7 Measures of Central Tendency · 7.1 Measures of central tendency
7.1.2 Median, mode and partition values (quartiles, deciles, percentiles) for raw and grouped data
The median splits ordered data into two equal halves, quartiles, deciles and percentiles split it into 4, 10 and 100 equal parts, and the mode is the value that occurs most often; for grouped data each is found by interpolation inside its class.
- APPSC Paper I
- Advanced level
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Core facts
- Median, quartiles, deciles, percentiles are positional averages (partition values); data must be arranged in order first.
- Number of dividing points: quartiles 3 (Q₁, Q₂, Q₃), deciles 9, percentiles 99.
- Equalities: Median = Q₂ = D₅ = P₅₀; Q₁ = P₂₅; Q₃ = P₇₅; D₁ = P₁₀.
- Individual and discrete series (textbook convention):
- median = size of the ((N + 1)/2)th item;
- Q₁ = ((N + 1)/4)th item, Q₃ = (3(N + 1)/4)th item;
- Dᵢ = (i(N + 1)/10)th item, Pᵢ = (i(N + 1)/100)th item;
- in a discrete series, find the first cumulative frequency (cf) that equals or exceeds the position.
- Continuous series (use N/2, N/4 … without the +1):
- Median = l + [(N/2 − c) ÷ f] × h;
- Q₁ = l + [(N/4 − c) ÷ f] × h, Q₃ = l + [(3N/4 − c) ÷ f] × h;
- Dᵢ = l + [(iN/10 − c) ÷ f] × h, Pᵢ = l + [(iN/100 − c) ÷ f] × h;
- l = lower limit of the class containing the value, c = cf of the class before it, f = its frequency, h = class width.
- Mode (Z): the value with the highest frequency.
- continuous series: Z = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h, also written l + [Δ₁ ÷ (Δ₁ + Δ₂)] × h with Δ₁ = f₁ − f₀, Δ₂ = f₁ − f₂;
- f₁ = modal class frequency, f₀ = class before, f₂ = class after.
- Empirical relation (Karl Pearson), moderately skewed data: Mode = 3 Median − 2 Mean.
- Graphs: the median is the x-value where the less-than and more-than ogives cross (or where N/2 meets the less-than ogive); the mode is located from the histogram.
- Median and mode can be found for open-end classes; the mean cannot without assumptions.
- Σ|x − A| is least when A = median.
Example 1. Find the median of 15, 22, 8, 30, 19, 27, 11. Solution. Sorted: 8, 11, 15, 19, 22, 27, 30. N = 7, the 4th item. Answer: 19
Example 2. Find the median of 5, 12, 7, 20, 15, 9. Solution. Sorted: 5, 7, 9, 12, 15, 20. The 3.5th item: (9 + 12)/2. Answer: 10.5
Explained
1. Quartiles of raw data.
Example 3. For 3, 5, 7, 8, 10, 12, 14, 15, 18, 20, 22 find Q₁, the median and Q₃. Solution. N = 11 (already sorted). Q₁ = (12/4)th = 3rd item; median = 6th item; Q₃ = (36/4)th = 9th item. Answer: Q₁ = 7, median = 12, Q₃ = 18
2. Discrete series.
Example 4. Find the median, Q₁, Q₃ and mode.
| x | 10 | 20 | 30 | 40 | 50 |
|---|---|---|---|---|---|
| f | 4 | 7 | 12 | 9 | 8 |
| cf | 4 | 11 | 23 | 32 | 40 |
Solution. N = 40.
- Median: (41/2) = 20.5th item; first cf ≥ 20.5 is 23 → 30.
- Q₁: (41/4) = 10.25th item; first cf ≥ 10.25 is 11 → 20.
- Q₃: 3 × 41/4 = 30.75th item; first cf ≥ 30.75 is 32 → 40.
- Mode: highest frequency 12 → 30.
3. Continuous series: one table, every measure.
Example 5.
| Class | f | cf |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 8 | 13 |
| 20–30 | 15 | 28 |
| 30–40 | 20 | 48 |
| 40–50 | 12 | 60 |
| 50–60 | 10 | 70 |
N = 70 (the mean of this table is 33).
| Measure | Position | Class (l, c, f) | Value |
|---|---|---|---|
| Median | N/2 = 35 | 30–40 (30, 28, 20) | 30 + (7/20) × 10 = 33.5 |
| Q₁ | N/4 = 17.5 | 20–30 (20, 13, 15) | 20 + (4.5/15) × 10 = 23 |
| Q₃ | 3N/4 = 52.5 | 40–50 (40, 48, 12) | 40 + (4.5/12) × 10 = 43.75 |
| D₇ | 7N/10 = 49 | 40–50 (40, 48, 12) | 40 + (1/12) × 10 = 40.83 |
| P₉₀ | 90N/100 = 63 | 50–60 (50, 60, 10) | 50 + (3/10) × 10 = 53 |
Mode: modal class 30–40 (f₁ = 20, f₀ = 15, f₂ = 12). Z = 30 + [5 ÷ (40 − 27)] × 10 = 30 + 3.85. Answer: about 33.85 Check: 3 × 33.5 − 2 × 33 = 34.5, close to 33.85, as expected for nearly symmetric data.
