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Question
Find lower quartile, upper quartile, 7th decile, 5th decile and 60th 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 |
Solution
| 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 30th 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 90th 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
7th decile = D7 = size of `("7N"/10)^"th"` value
= size of `((7 xx 120)/10)^"th"` value
= size of 84th 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
5th decile = D5 = size of `("5N"/10)^"th"` value
= size of `((5 xx 120)/10)^"th"` value
= size of 60th 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 72th 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
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