Median, quartiles and mode
Business Statistics · Unit II · Topic 5 of 15 · 6 min read
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1. Median
The median is the value of the middle item when the data are arranged in order of size. It divides the series into two equal halves.
- Individual series: arrange in order; median = size of the ((n + 1)/2)th item.
- Discrete series: arrange x in order, find cumulative frequencies; median = size of the ((N + 1)/2)th item, read from the first c.f. that equals or exceeds it.
- Continuous series: find N/2, locate the median class (first c.f. ≥ N/2), and apply Median = L + ((N/2 − c.f.) / f) × h, where L = lower limit of the median class, c.f. = cumulative frequency of the class before it, f = frequency of the median class, h = width.
Worked: discrete (June 2025). x: 10, 15, 8, 20, 18 with f: 24, 6, 30, 16, 26. The x values are not in order, so rearrange first.
| x | f | c.f. |
|---|---|---|
| 8 | 30 | 30 |
| 10 | 24 | 54 |
| 15 | 6 | 60 |
| 18 | 26 | 86 |
| 20 | 16 | 102 |
(N + 1)/2 = 103/2 = 51.5th item. The first c.f. ≥ 51.5 is 54, against x = 10. Median = 10.
Worked: continuous (January 2026). Classes 0–25 to 125–150, f = 12, 25, 40, 37, 28, 18, c.f. = 12, 37, 77, 114, 142, 160 (table in the previous note). N/2 = 80; the first c.f. ≥ 80 is 114, so the median class is 75–100 (L = 75, c.f. = 77, f = 37, h = 25).
Median = 75 + (80 − 77)/37 × 25 = 75 + 2.03 = 77.03.
Worked: inclusive classes (100-mark January 2026). 10–19: 8, 20–29: 12, 30–39: 30, 40–49: 28, 50–59: 40. Convert to 9.5–19.5, 19.5–29.5 … c.f. 8, 20, 50, 78, 118. N/2 = 59, median class 39.5–49.5. Median = 39.5 + (59 − 50)/28 × 10 = 39.5 + 3.21 = 42.71. Using 40 instead of 39.5 as L is the usual mistake.
2. Quartiles (the June 2025 short note)
Quartiles divide the series into four equal parts. Q1 (lower quartile) has 25% of items below it; Q2 is the median; Q3 (upper quartile) has 75% below it. The formula is the median formula with N/4 or 3N/4 in place of N/2. Deciles (tenths) and percentiles (hundredths) follow the same pattern. Quartiles are the basis of the quartile deviation and of Bowley’s skewness.
January 2026 data. Q1: N/4 = 40, class 50–75 (c.f. before 37, f = 40): Q1 = 50 + (40 − 37)/40 × 25 = 50 + 1.875 = 51.88. Q3: 3N/4 = 120, class 100–125 (c.f. before 114, f = 28): Q3 = 100 + (120 − 114)/28 × 25 = 100 + 5.36 = 105.36.
3. Mode
The mode is the value that occurs most often — the point of maximum concentration. In a discrete series it is the x with the highest frequency (if that is clear). In a continuous series, find the modal class (highest frequency) and apply
Mode = L + (f1 − f0) / (2f1 − f0 − f2) × h,
where f1 = modal frequency, f0 = frequency of the class before, f2 = frequency of the class after.
January 2026. Modal class 50–75 (f1 = 40, f0 = 25, f2 = 37). Mode = 50 + (40 − 25)/(80 − 25 − 37) × 25 = 50 + 15/18 × 25 = 50 + 20.83 = 70.83.
Answer for January 2026 Q.2(a): Mean 77.81, Median 77.03, Mode 70.83. Mean > median > mode, so the distribution is slightly positively skewed.
4. More worked sets from the papers
Discrete marks (100-mark January 2026). Marks 20, 30, 40, 50, 60, 70 with f = 8, 12, 20, 10, 6, 4 (N = 60). Σfx = 160 + 360 + 800 + 500 + 360 + 280 = 2,460; mean = 41. c.f. 8, 20, 40, 50, 56, 60; (N + 1)/2 = 30.5th item lies at c.f. 40, so median = 40. Highest frequency 20 at 40, so mode = 40.
Below table (January 2025). Frequencies 2, 3, 4, 3, 2, 1, 4 for 0–10 to 60–70; mean = 35 (previous note). c.f. 2, 5, 9, 12, 14, 15, 19; N/2 = 9.5, median class 30–40: Median = 30 + (9.5 − 9)/3 × 10 = 31.67. The highest frequency (4) occurs twice, at 20–30 and 60–70: the series is bimodal and the mode is ill-defined. Use Mode = 3 Median − 2 Mean = 3(31.67) − 2(35) = 95 − 70 = 25.
Above table (August 2024). Frequencies 3, 5, 7, 10, 12, 15, 12, 6, 2, 8 for 0–10 to 90–100; c.f. 3, 8, 15, 25, 37, 52 … N/2 = 40, median class 50–60: Median = 50 + (40 − 37)/15 × 10 = 52. Modal class 50–60 (f0 = 12, f2 = 12): Mode = 50 + 3/6 × 10 = 55. (Mean = 51.75.)
Weights (January 2026 alternative). Weights 58, 60, 61, 62, 63, 64, 65, 66 with 4, 12, 24, 32, 32, 16, 8, 2 students (N = 130). Σfx = 8,092, mean = 62.25. c.f. 4, 16, 40, 72 … the 65.5th item is 62, so median = 62. Frequencies 32 at both 62 and 63: bimodal, so Mode = 3(62) − 2(62.25) = 186 − 124.49 = 61.51.
5. Merits and demerits
Median. Merits: not affected by extreme values; can be found for open-end classes; can be located graphically (ogives); suitable for qualitative data that can be ranked (intelligence, honesty). Demerits: not based on all items; needs arranging; not capable of algebraic treatment (no combined median); more affected by sampling fluctuations than the mean.
Mode. Merits: simple and popular (the “most common” shoe size or wage); not affected by extremes; can be found graphically (histogram) and for open-end classes. Demerits: ill-defined when the series is bimodal or multimodal; not based on all items; not capable of algebraic treatment; depends heavily on the grouping chosen.
Empirical relation (Karl Pearson, for moderately skewed data): Mode = 3 Median − 2 Mean, or Mean − Mode = 3 (Mean − Median). In a symmetrical distribution all three coincide.
How this is examined
Mean, median and mode from one table: January 2026 (grouped and discrete), January 2025, August 2024, and the 100-mark January 2026 paper, which also set a median from inclusive classes. Discrete median: June 2025. “Write a short note on quartiles”: June 2025. Merits and demerits of mean, median and mode: June 2025 (ten marks). Show the c.f. column every time.
Now try the real questions: Business Statistics past question papers.