Question

Calculate Bowley’s coefficient of skewness Sk_b from the following data:

Marks above	0	10	20	30	40	50	60	70	80
No. of students	120	115	108	98	85	60	18	5	0

Answer 1

To calculate Bowley’s coefficient of skewness Skb, we construct the following table:

Marks above	No. of students ‘more than’ (c.f.)	Marks	Frequency (fi)	Less than cumulative frequency (c.f.)
0	120	0 – 10	5	5
10	115	10 – 20	7	12
20	108	20 – 30	10	22
30	98	30 – 40	13	35 ← Q1
40	85	40 – 50	25	60 ← Q2
50	60	50 – 60	42	102 ← Q3
60	18	60 – 70	13	115
70	5	70 – 80	5	120
80	0	80 – 90	0	120
		Total	120	–

Here, N = 120
Q₁ class = class containing the `("N"/4)^"th"` observation

∴ `"N"/4 = 120/4` = 30
Cumulative frequency which is just greater than (or equal to) 30 is 35.
∴ Q₁ lies in the class 30 – 40.
∴ L = 30, h = 10, f = 13, c.f. = 22

∴ Q₁ = `"L" + "h"/"f"("N"/4 - "c.f.")`

= `30 + 10/13(30 - 22)`

= `30 + 10/13(8)`

= 30 + 6.1538
∴ Q₁ = 36.1538

Q₂ class = class containing the `("N"/2)^"th"` observation

∴ `"N"/2 = 120/2` = 60
Cumulative frequency which is just greater than (or equal to) 60 is 60.
∴ Q₂ lies in the class 40 – 50.
∴ L = 40, h = 10, f = 25, c.f. = 35

∴ Q₂ = `"L" + "h"/"f"("N"/2 - "c.f.")`

= `40 + 10/25 (60 - 35)`

= `40 + 10/25(25)`

∴ Q₂ = 50

Q₃ class = class containing the `((3"N")/4)^"th"` observation

∴ `(3"N")/4 = (3 xx 120)/4` = 90
Cumulative frequency which is just greater than (or equal to) 90 is 102.
∴ Q₃ lies in the class 50 – 60.
∴ L = 50, h = 10, f = 42, c.f. = 60

∴ Q₃ = `"L" + "h"/"f"((3"N")/4 - "c.f.")`

= `50 + 10/42(90 - 60)`

= `50 + 10/42(30)`

= 50 + 7.1429
∴ Q₃ = 57.1429
Bowley’s coefficient of skewness:

Sk_b = `("Q"_3 + "Q"_1 - 2"Q"_2)/("Q"_3 - "Q"_1)`

= `(57.1429 + 36.1538 - 2(50))/(57.1429 - 36.1538)`

= `(93.2967 - 100)/(20.9891)`

= `(-6.7033)/(20.9891)`

∴ Sk_b = – 0.3194