Class 11 Statistics Notes · GSEB

Karl Pearson's Method

Karl Pearson's Method — measure skewness from mean, median, mode and standard deviation, choosing the right formula for well-defined or ill-defined modes. GSEB Class 11 Statistics notes with a decision-tree lab and solved sums.

Last updated: 5 Oct 2026

Notes

Why Mean − Mode?

In a skewed distribution the three averages refuse to sit together: the median always lands between the mean and the mode. Because the median splits the tail-side pull from the hump, the distance mean − mode spans the whole lean of the distribution — which is exactly what an absolute measure of skewness should capture.

Absolute measure Sₖ (data units)

S_k = \bar{x} - M_o

Coefficient j (unit-free)

j = \frac{\bar{x} - M_o}{s}

Two questions, two answers

Karl Pearson's method uses the mean, the mode (or median) and the standard deviation — dividing Sₖ by s strips the units off, so ₹-based and marks-based distributions can finally be ranked on one scale.

Zerodha, real numbers

Monthly brokerage paid by 500 active traders: most traders pay ₹0–₹400 (equity delivery is free), while a thin tail of F&O traders pays ₹4,000–₹40,000. Mean ≈ ₹1,850 sits far above the median ≈ ₹620 — the mean is dragged right by that tail, and x̄ − M₀ (or 3(x̄ − M)) measures exactly that drag.

Reading the sign

x̄ above M₀ → positive skewness (right tail longer) · x̄ below M₀ → negative skewness (left tail longer).

Units stay, until you divide

Sₖ = +3.8 units for the truck data — the moment you divide by s it becomes the pure number j = +0.20.

Pro tip: work the four moves in order — branch, table, totals, substitute.

When the Mode Misbehaves

If the mode is ill-defined — two classes tied at the top, the maximum frequency sitting in the first or last class, unequal class lengths, bimodal data, or the mode simply not given — M₀ cannot be trusted. Karl Pearson's fix is an empirical relation that rebuilds the mode from the median:

Empirical mode

M_o = 3M - 2\bar{x}

Absolute measure

\bar{x} - M_o = 3(\bar{x} - M) \;\Rightarrow\; S_k = 3(\bar{x} - M)

Coefficient — ill-defined mode

j = \frac{3(\bar{x} - M)}{s}

Empirical relation, plain

\bar{x} - M_o = 3(\bar{x} - M)

The empirical relation in words

Mean − Mode = 3(Mean − Median) — so substituting M₀ = 3M − 2x̄ into Sₖ = x̄ − M₀ collapses beautifully into Sₖ = 3(x̄ − M). Same method, one substitution, no mode needed.

Negative example — using the wrong branch

A student meets Ill 7 (maximum frequency in the lastclass) and still writes j = (x̄ − M₀)/s by forcing a group-mode formula on the edge class. The result (−0.94) contradicts the book's −0.68. The branch you pick decides the number — check the mode before you divide.

Sign Rules

Sₖ > 0 → positive

x̄ > M₀ (or x̄ > M) — the right tail is elongated. Ill 1: Sₖ = 163.8 − 160 = +3.8 units, j = +0.20.

Sₖ < 0 → negative

x̄ < M₀ (or x̄ < M) — the left tail is elongated. Ill 3: Sₖ = 43.3 − 43.91 = −0.61 items, j = −0.05.

Sₖ = 0 → symmetric

x̄ = M₀ = M — the seesaw balances, both tails match, j = 0.00.

Same rule, both formulas

The sign of Sₖ decides the type of skewness whether you used x̄ − M₀ or 3(x̄ − M) — the factor 3 is always positive, so it can never flip the sign.

Worked Walkthrough — Illustration 1

Fifty trucks, nine load sizes. The mode sits alone at x = 160, so this takes the well-defined branch — walk the five moves the exam expects.

Solved Example

Problem

Units transported by 50 trucks from the railway yard to factories in a day — find skewness and its coefficient by Karl Pearson's method.

Units x	120	130	140	150	160	170	180	190	200
Trucks f	2	3	4	5	11	9	9	6	1

Solution

Sₖ = 3.8 units (positive skewness) · j ≈ +0.20 — free of units, so it can be compared with any other distribution

Which Formula? — Decide Before You Compute

The branch choice is the most-missed decision in this chapter. Probe the data first, then let the tree pick the formula for you.

Method Chooser — which Karl Pearson formula?

Pick the data situation you were handed — the branch it belongs to lights up.

Ill 150 trucks — one clear peak at 160Ill 2Tax, equal classes — modal class 40–60Ill 3Rejected items, inclusive classesIll 4Transport cost — unequal class lengthsIll 5Bulb life — two classes share f = 76Ill 6Ad cost — mixed discrete + continuousIll 7Wages — max frequency in the LAST class

Choose one of the seven situations above — the formula card, the absolute measure and the j ruler all follow your pick.

Try It: StatsProblemLab — the chapter's question generator

Load a book illustration or paste your own frequencies, pick a method, then pass each checkpoint — the next step unlocks only when the current one checks out.

This is the one generator for the whole Skewness chapter— the other topic pages (Bowley's Method, Comparison of Methods) link back here whenever you want fresh practice.

Stats Problem Lab — Units transported by 50 trucksKarl Pearson — well-defined mode

1 · Pick the method, 2 · pick your data

n = 50x̄ · M · M₀ · s · Q₁ · Q₃ appear once you substitute themmode: well-defined

Method

Karl Pearson — well-defined mode

Data type

Discrete frequency

Units

units

Live probe — is the mode well-defined?

Yes: a single interior class carries the highest frequency, class lengths match, and there is one peak only → j = (x̄ − M₀)/s.

The distribution is loaded (9 rows). Type n = 50 after adding the f column yourself:

n (total frequency)

Tail Length Decides the Degree

Two curves can peak in the same place and still carry very different skewness — the elongated tail is the whole story.

Tail Length Comparator — same peak, longer tail

Distribution Aright tail — least

elongation metershort

Distribution Bright tail — medium

elongation metermedium

Distribution Cright tail — most

elongation meterlong

All three curves peak in almost the same place — only the elongation changes. C is the most elongated, so it carries the biggest |j|; A the least. The tail alone decides the degree of skewness, not the sign.

✗ Wrong way — reading the sign as size

“Cricketer A has j = −0.42 and cricketer B has +0.77, so A must be less skewed — negative is smaller.” The minus sign only points at the side of the tail.

✓ Right way — compare |j|

|−0.42| = 0.42 and |+0.77| = 0.77 → 0.77 > 0.42, so B is more skewed. Ignore the sign, rank the magnitudes (Illustrations 9 & 10).

How Far Can j Go?

Three range statements — know which one the question wants

j = 3(x̄ − M)/s ranges from −3 to 3 — use it when the mode is ill-defined or there are multiple modes.
In practice (sample data), j lies between −1 and 1.
N. L. Johnson (1951) proved the theoretical bound for unimodal data: −√3 … √3 ≈ −1.73 to 1.73 for j = (x̄ − M₀)/s.

The Supporting Machinery (reused from earlier chapters)

Karl Pearson's method is only as good as the four averages it borrows — every one of these you already know.

Group mode

M₀ = L + (fₘ − f₁) / (2fₘ − f₁ − f₂) × c

modal class only

Median

M = L + (n/2 − cf) / f × c

median class only

Step-deviation mean

x̄ = A + (Σfd / n) × c

assumed mean A

Standard deviation

s = √(Σfd²/n − (Σfd/n)²) × c

the divisor c

Pick A near the middle

The assumed mean A only shifts the arithmetic — the final x̄ and s never change with it. Choose the mid-value of the modal (or central) class and the d column stays small enough to add in your head.

Which Distribution Is More Skewed?

The sign tells you the side; the magnitude tells you the degree. Whenever a question asks which of two distributions is more skewed — or closer to symmetry — compare |j|, ignoring the sign.

✗ Negative example — comparing signed numbers

“Merchant A = −0.36 and merchant B = +0.94; since −0.36 < 0.94, A is more skewed.” Wrong — that ranks direction, not degree, and would fail Ill 10.

✓ Right way — rank |j|

|−0.36| = 0.36 < |+0.94| = 0.94 → B is more skewed, A is closer to symmetry. Smallest |j| = closest to symmetry (Illustration 10).

Inclusive Classes → Class Boundaries First

Classes written 11–20, 21–30… leave a 1-unit gap between them. Before any median or group-mode formula touches L, convert to boundaries by subtracting/adding 0.5 — the mid-values do not move, but L does.

