Arithmetic, weighted, geometric and harmonic means…

Core facts

Arithmetic mean (AM), written x̄:
- individual series: x̄ = Σx ÷ n;
- discrete series: x̄ = Σfx ÷ N, N = Σf;
- continuous series: x = class mid-value, same formula;
- short-cut (assumed mean A, d = x − A): x̄ = A + Σfd ÷ N;
- step deviation (d′ = (x − A)/h): x̄ = A + (Σfd′ ÷ N) × h.
Properties of AM:
- Σ(x − x̄) = 0: deviations from the mean add up to zero;
- Σ(x − A)² is least when A = x̄;
- combined mean x̄₁₂ = (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂);
- change of origin and scale: if u = (x − A)/h then x̄ = A + h ū.
Weighted AM: x̄w = Σwx ÷ Σw. With equal weights it equals the simple AM.
Geometric mean (GM): the nth root of the product of n values.
- GM = (x₁ × x₂ × … × xₙ)^(1/n);
- by logarithms: GM = antilog (Σ log x ÷ n); grouped: antilog (Σf log x ÷ N);
- weighted: antilog (Σw log x ÷ Σw).
Harmonic mean (HM): the reciprocal of the AM of reciprocals.
- HM = n ÷ Σ(1/x); grouped: HM = N ÷ Σ(f/x).
Relation: for positive values AM ≥ GM ≥ HM; all three are equal only when every value is the same.
Two positive numbers a and b: AM = (a + b)/2, GM = √(ab), HM = 2ab/(a + b), and GM² = AM × HM.
Uses: AM for ordinary averages; GM for average rates of growth, ratios, percentages and index numbers; HM for average speed over equal distances and average price when equal amounts of money are spent.
GM and HM cannot be used when any value is zero; GM is not used with negative values.

Example 1. Find the AM of 12, 18, 25, 30 and 15. Solution. Σx = 100, n = 5. Answer: 20

Example 2. Find the GM of 4, 9 and 6. Solution. Product = 216; cube root of 216. Answer: 6

Example 3. Find the HM of 2, 3 and 6. Solution. Σ(1/x) = 1/2 + 1/3 + 1/6 = 1. HM = 3 ÷ 1. Answer: 3

Explained

1. AM of a continuous series by step deviation. Choose A as a central mid-value and h as the class width so that d′ becomes small whole numbers.

Example 4. Find the mean.

Class	Mid-value x	f	d′ = (x − 25)/10
0–10	5	6	−2
10–20	15	10	−1
20–30	25	14	0
30–40	35	12	1
40–50	45	8	2

Solution. fd′ = −12, −10, 0, 12, 16, so Σfd′ = 6 and N = 50. x̄ = 25 + (6 ÷ 50) × 10 = 25 + 1.2. Answer: 26.2 (direct method: Σfx = 1,310; 1,310 ÷ 50 = 26.2.)

2. Weighted mean. Use it when items are not equally important (subjects with different credits, commodities in an index).

Example 5. A student scores 60, 75 and 80 in three papers with weights 2, 3 and 5. Find the weighted mean. Solution. Σwx = 120 + 225 + 400 = 745; Σw = 10. Answer: 74.5 (the simple mean would be 71.67).

3. Combined mean.

Example 6. Section A has 40 students with mean 52; section B has 60 students with mean 47. Find the mean of all 100. Solution. (40 × 52 + 60 × 47) ÷ 100 = (2,080 + 2,820) ÷ 100. Answer: 49

4. Corrected mean. Correct the total, then divide again.

Example 7. The mean of 50 observations is 36. Later it was found that 43 was wrongly copied as 34. Find the correct mean. Solution. Wrong total 1,800. Correct total = 1,800 − 34 + 43 = 1,809. 1,809 ÷ 50. Answer: 36.18

5. GM for average growth. Growth rates multiply, so their average is the GM of the growth factors (1 + r), not the AM of the rates.

Example 8. Production rose by 8%, 12% and 25% in three successive years. Find the average annual rate of growth. Solution. Factors 1.08, 1.12, 1.25; product = 1.512. GM = 1.512^(1/3) = 1.1478 (log route: Σ log = 0.17955, ÷ 3 = 0.05985, antilog = 1.1478). Answer: about 14.78% a year (the AM of the rates, 15%, overstates it).

6. HM for average speed.

Example 9. A car goes from A to B at 60 km/h and returns at 40 km/h. Find the average speed for the round trip. Solution. Equal distances, so use HM: 2 ÷ (1/60 + 1/40) = 2 ÷ (5/120). Answer: 48 km/h (not 50).

Example 10. A buyer spends ₹100 on rice at ₹4 per kg and another ₹100 at ₹5 per kg. Find the average price per kg. Solution. Equal money spent, so HM: 2 ÷ (1/4 + 1/5) = 40/9. Answer: about ₹4.44 per kg (₹200 buys 25 + 20 = 45 kg).

