Moments, skewness (Karl Pearson and Bowley)…

Core facts

Skewness = lack of symmetry. In a symmetric distribution mean = median = mode.
Positive (right) skew: long tail to the right; mean > median > mode.
Negative (left) skew: long tail to the left; mean < median < mode.
Karl Pearson's coefficient of skewness: Sk = (Mean − Mode)/σ; using Mode = 3 Median − 2 Mean, Sk = 3(Mean − Median)/σ. Limits of the median form: −3 to +3.
Bowley's coefficient (quartile measure): Sk_B = (Q3 + Q1 − 2 Median)/(Q3 − Q1); limits −1 to +1.
Kelly's coefficient (percentile measure): (P90 + P10 − 2 Median)/(P90 − P10).
Moments about the mean (central moments): μr = Σf(x − x̄)^r/N. μ1 = 0, μ2 = σ².
β1 = μ3²/μ2³ (skewness), β2 = μ4/μ2² (kurtosis); γ1 = √β1 with the sign of μ3, γ2 = β2 − 3.
Kurtosis = peakedness (Karl Pearson, 1905):
- β2 = 3: mesokurtic (normal);
- β2 > 3: leptokurtic (more peaked);
- β2 < 3: platykurtic (flat-topped).
Normal distribution: bell-shaped, symmetric about μ, mean = median = mode = μ, β1 = 0, β2 = 3, total area 1.
Areas: μ ± σ covers 68.27%, μ ± 2σ 95.45%, μ ± 3σ 99.73%; μ ± 1.96σ covers 95% and μ ± 2.58σ covers 99%.
Standard normal variate: Z = (X − μ)/σ, with mean 0 and SD 1.

Explained

1. Tests of skewness. A distribution is skewed if (a) mean, median and mode differ; (b) the quartiles are not equidistant from the median (Q3 − Md ≠ Md − Q1); (c) the curve has a longer tail on one side; (d) the total frequencies on either side of the mode are not equal.

Empirical relation (Karl Pearson) for moderately skewed data: Mode = 3 Median − 2 Mean, that is, Mean − Mode = 3(Mean − Median).

2. Karl Pearson's coefficient.

Example 1. Mean 50, median 48, SD 10. Find Karl Pearson's coefficient of skewness. Solution. Sk = 3(50 − 48)/10 = 6/10. Answer: +0.6 (positive skew)

Example 2. Mean 45, mode 50, SD 8. Solution. Sk = (45 − 50)/8. Answer: −0.625 (negative skew)

3. Bowley's coefficient. Useful for open-end classes and when the mode is ill-defined, since it uses only Q1, median and Q3. It ignores the 25% of values at each end.

Example 3. Q1 = 20, median = 30, Q3 = 50. Find Bowley's coefficient. Solution. Sk_B = (50 + 20 − 60)/(50 − 20) = 10/30. Answer: +0.33 (Q3 is farther from the median than Q1)

4. Both measures on a grouped table.

Marks	0–10	10–20	20–30	30–40	40–50
Students	5	8	12	10	5

From the dispersion note (same table): mean = 25.5, σ ≈ 12.03, Q1 = 16.25, Q3 = 35.

Example 4. Find Karl Pearson's (mode form) and Bowley's coefficients. Solution.

Cumulative frequencies 5, 13, 25, 35, 40; N/2 = 20 lies in 20–30. Median = 20 + ((20 − 13)/12) × 10 ≈ 25.83.
Modal class 20–30 (f1 = 12, f0 = 8, f2 = 10). Mode = 20 + (12 − 8)/(2 × 12 − 8 − 10) × 10 = 20 + (4/6) × 10 ≈ 26.67.
Karl Pearson: Sk = (25.5 − 26.67)/12.03 ≈ −0.097.
Bowley: Sk_B = (35 + 16.25 − 2 × 25.83)/(35 − 16.25) = (51.25 − 51.67)/18.75 ≈ −0.022.

