Finding the busiest class, then placing the mode inside it

What to assume they know

• Grouping loses the raw values, so we stand the class mark in for them — that grouping keeps counts and discards values
• The mode of ungrouped data as the value with the largest frequency, from Class IX
• Reading a frequency table and identifying its largest entry
• Substituting into an algebraic formula with several named symbols
• Dividing a segment internally in a given ratio
• Class limits, class size, and what makes classes continuous

What they should be able to do

• Find the mode of an ungrouped data set by tabulating frequencies
• Identify the modal class of a grouped distribution
• Substitute correctly into the mode formula, naming each symbol against the right class
• Rewrite the formula as a ratio and predict which way the mode will lean before computing
• Show that the mode sits at the mid-point of the modal class exactly when the two neighbouring frequencies are equal
• Compute and interpret the mode alongside the mean for the same data
• State the conditions the mode formula requires, and recognise a table that fails them

Where it usually goes wrong

• "The mode is the largest frequency." The largest frequency in Example 6 is 7; the mode is 52. One is a count of students, the other a mark. They are not even in the same units.
• "The mode is the modal class's class mark." Only when the two neighbours are equally busy. In Example 6 the class mark of 40–55 is 47.5 and the mode is 52; in Example 5 the class mark is 4 and the mode is 3.286.
• "The mode always comes out below the mean." Example 6 gives 52 against 62, but Exercise 13.2 Q1 gives 36.82 against 35.375 and Q4 gives 30.625 against 29.21. The chapter's own first Remark says as much, and its own exercises prove it both ways.
• "Once I have the modal class I am nearly done." You have done half the work, and the half that is left depends on data you have not looked at yet — the two neighbouring classes.
• "Any table can go into the formula." It needs classes of one common width, and it needs them continuous. The chapter declines to handle unequal widths at all, and its closing note insists on continuity first. Exercise 13.1 Q5's classes, printed 50–52, 53–55 and so on, would have to be corrected before any mode could be computed from them.
• "An empty neighbouring class breaks the formula." It does not — a zero simply makes that excess as large as it can be, pulling the mode away from the empty side. Exercise 13.2 Q4 has two empty classes and computes cleanly.
• "Grouped data has one mode the way ungrouped data does." The chapter says plainly that grouped data can be multimodal and restricts itself to single-peaked cases. The formula will return a number whatever you feed it; it is the interpretation that needs the restriction.

Questions to check understanding

• Find the mode of a small ungrouped data set
• Identify the modal class and compute the mode of a grouped distribution
• Compute both mode and mean for one distribution and interpret the pair in context — the form Exercise 13.2 uses in questions 1 and 4
• Given a distribution, state which measure a stated purpose calls for
• Convert inclusive classes to continuous ones before finding a mode
• Explain why the mode of grouped data is not simply the largest frequency
• Predict, from the two neighbouring frequencies alone, which half of the modal class the mode will fall in

Examples worth working on the board

Values marked verified are worked out here on the chapter's printed data.

• Example 4, the ungrouped case (p. 183). Ten matches, wickets per match: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3. Verified: tabulating gives 0 once, 1 once, 2 three times, 3 twice, 4 once, 5 once, 6 once — ten in all — so the mode is 2. Nothing is constructed here; the answer is visible in the table.
• Example 5 (pp. 184–185). Twenty households by family size: classes 1–3, 3–5, 5–7, 7–9, 9–11 with counts 7, 8, 2, 2, 1. Verified: the counts total 20; the largest is 8, making 3–5 the modal class, with l = 3, h = 2, the modal count 8, the count below 7 and the count above 2. The mode is 3 + 2 × (8 − 7) ÷ (16 − 7 − 2) = 3 + 2 ÷ 7 = 3.286 to three places.
• Example 5 read as a ratio — the section-7 and section-8 payload, and not in the book. Verified: the two excesses are 8 − 7 = 1 above the class below, and 8 − 2 = 6 above the class above. The mode divides the class 3–5 in the ratio 1 : 6, landing one seventh of the way along a class two units wide — that is 0.286 of a unit past the lower limit, hard against the boundary it shares with the crowded class of seven households.
• Example 6 (p. 185). The Example 1 marks again, from Table 13.3: classes 10–25, 25–40, 40–55, 55–70, 70–85, 85–100 with counts 2, 3, 7, 6, 6, 6. Verified: the largest count is 7, making 40–55 the modal class, with l = 40, h = 15, the modal count 7, the count below 3 and the count above 6. The mode is 40 + 15 × (7 − 3) ÷ (14 − 3 − 6) = 40 + 15 × 4 ÷ 5 = 52 exactly.
• Example 6 read as a ratio. Verified: the excesses are 7 − 3 = 4 and 7 − 6 = 1, so the mode divides 40–55 in the ratio 4 : 1 — four fifths of the way along, at 52, hard against the boundary it shares with the busy class of six. Set this beside Example 5 and the mechanism is unmistakable: in Example 5 the busier neighbour was below and the mode sank; here it is above and the mode rose. Same formula, opposite outcomes, and the reason is visible.
• The symmetric case — section 10, and added here. Verified algebraically: when the two neighbouring counts are equal, the two excesses are equal, the ratio is 1 : 1, and the mode is the mid-point of the modal class, which is its class mark. So the mode equals the class mark exactly when the distribution is locally balanced — and never otherwise. Worth constructing a two-line demo table for, since no example in the chapter has this shape.
• Mode against mean (p. 185). Verified: on the same thirty scripts the mode is 52 and the mean is 62. The chapter reads these as answering different questions — where most students clustered, against what a student scored on average — and its first Remark warns that the mode is not always the smaller of the two.
• Exercise 13.2, worked inputs (pp. 186–187). All values below are added here.
  • Q1, patients by age, classes 5–15 to 55–65, counts 6, 11, 21, 23, 14, 5. Verified: modal class 35–45; mode 35 + 10 × 2 ÷ 11 = 36.82 years. Mean by class marks 10, 20, 30, 40, 50, 60: products 60, 220, 630, 920, 700, 300, total 2830 over 80 patients, giving 35.375 years. Here the mode exceeds the mean, the reverse of Example 6 — the two directions the chapter's Remark 1 promises, both supplied by the chapter's own data.
  • Q2, lifetimes of 225 components, counts 10, 35, 52, 61, 38, 29 across 0–20 to 100–120. Verified: modal class 60–80; mode 60 + 20 × 9 ÷ 32 = 65.625 hours.
  • Q3, 200 families by monthly expenditure, counts 24, 40, 33, 28, 30, 22, 16, 7 across ₹1000–1500 to ₹4500–5000. Verified: modal class 1500–2000; mode 1500 + 500 × 16 ÷ 23 = ₹1847.83. Mean by step-deviation with a = 2750, h = 500: u values −3, −2, −1, 0, 1, 2, 3, 4, products −72, −80, −33, 0, 30, 44, 48, 28, total −35, giving 2750 − 87.5 = ