Lesson 11: Statistics — Mean, Median, Mode

Statistics in Class X means one big job: extracting a mean, median and mode from *grouped* data, where individual values are lost inside class intervals. NCERT teaches three methods for the mean (direct, assumed-mean, step-deviation — all giving the same answer), the grouped median formula with cumulative frequencies, and the grouped mode formula. The data in this lesson is used consistently across all three measures so you can see how they relate.

The working data set

Marks obtained by 40 students in a test (out of 100):

Class marks (midpoints) xᵢ: 10, 30, 50, 70, 90.

Mean: three roads, one answer

Direct method: x̄ = Σfᵢxᵢ / Σfᵢ. Here Σfᵢxᵢ = 6(10) + 10(30) + 14(50) + 6(70) + 4(90) = 60 + 300 + 700 + 420 + 360 = 1840, and Σfᵢ = 40, so x̄ = 46.

Assumed-mean method: pick a = 50, set dᵢ = xᵢ − a: −40, −20, 0, 20, 40. Σfᵢdᵢ = −240 − 200 + 0 + 120 + 160 = −160. x̄ = a + Σfᵢdᵢ/Σfᵢ = 50 − 4 = 46 ✓.

Step-deviation method: with class width h = 20, uᵢ = dᵢ/h: −2, −1, 0, 1, 2. Σfᵢuᵢ = −12 − 10 + 0 + 6 + 8 = −8. x̄ = a + h × (Σfᵢuᵢ/Σfᵢ) = 50 + 20 × (−8/40) = 50 − 4 = 46 ✓. All three agree because they are algebraically identical; choose step-deviation when the numbers are large.

Median: cumulative frequencies first

Cumulative frequencies: 6, 16, 30, 36, 40. N/2 = 20. The median class is the first whose cumulative frequency exceeds 20: 40–60 (cf jumps from 16 to 30). Then: Median = l + [(N/2 − cf)/f] × h = 40 + [(20 − 16)/14] × 20 = 40 + 40/7 ≈ 45.71.

Mode: the modal class formula

The modal class is 40–60 (highest frequency, f₁ = 14), with neighbours f₀ = 10 and f₂ = 6. Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 40 + [(14 − 10)/(28 − 10 − 6)] × 20 = 40 + (4/12) × 20 ≈ 46.67.

NCERT also records the empirical relationship: Mode ≈ 3 Median − 2 Mean. Check: 3(45.71) − 2(46) = 137.13 − 92 = 45.13 — close to 46.67 but not exact; the relation is an observed approximation for moderately skewed data, not a theorem. Use it only when a question explicitly invokes it, typically to find the third measure from two given ones.

Worked example (NCERT exemplar style): the missing frequency

The mean of a distribution is known instead of a frequency. Suppose classes 0–20, 20–40, 40–60 have frequencies 5, x, 10 and class marks 10, 30, 50, and the mean is 32. Then Σf x = 50 + 30x + 500 = 550 + 30x and Σf = 15 + x. So (550 + 30x)/(15 + x) = 32 → 550 + 30x = 480 + 32x → 70 = 2x → x = 35. Setting up Σfᵢxᵢ/Σfᵢ and solving the linear equation is the entire method.

Competency application: correcting a wrong entry

A teacher computes the mean of 20 observations as 30. Later she finds one observation recorded as 54 should have been 45. Corrected sum = 20 × 30 − 54 + 45 = 591, so the corrected mean is 591/20 = 29.55. Data-cleaning problems like this test whether you understand mean as *total ÷ count* rather than as a memorised formula — always return to the total.

Common mistakes

• Using class limits instead of class marks in the mean. The representative of 20–40 is 30, not 20 or 40.
• Picking the median class by frequency instead of cumulative frequency. The largest fᵢ class is the *modal* class; the median class is found from cf crossing N/2.
• Swapping cf and f in the median formula. cf is the cumulative frequency *before* the median class (16 here), f is the median class's own frequency (14).
• Applying the empirical relation as if exact, then "correcting" a computed mode to match it. Compute each measure by its own formula; the relation is a check, rough at best.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. The median of the data 12, 15, 11, 18, 14 is: (a) 12 (b) 13 (c) 14 (d) 15. Answer: (c) 14. Sorted: 11, 12, 14, 15, 18; the middle (3rd) value is 14.