4. Empirical relation.
Example 6. In a moderately skewed distribution the mean is 52 and the mode is 46. Find the median. Solution. Median = (2 Mean + Mode) ÷ 3 = (104 + 46) ÷ 3. Answer: 50
5. Mode when the modal class is the first class. Take the frequency of the missing class before it as 0.
Example 7. Classes 0–20, 20–40, 40–60 with frequencies 30, 20, 10. Find the mode. Solution. f₁ = 30, f₀ = 0, f₂ = 20. Z = 0 + [30 ÷ (60 − 0 − 20)] × 20. Answer: 15
| Feature | Median | Mode |
|---|---|---|
| Type | positional | most frequent value |
| Open-end classes | can be found | can be found |
| Best for | skewed data (incomes, wages) | most common size, fashion, qualitative data |
| Weakness | ignores the size of extreme items | may be ill-defined or bimodal |
Deeper
1. Inclusive classes. Convert to exclusive (true) class limits before using any formula: subtract half the gap (0.5 here) from lower limits and add it to upper limits.
Example 8. Find the median and mode.
| Class | 10–19 | 20–29 | 30–39 | 40–49 |
|---|---|---|---|---|
| f | 6 | 14 | 10 | 5 |
Solution. True classes 9.5–19.5, 19.5–29.5, … N = 35, N/2 = 17.5, median class 19.5–29.5 (c = 6, f = 14).
- Median = 19.5 + (11.5/14) × 10 = 27.71.
- Mode = 19.5 + [8 ÷ (28 − 6 − 10)] × 10 = 26.17.
Using 20 instead of 19.5 as l is the usual mistake and gives a wrong answer.
2. Missing frequency from a given median.
Example 9. The median of the distribution is 27. Find p.
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| f | 5 | p | 20 | 15 | 5 |
Solution. N = 45 + p. The median 27 lies in 20–30, so c = 5 + p, f = 20. 27 = 20 + [((45 + p)/2 − (5 + p)) ÷ 20] × 10 gives 14 = (35 − p)/2, so p = 7 (N = 52; check: 20 + (26 − 12)/20 × 10 = 27). Answer: 7
3. Ill-defined mode. When two or more values share the highest frequency, or the highest frequency is near the ends, the mode is located by the grouping method (frequencies added in twos and threes, the value occurring most often in the grouping table is the mode) or, failing that, by the empirical relation.
4. Unequal class widths. Before applying the mode formula, adjust frequencies to a common width (frequency density), since a wider class collects more items only because it is wider.
5. Graphical location.
- Less-than ogive: plot cf against upper limits; read the x-value at height N/2 for the median, N/4 for Q₁, 3N/4 for Q₃.
- Both ogives cross at the median.
- Histogram: join the top-right corner of the modal bar to the top-right corner of the bar before it, and the top-left corner of the modal bar to the top-left corner of the bar after it; drop a perpendicular from where these lines cross to the x-axis; its foot is the mode.
6. Uses of partition values. Quartiles give the quartile deviation (Q₃ − Q₁)/2 and Bowley's skewness; percentiles are used for ranks in competitive exams and growth charts (the 90th percentile means 90% scored at or below that value).
Common traps
- ✗ "Median of 15, 22, 8, 30, 19 is 8, the middle value as written." ✓ Arrange in order first: median 19.
- ✗ "In the grouped median formula, c is the cf of the median class." ✓ c is the cf of the class before it.
- ✗ "Mode = 3 Mean − 2 Median." ✓ Mode = 3 Median − 2 Mean.
- ✗ "There are 4 quartiles and 10 deciles." ✓ 3 quartiles, 9 deciles, 99 percentiles.
- ✗ "For a continuous series the median class is found at (N + 1)/2." ✓ Use N/2 for continuous series.
- ✗ "With inclusive classes 20–29, l = 20." ✓ Use the true limit 19.5.
- ✗ "The median is located where the less-than ogive meets the x-axis." ✓ It is where the two ogives intersect.
Quick revision
- Arrange data first: positional averages.
- Raw/discrete: median (N + 1)/2 th, Q₁ (N + 1)/4 th, Q₃ 3(N + 1)/4 th item.
- Continuous: l + [(position − c)/f] × h, position N/2, N/4, 3N/4, iN/10, iN/100.
- Median = Q₂ = D₅ = P₅₀.
- 3 quartiles, 9 deciles, 99 percentiles.
- Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h.
- Mode = 3 Median − 2 Mean.
- Ogives cross at the median; histogram gives the mode.
- Inclusive classes: convert to true limits.
- Median and mode work with open-end classes.
Practise this topic. Questions on Median, mode, quartiles, deciles & percentiles, with answers explained.
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