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Question
Find Q1, D6, and P78 for the following data:
| C.I. | 8 – 8.95 | 9 – 9.95 | 10 – 10.95 | 11 – 11.95 | 12 – 12.95 |
| f | 5 | 10 | 20 | 10 | 5 |
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Solution
Since the given data is not in the form of a continuous frequency distribution, we have to convert it into that form by subtracting 0.025 from the lower limit and adding 0.025 to the upper limit of each class interval.
∴ the class intervals will be 7.975 – 8.975, 8.975 – 9.975, etc.
We construct the less than cumulative frequency table as given below:
| C.I. | f | Less than cumulative frequency (c.f.) |
| 7.975 – 8.975 | 5 | 5 |
| 8.975 – 9.975 | 10 | 15 ← Q1 |
| 9.975 – 10.975 | 20 | 35 ← D6 |
| 10.975 – 11.975 | 10 | 45 ← P78 |
| 11.975 – 12.975 | 5 | 50 |
| Total | 50 |
Here, N = 50
Q1 class = class containing `("N"/4)^"th"` observation
∴ `"N"/4 = 50/4` = 12.5
Cumulative frequency which is just greater than (or equal) to 12.5 is 15.
∴ Q1 lies in the class 8.975 – 9.975
∴ L = 8.975, h = 1, f = 10, c.f. = 5
Q1 = `"L"+"h"/"f"("N"/4-"c.f.")`
= `8.975 + 1/10 (12.5 - 5)`
= 8.975 + 0.1(7.5)
= 8.975 + 0.75
= 9.725
D6 class = class containing `((6"N")/10)^"th"` observation
∴ `(6"N")/10 = (6 xx 50)/10` = 30
Cumulative frequency which is just greater than (or equal) to 30 is 35.
∴ D6 lies in the class 9.975 – 10.975
∴ L = 9.975, h = 1, f = 20, c.f. = 15
D6 = `"L"+"h"/"f"((6"N")/10-"c.f.")`
= `9.975 + 1/20 (30 - 15)`
= 9.975 + 0.05(15)
= 9.975 + 0.75
= 10.725
P78 class = class containing `((78"N")/100)^"th"` observation
`(78"N")/100 = (78 xx 50)/100` = 39
Cumulative frequency which is just greater than (or equal) to 39 is 45.
∴ P78 lies in the class 10.975 – 11.975
∴ L = 10.975, h = 1, f = 10, c.f. = 35
P78 = `"L"+"h"/"f"((78"N")/10-"c.f.")`
= `10.975 + 1/10 (39 - 35)`
= 10.975 + 0.1(4)
= 10.975 + 0.4
= 11.375
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