Class (inclusive)	Boundary	Mid-value x	Effect on L
11 – 20	10.5 – 20.5	15.5	modal class L = 40.5, not 41
41 – 50	40.5 – 50.5	45.5	modal class carries fₘ = 39

Using L = 41 → M₀ = 44.41 (off by half a class) Boundaries → M₀ = 43.91 ✓ (Ill 3)

Only L changes

Mean and SD use mid-values, which are identical either way — that is why Ill 3 still gets x̄ = 43.3 and s = 11.96 with or without the conversion. The mode formula is the one that needs the boundaries.

Worked Example Gallery — one per sum type

Karl Pearson — solved practicals

10 problems

1 · Unimodal, discrete (mode well-defined)

Full question

The following data related to units transported by 50 trucks from railway yard to different factories on a day. Find the skewness and its coefficient using Karl Pearson's method from these data.

No. of units transported	120	130	140	150	160	170	180	190	200
No. of trucks	2	3	4	5	11	9	9	6	1

x	f	d = x − 160	fd	fd²
120	2	−40	−80	3200
130	3	−30	−90	2700
140	4	−20	−80	1600
150	5	−10	−50	500
160	11	0	0	0
170	9	10	90	900
180	9	20	180	3600
190	6	30	180	5400
200	1	40	40	1600
Total	n = 50		190	19500

x̄ = 160 + 190/50 = 163.8 units · M₀ = 160 (f = 11, alone, interior)

s = √(19500/50 − (190/50)²) = √(390 − 14.44) = 19.38 units

Sₖ = 163.8 − 160 = 3.8 units · j = 3.8 / 19.38 = +0.20

Positive, mild — |j| < 0.30 means close to symmetry.

2 · Equal class intervals

Full question

Find the skewness and coefficient of skewness by Karl Pearson's method from the following data about annual tax of 100 companies.

Annual tax (lakh ₹)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
No. of companies	5	20	40	25	6	4

Class	f	Mid x	d = (x−50)/20	fd	fd²
0 – 20	5	10	−2	−10	20
20 – 40	20	30	−1	−20	20
40 – 60	40	50	0	0	0
60 – 80	25	70	1	25	25
80 – 100	6	90	2	12	24
100 – 120	4	110	3	12	36
Total	n = 100			19	125

x̄ = 50 + (19/100) × 20 = 53.8 lakh ₹

M₀ = 40 + (40 − 20)/(80 − 20 − 25) × 20 = 40 + 400/35 = 51.43 lakh ₹

s = √(1.25 − 0.0361) × 20 = 22.04 lakh ₹

Sₖ = 53.8 − 51.43 = 2.37 lakh ₹ · j = 2.37 / 22.04 = +0.11

Small j → the book calls it "almost symmetric".

3 · Inclusive classes → convert to boundaries

Full question

A factory has 100 machines for production. The following data are obtained about rejected items during the production process. Find the skewness and its coefficient using Karl Pearson's method from these data.

No. of rejected items	11 - 20	21 - 30	31 - 40	41 - 50	51 - 60	61 - 70	71 - 80
No. of machines	2	12	25	39	12	9	1

Class	f	Mid x	d = (x−45.5)/10	fd	fd²
11 – 20	2	15.5	−3	−6	18
21 – 30	12	25.5	−2	−24	48
31 – 40	25	35.5	−1	−25	25
41 – 50	39	45.5	0	0	0
51 – 60	12	55.5	1	12	12
61 – 70	9	65.5	2	18	36
71 – 80	1	75.5	3	3	9
Total	n = 100			−22	148

Inclusive → modal class 41–50 becomes 40.5 – 50.5, so L = 40.5

x̄ = 45.5 + (−22/100) × 10 = 43.3 · M₀ = 40.5 + 14/41 × 10 = 43.91

s = √(1.48 − 0.0484) × 10 = 11.96 · Sₖ = 43.3 − 43.91 = −0.61 items

j = −0.61 / 11.96 = −0.05

Negative but almost zero — closer to symmetry than the truck data (|−0.05| < |+0.20|).

4 · Unequal classes → median form

Full question

The average monthly transportation cost of 200 families in a city in the year 2014 is as follows. Find skewness and coefficient of skewness using Karl Pearson's method.

