Measures of Central Tendency and Partition Values

A single number that stands for a whole distribution is an average. This chapter teaches the three averages the syllabus names — mean, median and mode — for raw, discrete and grouped data, their key properties, the weighted and combined mean, the geometric and harmonic means that appear alongside them, and the partition values: quartiles, deciles and percentiles. Most Paper-II questions here are short computations, so the chapter puts the formulas and the class-boundary traps side by side.

What an average is for

An average answers 'what is a typical value?' The arithmetic mean balances the data like a see-saw: the deviations on either side cancel. The median is the middle value once the data are arranged, so half lie below it and half above. The mode is the value that occurs most often. On a symmetric distribution all three coincide; on a skewed one they separate, and that separation is itself information (Chapter 4).

Choosing an average
Average	Uses every value?	Hurt by extreme values?	Open-end classes?
Mean	yes	yes	cannot be found
Median	no, only position	no	can be found
Mode	no	no	can be found

The arithmetic mean and its properties

For raw data x̄ = Σx ÷ n; for a frequency distribution x̄ = Σfx ÷ N, with x the mid-value of each class. The assumed-mean (short-cut) method takes d = x − A and uses x̄ = A + Σfd ÷ N; the step-deviation method divides further by the width h: u = (x − A) ÷ h and x̄ = A + h·Σfu ÷ N.

Worked grouped data (N = 50)
Class	f	Mid-value x	fx	Cumulative f
0–10	5	5	25	5
10–20	8	15	120	13
20–30	15	25	375	28
30–40	12	35	420	40
40–50	10	45	450	50

Σfx = 1390, so x̄ = 1390 ÷ 50 = 27.8. Three properties are asked again and again:
(1) the sum of deviations from the mean is zero, Σ(x − x̄) = 0;
(2) the sum of squared deviations is least when taken about the mean;
(3) the mean changes with origin and scale — if y = a + bx then ȳ = a + b·x̄.

The combined mean of two groups is x̄ = (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂). The weighted mean is Σwx ÷ Σw. For positive numbers the geometric mean is the n-th root of their product and the harmonic mean is n ÷ Σ(1/x); always AM ≥ GM ≥ HM, with equality only when all values are equal, and for two numbers GM² = AM × HM. The harmonic mean is the right average for speeds over equal distances.

Median and mode for grouped data

For raw data, arrange in order: with n odd the median is the ((n + 1)/2)-th value; with n even it is the mean of the (n/2)-th and (n/2 + 1)-th values. For grouped data first find the median class — the first class whose cumulative frequency reaches N/2 — then interpolate: Median = L + ((N/2 − c.f.) ÷ f) × h, where L is the lower boundary of the median class, c.f. the cumulative frequency before it, f its frequency and h its width.

In the worked table N/2 = 25. Cumulative frequencies are 5, 13, 28, …, so the median class is 20–30 with c.f. = 13 and f = 15: Median = 20 + (25 − 13)/15 × 10 = 28. For the mode, take the modal class (highest frequency, 20–30) and Mode = L + ((f₁ − f₀) ÷ (2f₁ − f₀ − f₂)) × h = 20 + (15 − 8)/(30 − 8 − 12) × 10 = 27. When the mode is ill-defined, the empirical relation Mode ≈ 3 Median − 2 Mean is used for a moderately skewed distribution.

⚠️ Exam trap

Use the lower boundary, not the lower limit, of the median or modal class. For an inclusive class 20–29, L is 19.5. Using 20 shifts every answer by 0.5 — and the wrong option is usually waiting for you.

Quartiles, deciles and percentiles

The median splits the data in two; partition values split it further. Three quartiles Q₁, Q₂, Q₃ cut it into four equal parts, nine deciles D₁…D₉ into ten, and ninety-nine percentiles P₁…P₉₉ into a hundred. So Q₂ = D₅ = P₅₀ = median, Q₁ = P₂₅ and Q₃ = P₇₅. For raw data, Qₖ is the (k(n + 1)/4)-th value in order. For grouped data the median formula is reused with kN/4, kN/10 or kN/100 in place of N/2.

In the worked table N/4 = 12.5 falls in 10–20 (c.f. before it 5, f = 8), so Q₁ = 10 + (12.5 − 5)/8 × 10 = 19.38; 3N/4 = 37.5 falls in 30–40 (c.f. 28, f = 12), so Q₃ = 30 + (37.5 − 28)/12 × 10 = 37.92. These feed directly into the quartile deviation of Chapter 3 and Bowley's skewness of Chapter 4.

🧠 Memory trick

Quartile ↔ 4, decile ↔ 10, percentile ↔ 100: the k-th one sits at position kN/4, kN/10, kN/100.