Q2. For the table in this lesson, the modal class is: (a) 20–40 (b) 40–60 (c) 60–80 (d) 0–20. Answer: (b) 40–60, having the maximum frequency, 14.

Section B — Short answer (3 marks) Q3. The mean of 20 observations was calculated as 30. It was later discovered that one observation, 54, was wrongly recorded instead of 45. Find the correct mean. Marking scheme: 1 mark for the original total = 20 × 30 = 600; 1 mark for the corrected total = 600 − 54 + 45 = 591; 1 mark for the correct mean = 591/20 = 29.55.

Section C — Case-based long answer (5 marks) Q4. For the marks distribution of 40 students given in this lesson (frequencies 6, 10, 14, 6, 4 over classes 0–20, …, 80–100): (i) Compute the mean by the step-deviation method, taking a = 50 and h = 20. (3 marks) (ii) State the median class and the modal class. (1 mark) (iii) The school sets a "typical performance" badge at the mode. A student argues the median is a fairer typical value here. Give one mathematical reason supporting either claim. (1 mark) Marking scheme: (i) uᵢ values −2, −1, 0, 1, 2 (1 mark); Σfᵢuᵢ = −8 (1 mark); x̄ = 50 + 20 × (−8/40) = 46 (1 mark). (ii) Median class 40–60 (cf crosses 20 there); modal class 40–60 (1 mark for both). (iii) Any sound reason, e.g.: the median (45.71) is unaffected by how extreme the top and bottom scores are, while the mode (46.67) simply names the most crowded class — or conversely the mode reflects the largest single group of students (14 of 40) (1 mark).

Going deeper: two missing frequencies

Some problems hide *two* frequencies and give two conditions — the total count and the mean. The values 10, 20, 30, 40, 50 have frequencies 4, x, 12, y, 5 respectively, and the mean of the distribution is 29.25. Find x and y. Total frequency: 4 + x + 12 + y + 5 = 40, so x + y = 19. Σfᵢxᵢ = 40 + 20x + 360 + 40y + 250 = 650 + 20x + 40y. Mean condition: 650 + 20x + 40y = 29.25 × 40 = 1170, so 20x + 40y = 520, i.e. x + 2y = 26. Subtracting the first equation from the second: y = 7, and then x = 12. Two unknowns, two equations — statistics has quietly become Lesson 3 (simultaneous equations). That crossover is why CBSE loves this problem type: it examines two chapters with one question.

Going deeper: ogives and the graphical median

Plotting cumulative frequency against class boundaries gives an ogive. The "less than" ogive (cf plotted at upper boundaries) and the "more than" ogive (at lower boundaries) cross at a point whose x-coordinate is the median — for this lesson's data, both constructions point to ≈ 45.7, matching the formula's 45.71. NCERT asks for ogives mainly as a construction and reading exercise: the median from a graph is approximate, and examiners accept a small tolerance band. The formula value remains the authority when both are computed — if your graph disagrees wildly, the graph is wrong.

Going deeper: which average, when?

• Mean uses every value: right for totals and further algebra, fragile under extreme values (one 100 in a class of low scores drags it up).
• Median uses position only: robust for incomes, house prices, and anything skewed — half the data on each side by construction.
• Mode is the only sensible "average" for categories: the most common shoe size, the most popular bus route.

A one-mark competency favourite: "A factory reports the average wage using the *mean*; workers claim the *median* is lower and more honest. Explain." — the directors' salaries inflate the mean; the median reports the middle worker. The mathematics of this lesson is computation; the *judgement* of this paragraph is what competency questions grade.

Final drills

Drill 1 — a quick direct-method mean. Marks classes 0–10, 10–20, 20–30, 30–40, 40–50 have frequencies 2, 5, 8, 4, 1. Class marks: 5, 15, 25, 35, 45. Σfᵢxᵢ = 10 + 75 + 200 + 140 + 45 = 470; Σfᵢ = 20; mean = 470/20 = 23.5. On small tables the direct method beats the machinery — choose the method by the size of the numbers, not by habit.

Drill 2 — the empirical relation, forward and backward. If the median of a distribution is 30.6 and its mean is 28.4, estimate the mode: Mode ≈ 3(30.6) − 2(28.4) = 91.8 − 56.8 = 35.0. Backwards: given mode 35 and median 30.6, the implied mean is (3 × 30.6 − 35)/2 = 56.8/2 = 28.4 ✓. The relation is one equation linking three quantities — any one follows from the other two, which is exactly how it is examined.