Class 11 Statistics Notes · GSEB

Bowley's Method

Bowley's Method — measure skewness from quartiles alone, the only formula that works with open-ended classes. GSEB Class 11 Statistics notes with the sign rule, guaranteed range and a seesaw lab.

Last updated: 5 Oct 2026

Notes

The Assumption Behind Bowley

Prof. A. L. Bowley measured skewness entirely from positions — the quartiles and the median. His whole method rests on one observation about skewed data: Q₁ and Q₃ are not equidistant from the median in a skewed distribution, even though exactly 25% of the data lies on each side of them.

Absolute measure Sₖ

S_k = (Q_3 - M) - (M - Q_1) = Q_3 + Q_1 - 2M

Coefficient j

j = \frac{(Q_3 - M) - (M - Q_1)}{(Q_3 - M) + (M - Q_1)} = \frac{Q_3 + Q_1 - 2M}{Q_3 - Q_1}

Read it as two arms

Sₖ = (right arm) − (left arm)The right arm runs median → Q₃, the left arm median → Q₁. Subtracting them says which side is stretched; dividing by their sum (Q₃ − Q₁) turns it into a unit-free coefficient.

Sign Logic — Compare Q₃ + Q₁ with 2M

Q₃ + Q₁ > 2M

Right arm longer → Sₖ > 0 → positive skewness. Ill 12: 13 + 5 = 18 > 12.

Q₃ + Q₁ < 2M

Left arm longer → Sₖ < 0 → negative skewness. Ill 13: 766.17 + 423.03 = 1189.20 < 1294.24.

Q₃ + Q₁ = 2M

Arms equal → Sₖ = 0 → symmetric, the seesaw sits level.

The zero test

j = 0 ⇔ Q₃ + Q₁ = 2M — no formula needed, one multiplication decides it.

✗ Negative example — skipping the arrangement

Feed Ill 12's rainfall rows in the order they are printed and the cf chain lies: you get Q₁ = 6, M = 13, Q₃ = 5 — a Q₃ below Q₁, and j = +15. A coefficient outside −1…+1 is the range itself telling you the data was never sorted.

✓ Right way — arrange, then cf

Sort the rainfall values ascending, build cf (8, 18, 23, 26, 29, 31), then read the 8th, 16th and 24th observations → Q₁ = 5, M = 6, Q₃ = 13 → j = +0.75, comfortably inside the range.

Lab: Tilt the Quartile Seesaw

Drag the quartiles (or snap to a preset) and watch the beam tilt — the sign of j is literally the direction the seesaw falls.

Quartile Seesaw — the median is the fulcrum

First quartile Q₁5Median M (fulcrum)6Third quartile Q₃13

j = (Q₃ + Q₁ − 2M) / (Q₃ − Q₁) = (13 + 5 − 2 × 6) / (13 − 5) = 6 / 8 = +0.75

positive skewnessSₖ = 6the Q₃ arm is longer → the right pan sinks → positive

Live — every number above feeds this formula

j = \frac{Q_3 + Q_1 - 2M}{Q_3 - Q_1}

Level the seesaw (j = 0.00) and you have a symmetric distribution: the quartiles sit equally far from the median, Q₃ + Q₁ = 2M exactly.

Where Karl Pearson Gives Up — Open-Ended Classes

An open-ended class has no limit on one side, so it has no mid-value — and without mid-values there is no mean, no Σfd, no standard deviation.

The Open-Ended Gate — sales of 500 companies (Illustration 14)

Sales (thousand tonnes)	No. of companies f	cf
Less than 4	26	26
4 – 7	119	145
7 – 10	198	343
10 – 13	86	429
13 – 16	39	468
16 – 20	20	488
20 and above	12	500
Total	n = 500

LOCKED

Karl Pearson

j = \frac{\bar{x} - M_o}{s} \quad \text{or} \quad \frac{3(\bar{x} - M)}{s}

The mean and the SD need a mid-value for everyclass — “Less than 4” and “20 and above” have no upper/lower limit, so x̄ and s simply cannot be computed.

Bowley

ONLY METHOD

j = \frac{Q_3 + Q_1 - 2M}{Q_3 - Q_1}

Quartiles and the median are positional — they only need the cf column, which every open-ended table still has.

Prompt worth memorising

Only method for open-ended classes · uses positional averages only. The moment you see “Less than …” or “… and above”, reach for quartiles.

The Guaranteed Range −1 ≤ j ≤ 1

Bowley's coefficient can never escape the unit interval

|(Q₃ − M) − (M − Q₁)| / [(Q₃ − M) + (M − Q₁)] ≤ 1

Both arms (Q₃ − M) and (M − Q₁) are positive, and the difference of two real numbers can never exceed their sum — so the fraction is always ≤ 1. Hence Range: −1 ≤ j ≤ 1 (always).

