Dispersion: range, quartile deviation, mean…

Core facts

Dispersion = the degree of scatter of values about a central value. Two series can have the same mean and very different spread.
Absolute measures are in the units of the data: range, quartile deviation (QD), mean deviation (MD), standard deviation (SD).
Relative measures are pure numbers (coefficients), used to compare series in different units or with different averages.

Measure	Formula	Relative measure
Range	R = L − S	(L − S)/(L + S)
Quartile deviation	(Q3 − Q1)/2	(Q3 − Q1)/(Q3 + Q1)
Mean deviation	Average absolute deviation from A	MD/A (A = mean or median)
Standard deviation	√(Σf(x − x̄)²/N)	CV = (σ/x̄) × 100

Interquartile range = Q3 − Q1; QD is half of it, so it is also called the semi-interquartile range.
Mean deviation can be taken about the mean, median or mode; it is least when taken about the median.
Standard deviation (σ) was introduced by Karl Pearson (1893); it is the root-mean-square deviation from the mean.
Variance = σ². Shortcut: σ² = Σx²/n − x̄² (individual) or Σfx²/N − x̄² (frequency).
Coefficient of variation (Karl Pearson): CV = σ/x̄ × 100. Lower CV = more consistent / more uniform / more stable.
Effect of origin and scale: adding or subtracting a constant leaves range, QD, MD and SD unchanged; multiplying by k multiplies them by |k| and the variance by k².
SD of the first n natural numbers = √((n² − 1)/12).
Lorenz curve is a graphic measure of dispersion (of income or wealth); the line of equal distribution is the 45° diagonal.

Explained

1. Range. Range = largest − smallest value. For a grouped table, take the upper limit of the highest class minus the lower limit of the lowest class. It is quick, but depends only on the two extremes, so one outlier changes it. Used in quality control charts, stock price and temperature reports.

Example 1. For 5, 8, 12, 15, 20 find the range and its coefficient. Solution. R = 20 − 5 = 15. Coefficient = 15/(20 + 5) = 15/25. Answer: 15; 0.6

2. Quartile deviation. It ignores the lowest 25% and highest 25% of values, so it is not affected by extreme values and can be found for open-end classes. In individual and discrete series, Q1 = size of the (n + 1)/4 th item and Q3 = size of the 3(n + 1)/4 th item after arranging in order.

Example 2. Find QD and its coefficient for 20, 28, 40, 12, 30, 15, 50, 33, 25, 45, 18. Solution. In order: 12, 15, 18, 20, 25, 28, 30, 33, 40, 45, 50; n = 11. Q1 = (11 + 1)/4 = 3rd item = 18. Q3 = 3 × 12/4 = 9th item = 40. QD = (40 − 18)/2 = 11. Coefficient = (40 − 18)/(40 + 18) = 22/58. Answer: QD 11; coefficient ≈ 0.379

For a continuous series, Q1 = L + ((N/4 − cf)/f) × h and Q3 = L + ((3N/4 − cf)/f) × h, where L is the lower limit of the quartile class, cf the cumulative frequency before it, f its frequency and h its width.

Example 3. Find QD and its coefficient.

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

Solution. N = 40; cumulative frequencies 5, 13, 25, 35, 40. Q1: N/4 = 10th item lies in 10–20. Q1 = 10 + ((10 − 5)/8) × 10 = 16.25. Q3: 3N/4 = 30th item lies in 30–40. Q3 = 30 + ((30 − 25)/10) × 10 = 35. QD = (35 − 16.25)/2 = 9.375. Coefficient = 18.75/51.25. Answer: QD 9.375; coefficient ≈ 0.366

3. Mean deviation. The average of the absolute deviations (signs ignored) from the mean or median. MD = Σ|x − A|/n, or Σf|x − A|/N for a frequency table (with mid-points for classes).