Answer: about −0.10 and −0.02; a very slight negative skew

5. Moments.

Raw moments about an arbitrary origin A: μ′r = Σf(x − A)^r/N. About zero they are written v_r (v1 = mean).
Converting raw to central moments (μ′1 = x̄ − A):
- μ2 = μ′2 − μ′1²
- μ3 = μ′3 − 3μ′2μ′1 + 2μ′1³
- μ4 = μ′4 − 4μ′3μ′1 + 6μ′2μ′1² − 3μ′1⁴
First moment about the mean is always 0; second is the variance; third measures skewness; fourth measures kurtosis.

Example 5. The first four moments about 4 are 1, 4, 10 and 45. Find the mean, the central moments and comment on skewness and kurtosis. Solution. Mean = 4 + 1 = 5. μ2 = 4 − 1 = 3. μ3 = 10 − 3(4)(1) + 2(1) = 0. μ4 = 45 − 4(10)(1) + 6(4)(1) − 3(1) = 26. β1 = 0; β2 = 26/9 ≈ 2.89. Answer: symmetric (β1 = 0) and slightly platykurtic (β2 < 3)

6. The normal curve. Probability density f(x) = (1/(σ√(2π))) e^(−(x − μ)²/(2σ²)), −∞ < x < ∞. It was derived by De Moivre (1733) and developed by Laplace and Gauss, so it is also called the Gaussian curve. Properties:

bell-shaped, unimodal, symmetric about x = μ;
mean = median = mode = μ; quartiles at μ ± 0.6745σ;
the curve never touches the x-axis (asymptotic);
points of inflexion at μ ± σ;
maximum ordinate at x = μ, equal to 1/(σ√(2π)) ≈ 0.3989/σ;
odd central moments are 0; μ2 = σ², μ4 = 3σ⁴, so β1 = 0 and β2 = 3;
QD ≈ 0.6745σ (about 2σ/3) and MD ≈ 0.7979σ (about 4σ/5);
determined completely by its two parameters, μ and σ.

Example 6. Marks are normal with mean 60 and SD 10. What share of students score between 50 and 70? Above 80? Solution. 50 and 70 are μ ± σ: 68.27%. 80 is μ + 2σ; area beyond z = 2 is (1 − 0.9545)/2 ≈ 0.0228. Answer: about 68.3%; about 2.3%

Deeper

1. Using the normal table. Tables give the area from 0 to z. Useful values: z = 0.5 → 0.1915; z = 1 → 0.3413; z = 1.5 → 0.4332; z = 1.96 → 0.4750; z = 2 → 0.4772; z = 2.58 → 0.4951; z = 3 → 0.49865.

Example 7. Marks of 1,000 students are normal with mean 60 and SD 10. How many score above 75? Solution. z = (75 − 60)/10 = 1.5. Area above = 0.5 − 0.4332 = 0.0668. 0.0668 × 1,000 ≈ 66.8. Answer: about 67 students

Example 8. In the same marks distribution, how many of 500 students score between 50 and 65? Solution. z = −1 and z = 0.5. Area = 0.3413 + 0.1915 = 0.5328; × 500 ≈ 266.4. Answer: about 266 students

2. Moments of an individual series.

Example 9. For 2, 3, 7, 8, 10 find μ2, μ3, μ4, β1 and β2. Solution. x̄ = 30/5 = 6; deviations −4, −3, 1, 2, 4. μ2 = (16 + 9 + 1 + 4 + 16)/5 = 46/5 = 9.2. μ3 = (−64 − 27 + 1 + 8 + 64)/5 = −18/5 = −3.6. μ4 = (256 + 81 + 1 + 16 + 256)/5 = 610/5 = 122. β1 = (−3.6)²/9.2³ = 12.96/778.688 ≈ 0.0166; β2 = 122/9.2² = 122/84.64 ≈ 1.44. Answer: slight negative skew (μ3 < 0); platykurtic (β2 < 3)

3. Effect of origin and scale. Central moments do not change with origin; under y = (x − A)/h, μr(x) = h^r μr(y). Hence β1 and β2 are pure numbers, independent of both origin and scale.

