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Question
Calculate D4 and P48 from the following data:
| Mid value | 2.5 | 7.5 | 12.5 | 17.5 | 22.55 | Total |
| Frequency | 7 | 18 | 25 | 30 | 20 | 100 |
Solution
The difference between any two consecutive mid values is 5, the width of class interval = 5
∴ Class interval with mid-value 2.5 is 0 - 5 Class interval with mid-value 7.5 is 5 - 10, etc.
We construct the less than cumulative frequency table as given below:
| Class Interval | Frequency (f) |
Less than cumulative frequency (c.f.) |
| 0 - 5 | 7 | 7 |
| 5 - 10 | 18 | 25 |
| 10 - 15 | 25 | 50 ← D4, P48 |
| 15 - 20 | 30 | 80 |
| 20 - 25 | 20 | 100 |
| Total | 100 |
Here, N = 100
D4 class = class containing `((4"N")/10)^"th"` observation
∴ `(4"N")/(10) = (4 xx 100)/(10)` = 40
Cumulative frequency which is just greater than (or equal to) 40 is 50.
∴ D4 lies in the class 10 - 15
∴ L = 10, f = 25, c.f. = 25, h = 5
D4 = `"L"+"h"/"f"((4"N")/(10) - "c.f.")`
= `10 + (5)/(25)(40 - 25)`
= `10+1/5(15)`
= 10 + 3
∴ D4 = 13
P48 class = class containing `((48"N")/100)^"th"` observation
∴ `(48"N")/(100) = (48 xx 100)/(100)` = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P48 lies in the class 10 - 15
∴ L=10, f = 25, c.f. = 25, h = 5
P48 = `"L"+"h"/"f"((48"N")/(100) -"c.f.")`
= `10 + (5)/(25)(48 - 25)`
= `10 + (1)/(5)(23)`
= 10 + 4.6
∴ P48 = 14.6
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