Practical use: any Bowley answer outside −1…+1 is arithmetic gone wrong (like an unsorted table), never a genuine distribution.

The Quartile Machinery (from Chapter 3)

Bowley adds nothing new — it borrows the quartile formulas you already use, then does one subtraction.

Discrete Q₁ · M · Q₃

Q₁ = (n+1)/4 th value · M = (n+1)/2 th · Q₃ = 3(n+1)/4 th

read from cf

Grouped Q₁

Q₁ = L + (n/4 − cf) / f × c

quartile class

Grouped median

M = L + (n/2 − cf) / f × c

median class

Grouped Q₃

Q₃ = L + (3n/4 − cf) / f × c

quartile class

Inclusive classes again

Boundary conversion first (11–20 → 10.5–20.5), exactly as in Karl Pearson's method — only L of the quartile class changes; the cf column does not.

Percent Statements Are Quartiles in Disguise

Translation tip

25% of observations lie below Q₁ and 75% lie below Q₃— so any sentence of the form “x% scored less than …” is handing you a quartile.Swiggy's own delivery dashboard reads like this: 25% of orders arrive within 22 min, the median order takes 28 min, 75% within 38 min. Translate → Q₁ = 22, M = 28, Q₃ = 38 → Sₖ = 38 + 22 − 56 = 4 min, j = 4 / 16 = +0.25 — a mild right tail of late orders, nothing alarming.

Who Was Bowley?

A. L. Bowley (1869–1957)

Arthur Lyon Bowley — British statistician and economist, a founding figure of modern social statistics and a fellow of the Royal Statistical Society from 1893. His quartile-based measure survives because it asks so little of the data: three positions, one subtraction, and it works even when the mean refuses to.

Worked Walkthrough — Illustration 14 (the open-ended showcase)

The book's own open-ended problem — the one that proves why this method exists.

Solved Example

Problem

Find the coefficient of skewness for the frequency distribution of sales of 500 companies in the year 2014-2015 using an appropriate method.

Sales (thousand tonnes)	Less than 4	4–7	7–10	10–13	13–16	16–20	20 and above
No. of companies	26	119	198	86	39	20	12

Solution

Sₖ = +0.44 thousand tonnes · j ≈ +0.10 — positive skewness, very close to symmetry

Worked Example Gallery — one per sum type

Bowley — solved practicals

6 problems

1 · Discrete data — arrange it first!

Full question

The data about rainfall (in cm) at a place in a month during monsoon are obtained as follows. Find the skewness and coefficient of skewness using Bowley's method.

Rainfall (cm)	6	7	13	5	15	20
No. of days	10	5	3	8	3	2

Rainfall (cm) — arranged	f	cf
5	8	8
6	10	18
7	5	23
13	3	26
15	3	29
20	2	31
Total	n = 31

Q₁ = value of ((n+1)/4)th = 8th observation → cf 8 → Q₁ = 5 cm

M = value of ((n+1)/2)th = 16th observation → falls in cf 18 → M = 6 cm

Q₃ = value of 3((n+1)/4)th = 24th observation → falls in cf 26 → Q₃ = 13 cm

Sₖ = 13 + 5 − 2(6) = 18 − 12 = 6 cm · j = 6 / (13 − 5) = 6/8 = +0.75

Positive skewness — the long arm from the median out to Q₃ gives it away.

2 · Grouped data, inclusive classes

Full question

The frequency distribution of number of cheques received per day for clearing of 5 branches of a bank on 100 days in the year 2014 is as follows. Find the coefficient of skewness by Bowley's method using this distribution.

No. of cheques	0 – 199	200 – 399	400 – 599	600 – 799	800 – 999
No. of days	10	13	17	42	18

Class	f	cf
0 – 199	10	10
200 – 399	13	23
400 – 599	17	40
600 – 799	42	82
800 – 999	18	100
Total	n = 100

Inclusive → boundaries −0.5/+0.5. Positions n/4 = 25, n/2 = 50, 3n/4 = 75.

Q₁: class 400–599 (L = 399.5, cf = 23, f = 17, c = 200) → 399.5 + (2×200)/17 = 423.03

M: class 600–799 (L = 599.5, cf = 40, f = 42) → 599.5 + (10×200)/42 = 647.12

Q₃: same class (75th) → 599.5 + (35×200)/42 = 766.17

Sₖ = 766.17 + 423.03 − 2(647.12) = −105.04 · j = −105.04/343.14 = −0.31

Negative skewness — Q₃ sits closer to the median than Q₁ does.

3 · Open-ended classes — the showcase

Full question

Find the coefficient of skewness for the frequency distribution of sales of 500 companies in the year 2014-2015 using an appropriate method.