4. Sheppard's corrections (grouped data, equal width h, frequencies tapering to zero at both ends):

corrected μ2 = μ2 − h²/12;
μ3 needs no correction;
corrected μ4 = μ4 − (1/2)h²μ2 + (7/240)h⁴.

5. Other facts.

β1 ≥ 0 always (it is a square), so it cannot show the direction of skew; use γ1 or the sign of μ3.
β2 ≥ 1 for every distribution, and β2 ≥ β1 + 1.
Percentile coefficient of kurtosis κ = QD/(P90 − P10); κ ≈ 0.263 for the normal curve.
Normal approximation: the binomial tends to the normal when n is large and p is not near 0 or 1; the Poisson tends to the normal when its mean is large. The central limit theorem makes sample means approximately normal, which is why most large-sample tests use z.
Uses of the normal curve: sampling theory and tests of significance, quality control limits (μ ± 3σ), grading and standardising scores.

Common traps

✗ "In positive skewness, mean < median < mode." ✓ Positive skew: mean > median > mode.
✗ "Bowley's coefficient lies between −3 and +3." ✓ Bowley's lies between −1 and +1; Karl Pearson's median form lies between −3 and +3.
✗ "Karl Pearson's coefficient = (Mean − Median)/σ." ✓ It is (Mean − Mode)/σ or 3(Mean − Median)/σ.
✗ "Empirical relation: Mode = 3 Mean − 2 Median." ✓ Mode = 3 Median − 2 Mean.
✗ "Leptokurtic means β2 < 3." ✓ Leptokurtic (peaked) is β2 > 3; platykurtic (flat) is β2 < 3.
✗ "The first moment about the mean is the mean." ✓ It is 0; the second central moment is the variance.
✗ "β1 is negative for negatively skewed data." ✓ β1 is a square and never negative; the sign of μ3 (or γ1) shows direction.
✗ "Points of inflexion of the normal curve are at μ ± 2σ." ✓ They are at μ ± σ.
✗ "The normal curve meets the x-axis at μ ± 3σ." ✓ It never touches the axis; 99.73% of the area lies within μ ± 3σ.

Quick revision

Symmetric: mean = median = mode; positive skew: mean > median > mode.
Karl Pearson Sk = (Mean − Mode)/σ = 3(Mean − Median)/σ.
Bowley Sk = (Q3 + Q1 − 2Md)/(Q3 − Q1), between −1 and +1.
Mode = 3 Median − 2 Mean.
μ1 = 0, μ2 = σ²; β1 = μ3²/μ2³, β2 = μ4/μ2².
β2 = 3 meso, > 3 lepto, < 3 platy; γ2 = β2 − 3.
μ3 = μ′3 − 3μ′2μ′1 + 2μ′1³.
Normal: inflexion at μ ± σ; asymptotic; β1 = 0, β2 = 3.
Areas: 68.27%, 95.45%, 99.73%; 1.96σ → 95%, 2.58σ → 99%.
Z = (X − μ)/σ; QD ≈ 2σ/3, MD ≈ 4σ/5.

Go deeper: Mains writing (beyond what APPSC Paper I usually asks)

Mains writing Mains writing

"The normal distribution is the most important distribution in statistics." Discuss (framework):

Define the curve and its two parameters μ and σ.
List the properties: symmetry, mean = median = mode, inflexion at μ ± σ, asymptotic tails, β1 = 0, β2 = 3, the 68–95–99.7 area rule.
Give the reasons for its importance: many natural measurements (heights, errors of measurement) follow it; binomial and Poisson approach it; the central limit theorem; large-sample tests; ±3σ control charts.
Note limits: income, wealth and waiting-time data are skewed; check skewness and kurtosis before assuming normality.
Conclude: it is the benchmark against which skewness (β1 = 0) and kurtosis (β2 = 3) are measured.

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8.1.2 Moments, skewness (Karl Pearson and Bowley), kurtosis and the normal distribution

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