Example 4. For 5, 8, 12, 15, 20 find MD about the mean and its coefficient. Solution. x̄ = 60/5 = 12. |x − 12| = 7, 4, 0, 3, 8; total 22. MD = 22/5 = 4.4. Coefficient = 4.4/12. Answer: MD 4.4; coefficient ≈ 0.367

Example 5. Find MD about the median.

x	2	4	6	8	10
f	2	4	8	4	2

Solution. N = 20, median = 6 (the table is symmetric). Σf|x − 6| = 2(4) + 4(2) + 8(0) + 4(2) + 2(4) = 32. MD = 32/20. Answer: 1.6

4. Standard deviation and variance.

Individual series: σ = √(Σ(x − x̄)²/n) = √(Σx²/n − x̄²).
Assumed mean (short-cut) method: with d = x − A, σ = √(Σfd²/N − (Σfd/N)²).
Step deviation method: with d′ = (x − A)/h, σ = h × √(Σfd′²/N − (Σfd′/N)²).

Example 6. For 5, 8, 12, 15, 20 find variance, SD and CV. Solution. x̄ = 12. Squared deviations 49, 16, 0, 9, 64; total 138. σ² = 138/5 = 27.6; σ = √27.6 ≈ 5.254. CV = 5.254/12 × 100. Answer: variance 27.6; SD ≈ 5.25; CV ≈ 43.8%

Example 7. Find the mean, SD and CV for the marks table of Example 3 by the step deviation method. Solution. Mid-points 5, 15, 25, 35, 45; A = 25, h = 10, d′ = −2, −1, 0, 1, 2.

m	f	d′	fd′	fd′²
5	5	−2	−10	20
15	8	−1	−8	8
25	12	0	0	0
35	10	1	10	10
45	5	2	10	20

Σfd′ = 2, Σfd′² = 58, N = 40. x̄ = 25 + (2/40) × 10 = 25.5. σ = 10 × √(58/40 − (2/40)²) = 10 × √(1.45 − 0.0025) = 10 × √1.4475 ≈ 12.03. CV = 12.03/25.5 × 100. Answer: mean 25.5; SD ≈ 12.03; CV ≈ 47.2%

5. Coefficient of variation and consistency.

Example 8. Factory A pays an average daily wage of ₹300 with SD ₹15; factory B pays ₹500 with SD ₹20. Where are wages more uniform? Solution. CV(A) = 15/300 × 100 = 5%; CV(B) = 20/500 × 100 = 4%. Answer: factory B (lower CV), even though its SD is larger.

6. Properties of SD.

It is the most widely used measure: rigidly defined, based on all values, and suited to algebra (combined SD, sampling theory).
Σ(x − x̄)² is least when deviations are taken from the mean, so SD about the mean is the smallest root-mean-square deviation.
Independent of change of origin, not of scale.
σ = 0 only when all values are equal.

Deeper

1. Combined standard deviation of two groups. σ₁₂ = √[(n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)) / (n₁ + n₂)], where d₁ = x̄₁ − x̄₁₂ and d₂ = x̄₂ − x̄₁₂, and x̄₁₂ = (n₁x̄₁ + n₂x̄₂)/(n₁ + n₂).

Example 9. Section A: 50 students, mean 60, SD 8. Section B: 50 students, mean 50, SD 6. Find the combined SD. Solution. x̄₁₂ = (3,000 + 2,500)/100 = 55; d₁ = 5, d₂ = −5. σ₁₂² = [50(64 + 25) + 50(36 + 25)]/100 = (4,450 + 3,050)/100 = 75. Answer: σ₁₂ = √75 ≈ 8.66

2. Correcting a wrong observation.

Example 10. 20 observations have mean 40 and SD 5. One value, 43, was wrongly copied as 53. Find the correct mean and SD. Solution. Wrong Σx = 20 × 40 = 800; correct Σx = 800 − 53 + 43 = 790; correct mean = 39.5. Wrong Σx² = n(σ² + x̄²) = 20(25 + 1,600) = 32,500; correct Σx² = 32,500 − 2,809 + 1,849 = 31,540. Correct variance = 31,540/20 − 39.5² = 1,577 − 1,560.25 = 16.75. Answer: mean 39.5; SD = √16.75 ≈ 4.09

3. Mean deviation of a grouped table. For the marks table (x̄ = 25.5), Σf|m − 25.5| = 5(20.5) + 8(10.5) + 12(0.5) + 10(9.5) + 5(19.5) = 385, so MD about the mean = 385/40 = 9.625.