Sales (thousand tonnes)	Less than 4	4 – 7	7 – 10	10 – 13	13 – 16	16 – 20	20 and above
No. of companies	26	119	198	86	39	20	12

Sales	f	cf
Less than 4	26	26
4 – 7	119	145
7 – 10	198	343
10 – 13	86	429
13 – 16	39	468
16 – 20	20	488
20 and above	12	500
Total	n = 500

Open-ended → only Bowley can run (no mid-value for the end classes).

Q₁: 125th obs in 4–7 → 4 + (125 − 26)/119 × 3 = 6.50 thousand tonnes

M: 250th obs in 7–10 → 7 + (250 − 145)/198 × 3 = 8.59

Q₃: 375th obs in 10–13 → 10 + (375 − 343)/86 × 3 = 11.12

Sₖ = 11.12 + 6.50 − 17.18 = 0.44 · j = 0.44/4.62 = +0.10

Positive but tiny — the book calls the distribution close to symmetry.

4 · Sum / difference shortcut

Full question

For a frequency distribution of monthly overtime (in hours) of employees of a company, the difference between the two extreme quartiles is 50 and their sum is 218. If the median is 112, find the coefficient of skewness.

j = (Q₃ + Q₁ − 2M) / (Q₃ − Q₁) = (218 − 2 × 112) / 50

= (218 − 224) / 50 = −6 / 50 = −0.12

Negatively skewed — no need to separate Q₁ and Q₃ at all.

When the question hands you the sum and the difference, substitute straight away.

5 · Symmetric inverse — find the missing quartile

Full question

Mode of a symmetric frequency distribution is 84. If the first quartile is 68, find the third quartile.

Symmetric → x̄ = M = M₀, so M = 84 and j = 0.

0 = (Q₃ + 68 − 2 × 84) / (Q₃ − 68) → numerator must be 0

Q₃ + 68 − 168 = 0 → Q₃ = 100

Check: (100 + 68 − 168) / (100 − 68) = 0/32 = 0.00 ✓

Set j = 0 first, then solve — the denominator never matters when the numerator is zero.

6 · Percent statements → Q₁ / Q₃ in disguise

Full question

The following information is available about marks in a subject obtained by the students of a school in the annual examination. 25% students have scored less than 28 marks whereas, 75% students have scored less than 47 marks. If the coefficient of skewness for marks is 0.4, find the median.

25% below 28 → Q₁ = 28 · 75% below 47 → Q₃ = 47 · j = 0.4

0.4 = (47 + 28 − 2M) / (47 − 28) = (75 − 2M) / 19

0.4 × 19 = 75 − 2M → 7.6 = 75 − 2M → 2M = 67.4 → M = 33.7 marks

“25% below / 75% below” is the book translating quartiles into percentages.

Sₖ vs j — the Two Flavours of Bowley

Bowley's absolute measure and its coefficient
Aspect	Absolute measure Sₖ	Relative measure j
Formula	Sₖ = (Q₃ − M) − (M − Q₁) = Q₃ + Q₁ − 2M	j = (Q₃ + Q₁ − 2M) / (Q₃ − Q₁)
Divided by	nothing — keeps the data units (cm, ₹, marks)	the quartile range Q₃ − Q₁, which has the same units
Reads as	how many units the two arms differ by	what fraction of the middle 50% the difference covers
Range	unbounded — as big as the data itself	always −1 to +1
Compare two populations?	no — different units spoil it	yes — unit-free

Quick Reference

Words marked with a dotted underline carry the question they usually appear in — tap them to reveal it.

Rule	Statement
⭐ Absolute measure	Sₖ = Q₃ + Q₁ − 2M
Averages used	Bowley's method uses the quartiles and the median (positional averages)
⭐ Coefficient	j = (Q₃ + Q₁ − 2M) / (Q₃ − Q₁) — the difference over the sum of the arms
Negative case	Q₃ + Q₁ < 2M ⇒ negative
Range	Range: −1 ≤ j ≤ 1 (always)
Symmetry test	j = 0 ⇔ Q₃ + Q₁ = 2M — quartiles equidistant from the median
Open-ended data	Mean and SD cannot be computed for open-ended classes · Bowley is the only option
Units	The difference-of-quartiles form makes j unit-free

Key Takeaways

Bowley assumes the quartiles are not equidistant from the median — the size of that imbalance IS the skewness: Sₖ = Q₃ + Q₁ − 2M.
Sign test without any formula: Q₃ + Q₁ > 2M positive · < 2M negative · = 2M symmetric (j = 0.00).
j = (Q₃ + Q₁ − 2M)/(Q₃ − Q₁) is guaranteed to sit inside −1…+1 — outside that range means the data was never arranged or a cf chain broke.
Open-ended classes kill the mean and the SD but never the quartiles — that is why Bowley is the only method that survives them.
Percent statements ("25% below 28") are just quartiles wearing a percentage label — translate them before anything else.
So what? When Swiggy quotes "25% within 22 min · median 28 · 75% within 38", it is publishing Q₁, M and Q₃ — you can read the skew of their delivery times straight off the percentiles.