/100` = 72, pcf = 36, f = 43 and C = 10.

P₆₀ = `"L" +[("60N"/100 - "pcf")/"f"] xx "C"`

= `50 + ((72 - 36)/43) xx 10`

= `50 + 36/43 xx 10`

= 50 + 8.37

= 58.37