Average monthly transportation cost (thousand ₹)	1 - 3	4 - 6	7 - 9	10 - 13	14 - 16	17 - 19
No. of families	5	40	120	20	10	5

Class	f	Mid x	d = x − 8	fd	fd²	cf
1 – 3	5	2	−6	−30	180	5
4 – 6	40	5	−3	−120	360	45
7 – 9	120	8	0	0	0	165
10 – 13	20	11.5	3.5	70	245	185
14 – 16	10	15	7	70	490	195
17 – 19	5	18	10	50	500	200
Total	n = 200			40	1775

Unequal lengths → j = 3(x̄ − M)/s (median form, no mode)

x̄ = 8 + 40/200 = 8.2 · 100th obs in 7–9 (boundaries 6.5–9.5)

M = 6.5 + (100 − 45)/120 × 3 = 7.88 · s = √(8.875 − 0.04) = 2.97

Sₖ = 3(8.2 − 7.88) = 0.96 thousand ₹ · j = 0.96 / 2.97 = +0.32

The one unequal-class anchor in the chapter — j = +0.32 also loads in the StatsProblemLab presets.

5 · Bimodal (two classes tied at the top)

Full question

The following information is obtained for life (in complete hours) of 400 electric bulbs. Find the skewness and the coefficient of skewness by Karl Pearson's method.

Life of electric bulbs (complete hours)	4000-4199	4200-4399	4400-4599	4600-4799	4800-4999	5000-5199	5200-5399	5400-5599
No. of electric bulbs	14	46	58	76	70	76	40	20

Class	f	Mid x	d	fd	fd²	cf
4000 – 4199	14	4099.5	−3	−42	126	14
4200 – 4399	46	4299.5	−2	−92	184	60
4400 – 4599	58	4499.5	−1	−58	58	118
4600 – 4799	76	4699.5	0	0	0	194
4800 – 4999	70	4899.5	1	70	70	264
5000 – 5199	76	5099.5	2	152	304	340
5200 – 5399	40	5299.5	3	120	360	380
5400 – 5599	20	5499.5	4	80	320	400
Total	n = 400			230	1422

Two classes at f = 76 → multiple modes → median form

x̄ = 4699.5 + (230/400) × 200 = 4814.5 · 200th obs in 4800–4999

M = 4799.5 + (200 − 194)/70 × 200 = 4816.64 · s = √(3.2244) × 200 = 359.13

Sₖ = 3(4814.5 − 4816.64) = −6.42 hours · j = −6.42 / 359.13 = −0.02

Essentially symmetric — two equal humps, not a long tail.

6 · Mixed discrete + continuous

Full question

The following information is about the annual advertisement cost (in lakh ₹) of 30 companies. Find skewness and coefficient of skewness using it.

Annual advertisement cost (lakh ₹)	0	3	5	8	10 - 20	20 - 30	30 - 40
No. of companies	3	4	5	10	5	2	1

Class	f	Mid x	d = x − 15	fd	fd²	cf
0	3	0	−15	−45	675	3
3	4	3	−12	−48	576	7
5	5	5	−10	−50	500	12
8	10	8	−7	−70	490	22
10 – 20	5	15	0	0	0	27
20 – 30	2	25	10	20	200	29
30 – 40	1	35	20	20	400	30
Total	n = 30			−173	2841

Points + classes mixed → median form

x̄ = 15 + (−173/30) = 9.23 · 15th observation reads 8 from the cf column → M = 8

s = √(94.7 − 33.2544) = 7.84 · Sₖ = 3(9.23 − 8) = 3.69 lakh ₹

j = 3.69 / 7.84 = +0.47

Moderately skewed — the few big spenders pull the mean to the right of the median.

7 · Mode ill-defined (max frequency in the last class)

Full question

The frequency distribution of hourly wages paid to 600 workers is given below. Find the Karl Pearson's coefficient of skewness from this distribution.

Hourly wage (₹)	40 - 60	60 - 80	80 - 100	100 - 120	120 - 140	140 - 160	160 - 180	180 - 200	200 - 220	220 - 240
No. of workers (f)	10	12	16	20	50	60	72	100	120	140

Class	f	Mid x	d	fd	fd²	cf
40 – 60	10	50	−4	−40	160	10
60 – 80	12	70	−3	−36	108	22
80 – 100	16	90	−2	−32	64	38
100 – 120	20	110	−1	−20	20	58
120 – 140	50	130	0	0	0	108
140 – 160	60	150	1	60	60	168
160 – 180	72	170	2	144	288	240
180 – 200	100	190	3	300	900	340
200 – 220	120	210	4	480	1920	460
220 – 240	140	230	5	700	3500	600
Total	n = 600			1556	7020

Max f = 140 sits in the LAST class → mode ill-defined → median form

x̄ = 130 + (1556/600) × 20 = 181.87 · 300th obs in 180–200 → M = 180 + 60/100 × 20 = 192

s = √(11.7 − 6.7254) × 20 = 44.61 · Sₖ = 3(181.87 − 192) = −30.39 ₹

j = −30.39 / 44.61 = −0.68

Strong negative skew — most workers earn on the high side, a thin tail trails to low wages.