Exam traps and a quick revision

Σ(x − x̄) = 0 always; Σ(x − A)² is least when A = x̄; Σ|x − A| is least when A is the median.
If every value is multiplied by b and increased by a, the mean, median and mode all become a + b × (old value).
Median and mode can be found for open-end classes; the mean cannot.
Mode ≈ 3 Median − 2 Mean, for a moderately skewed distribution only.
AM ≥ GM ≥ HM; for two positive numbers GM² = AM × HM.

Key takeaways

The mean uses every value and balances the deviations; the median uses position; the mode uses frequency.
Grouped median = L + ((N/2 − c.f.) ÷ f) × h, with L the lower boundary of the class containing N/2.
Grouped mode = L + ((f₁ − f₀) ÷ (2f₁ − f₀ − f₂)) × h; empirically Mode ≈ 3 Median − 2 Mean.
Combined mean = (n₁x̄₁ + n₂x̄₂) ÷ (n₁ + n₂); AM ≥ GM ≥ HM.
Q₂ = D₅ = P₅₀ = median; partition values reuse the median formula at kN/4, kN/10, kN/100.

Practice questions (10)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The sum of the deviations of a set of values from their arithmetic mean is always:
1. a minimum but not zero
2. equal to the variance
3. zero
4. one
Show answer
Answer: C — zero
The mean is the balance point, so positive and negative deviations cancel exactly: Σ(x − x̄) = Σx − n·x̄ = 0. It is the sum of SQUARED deviations that is a minimum.
The sum of absolute deviations Σ|x − A| is least when A is the:
1. mean
2. geometric mean
3. median
4. mode
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Answer: C — median
Absolute deviations are minimised about the median; squared deviations are minimised about the mean. The two are a favourite pair to swap.
Which average can be found for a distribution with open-end classes such as 'below 10' and '60 and above'?
1. arithmetic mean
2. geometric mean
3. median
4. harmonic mean
Show answer
Answer: C — median
The median needs only the class containing N/2 and the frequencies before it, so open ends do not matter. Every mean needs the mid-value of every class, which an open-end class does not have.
Two sections have 30 and 20 students with mean marks 60 and 70. The combined mean is:
1. 62
2. 64
3. 65
4. 66
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Answer: B — 64
Combined mean = (30 × 60 + 20 × 70) ÷ 50 = (1800 + 1400) ÷ 50 = 64. The plain average of 60 and 70 is 65 and ignores the unequal group sizes.
In a moderately skewed distribution the mean is 40 and the median is 38. The approximate mode is:
1. 42
2. 39
3. 36
4. 34
Show answer
Answer: D — 34
Mode ≈ 3 × Median − 2 × Mean = 114 − 80 = 34. The mode lies on the far side of the median from the mean.
For the distribution 0–10 (5), 10–20 (8), 20–30 (15), 30–40 (12), 40–50 (10), the median is:
1. 30
2. 25
3. 28
4. 27
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Answer: C — 28
N = 50, N/2 = 25; cumulative frequencies 5, 13, 28 put the median in 20–30 with c.f. = 13, f = 15. Median = 20 + (25 − 13)/15 × 10 = 28. 27 is the mode of the same table.
For the same distribution 0–10 (5), 10–20 (8), 20–30 (15), 30–40 (12), 40–50 (10), the mode is:
1. 26
2. 25
3. 28
4. 27
Show answer
Answer: D — 27
Modal class 20–30 with f₁ = 15, f₀ = 8, f₂ = 12. Mode = 20 + (15 − 8)/(30 − 8 − 12) × 10 = 20 + 7 = 27. 25 is the class mid-value, which is not the mode.
For two positive numbers the arithmetic mean is 25 and the harmonic mean is 16. Their geometric mean is:
1. 41
2. 20
3. 400
4. 20.5
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Answer: B — 20
For two numbers GM² = AM × HM = 25 × 16 = 400, so GM = 20. Check: 20 lies between 16 and 25, as AM ≥ GM ≥ HM requires.
Which of the following is the same as the median?
1. the first quartile, Q₁
2. the fifth decile, D₅
3. the fifth percentile, P₅
4. the third quartile, Q₃
Show answer
Answer: B — the fifth decile, D₅
The median splits the data into halves, which is Q₂ = D₅ = P₅₀. P₅ is only the 5% point, a common misreading of 'fifth'.
Every observation of a data set is multiplied by 3 and then 5 is subtracted. If the old median was 12, the new median is:
1. 36
2. 7
3. 31
4. 41
Show answer
Answer: C — 31
A linear change y = 3x − 5 keeps the order of the values when the multiplier is positive, so the middle value transforms the same way: 3 × 12 − 5 = 31.