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

Answer 1

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