8 · Summary statistics only

Full question

The following figures are given to describe the changes in the market prices of shares of a company before and after their general body meeting. Take them into consideration to comment whether the proceedings of this meeting have affected the market prices of shares by computing coefficient of skewness.

Details	Before meeting	After meeting
No. of share transactions	6000	5800
Mean of prices of share (₹)	440	460
Median of prices of share (₹)	500	480
Standard deviation of prices of share (₹)	60	52

Mode not given → j = 3(x̄ − M)/s

Before: j = 3(440 − 500)/60 = −180/60 = −3.00

After: j = 3(460 − 480)/52 = −60/52 = −1.15

Both negative → both negatively skewed; |j| fell from 3.00 to 1.15, so the meeting pulled the distribution back towards symmetry (a partial effect).

This is the −3.00 anchor — the widest value the median-form coefficient can reach.

9 · Which distribution is more skewed?

Full question

The data about the number of runs scored by two cricketers in 10 matches are as follows. Use this information to determine which cricketer's game is more skewed.

Details	Cricketer A	Cricketer B
Mean x̄	50	35
Mode M₀	56	31
Standard deviation s	14.4	5.2

A: j = (50 − 56)/14.4 = −6/14.4 = −0.42

B: j = (35 − 31)/5.2 = 4/5.2 = +0.77

|−0.42| = 0.42 < |+0.77| = 0.77 → cricketer B is more skewed — ignore the signs, rank the magnitudes.

The signs point at opposite tails; the magnitudes carry the degree.

10 · Only mean and median given

Full question

The median of the distribution of marks in statistics obtained by students of a school is 72 and its mean is 75. Find the skewness for the marks obtained by the students from these data and state the type of skewness.

Median M = 72 and mean x̄ = 75.

Mode not given → use the empirical form directly on the absolute measure:

Sₖ = 3(x̄ − M) = 3(75 − 72) = 3 × 3 = 9 marks

Sₖ > 0 → positive skewness. (No s in the question, so only the absolute measure is reported.)

Units matter: the answer is "9 marks", not "9".

Quick Reference

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

Rule	Formula / statement
⭐ Well-defined mode	Sₖ = x̄ − M₀ · j = (x̄ − M₀)/s
Measures used	Karl Pearson's method uses the mean, the mode (or median) and the standard deviation
⭐ Ill-defined / multiple modes	M₀ = 3M − 2x̄ ⇒ Sₖ = 3(x̄ − M) · j = 3(x̄ − M)/s
Empirical relation	Empirical relation: Mean − Mode = 3(Mean − Median)
Sign	j > 0 positive · j < 0 negative · j = 0 symmetric
Ill-defined range	j = 3(x̄ − M)/s ranges from −3 to 3; use it when the mode is ill-defined or there are multiple modes
The two divisions	The well-defined formula divides x̄ − M₀ by s; the ill-defined form divides 3(x̄ − M) by s
Practice range	In practice (sample data), j lies between −1 and 1
Johnson bound	Johnson's theoretical bound for unimodal data is ±1.73
Compare degrees	Compare distributions by \|j\|, ignoring sign

Key Takeaways

Karl Pearson measures skewness as the gap between the averages: Sₖ = x̄ − M₀ when the mode is well-defined, Sₖ = 3(x̄ − M) when it is not.
Branch first, compute second — tie at the top, edge class, unequal lengths, bimodal or no mode means the median form; everything else means the mode form.
Sₖ carries the data units (₹, cm, hours); j = Sₖ ÷ s (or ÷ Q-spread) strips them so distributions can be ranked.
Ranges are not interchangeable: −1…+1 in practice, −3…+3 for the median form, Johnson ±1.73 for the unimodal form.
Rank skewness by |j| — the sign only names the elongated side, never the degree.
So what? When Zerodha reports a mean brokerage far above the median, it is telling investors that a minority of F&O traders drives the average — the skew, not the mean, is the story.