Question

Find Q₁, D_6, 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

Answer 1

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