7. The three means side by side.

Example 11. For 2, 3 and 6: AM = 11/3 = 3.67, GM = 36^(1/3) = 3.30, HM = 3. The order AM > GM > HM holds, as it must for unequal positive values.

Mean	Best for	Gives most weight to	Fails when
AM	ordinary totals shared equally	large values	extreme values distort it
GM	ratios, growth rates, index numbers	large values less than the AM does	any value is zero or negative
HM	speeds, rates per unit (equal amounts)	small values	any value is zero

Deeper

1. Why AM ≥ GM ≥ HM (two numbers). (√a − √b)² ≥ 0 gives a + b ≥ 2√(ab), so AM ≥ GM. Since HM = GM² ÷ AM and AM ≥ GM, HM ≤ GM. Equality holds only when a = b.

Example 12. The AM of two numbers is 25 and their HM is 16. Find the GM and the numbers. Solution. GM = √(25 × 16) = 20. So a + b = 50 and ab = 400, giving 40 and 10 (check: HM = 2 × 400 ÷ 50 = 16). Answer: GM 20; numbers 40 and 10

2. Grouped GM and HM.

Example 13. Find the AM, GM and HM.

x	5	10	20	40
f	2	3	3	2

Solution. N = 10.

AM = (10 + 30 + 60 + 80) ÷ 10 = 18.
log x = 0.6990, 1.0000, 1.3010, 1.6021; Σf log x = 11.5051; ÷ 10 = 1.15051; antilog = 14.14 (exactly √200).
Σ(f/x) = 0.4 + 0.3 + 0.15 + 0.05 = 0.9; HM = 10 ÷ 0.9 = 11.11.

Answer: AM 18, GM 14.14, HM 11.11

3. Weighted HM (unequal amounts). When different amounts are spent or distances differ, weight each rate by the amount or distance.

Example 14. ₹100 is spent on a commodity at ₹4 per kg and ₹200 at ₹5 per kg. Find the average price. Solution. Quantity = 25 + 40 = 65 kg for ₹300. Weighted HM = 300 ÷ (100/4 + 200/5). Answer: about ₹4.62 per kg

Example 15. A cyclist covers 30 km at 30 km/h and then 90 km at 45 km/h. Find the average speed. Solution. Time = 1 h + 2 h = 3 h for 120 km. Answer: 40 km/h

4. Missing frequency.

Example 16. The mean of the distribution is 25. Find p.

x	10	20	30	40
f	5	p	10	5

Solution. (50 + 20p + 300 + 200) ÷ (20 + p) = 25 gives 550 + 20p = 500 + 25p, so p = 10. Answer: 10

5. Useful properties of GM.

log GM = mean of the logs, so GM suits data that grow by multiplication.
GM of the ratios of two series = ratio of their GMs; this is why the GM index satisfies the time reversal test (see index numbers).
Multiplying every value by k multiplies the AM, GM and HM by k; adding k changes the AM by k but not the GM or HM by k.

6. Merits and limits of the AM (textbook list). Rigidly defined, based on all items, capable of algebraic treatment, least affected by sampling fluctuations; but distorted by extreme values, cannot be found for open-end classes without assumptions, and cannot be located by inspection or graph.

Common traps

✗ "Average growth rate = AM of the yearly rates." ✓ Use the GM of the factors (1 + r), then subtract 1.
✗ "Average speed over equal distances = (60 + 40)/2 = 50." ✓ Use the HM: 48 km/h.
✗ "HM ≥ GM ≥ AM." ✓ AM ≥ GM ≥ HM for positive values.
✗ "GM² = AM + HM." ✓ For two numbers GM² = AM × HM.
✗ "Σ(x − x̄)² is zero." ✓ Σ(x − x̄) = 0; Σ(x − x̄)² is the minimum sum of squares.
✗ "In step deviation, x̄ = A + Σfd′ ÷ N." ✓ Multiply by the class width: x̄ = A + (Σfd′ ÷ N) × h.
✗ "Combined mean = average of the two means." ✓ Weight by sizes: (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂).

Quick revision

AM = Σfx ÷ N; step deviation A + (Σfd′/N) × h.
Σ(x − x̄) = 0; Σ(x − A)² least at A = x̄.
Weighted AM = Σwx ÷ Σw; combined mean weights by n.
GM = (product)^(1/n) = antilog (Σf log x ÷ N).
HM = N ÷ Σ(f/x).
AM ≥ GM ≥ HM; equal only if all values equal.
Two numbers: GM² = AM × HM.
GM: growth rates, ratios, index numbers.
HM: speeds over equal distances, prices with equal spend.
Zero value: no GM or HM.

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7.1.1 Arithmetic, weighted, geometric and harmonic means and their relation

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Core facts

Explained

Deeper

Common traps

Quick revision