4. Useful inequalities and relations.

For any data, MD about the mean ≤ SD, and SD ≤ Range/2.
For a normal distribution, approximately QD = (2/3)σ and MD = (4/5)σ, so QD : MD : SD ≈ 10 : 12 : 15 (exact values 0.6745σ and 0.7979σ).
Variance of a sample: these notes, like most degree texts, divide by n. The unbiased sample estimate divides by n − 1; the question will say so when it is wanted.

5. Lorenz curve (M.O. Lorenz, 1905). Plot cumulative percentages of the variable (income) on the Y axis against cumulative percentages of the units (persons) on the X axis, both from 0 to 100. The diagonal is the line of equal distribution. The farther the curve sags below the diagonal, the greater the inequality. The Gini coefficient = area between the curve and the diagonal ÷ whole area under the diagonal (0 = perfect equality, 1 = perfect inequality).

6. Merits and demerits.

Measure	Merit	Demerit
Range	Simplest, quick	Only two extremes
QD	Not hit by extremes; open-end classes	Ignores half the data
MD	Uses all values; easy meaning	Ignores signs; poor algebra
SD	Uses all values; algebraic	Pulled by extremes; harder

Common traps

✗ "Coefficient of range = (L − S)/L." ✓ It is (L − S)/(L + S).
✗ "QD = Q3 − Q1." ✓ That is the interquartile range; QD = (Q3 − Q1)/2.
✗ "Mean deviation is least about the mean." ✓ MD is least about the median; the sum of squared deviations is least about the mean.
✗ "Higher SD always means less consistent." ✓ Compare CV when the means differ.
✗ "SD changes when 10 is added to every value." ✓ SD is independent of origin; it changes only with scale.
✗ "Multiplying by 3 triples the variance." ✓ Variance becomes 9 times; SD triples.
✗ "Variance = (Σx)²/n − x̄²." ✓ It is Σx²/n − x̄².
✗ "Step deviation: σ = √(Σfd′²/N − (Σfd′/N)²)." ✓ Multiply by the class width: σ = h × √(…).

Quick revision

Range = L − S; coefficient (L − S)/(L + S).
QD = (Q3 − Q1)/2; coefficient (Q3 − Q1)/(Q3 + Q1).
Q1 = (n + 1)/4 th item; Q3 = 3(n + 1)/4 th item.
MD = Σ|x − A|/n; least about the median.
σ² = Σx²/n − x̄²; step deviation σ = h√(Σfd′²/N − (Σfd′/N)²).
CV = σ/x̄ × 100; lower CV = more consistent.
Origin change: no effect; scale × k: SD × |k|, variance × k².
Combined σ² = [n₁(σ₁² + d₁²) + n₂(σ₂² + d₂²)]/(n₁ + n₂).
Normal: QD ≈ (2/3)σ, MD ≈ (4/5)σ.
Lorenz curve: graphic measure; farther from diagonal = more inequality.

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

Mains writing Mains writing

"Standard deviation is the best measure of dispersion." Discuss (framework):

Define dispersion and the need for it: an average alone hides how reliable it is.
List the requisites of a good measure (Yule): rigidly defined, easy, based on all values, amenable to algebra, little affected by sampling fluctuation and extremes.
Test each measure against them (table above); SD meets all except resistance to extreme values.
Note where others win: QD for open-end or skewed data (income), range in quality control.
Conclude: SD is the best general measure, and with the mean gives CV, the standard tool for comparing consistency.

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8.1.1 Dispersion: range, quartile deviation, mean deviation, standard deviation, variance and coefficient of variation

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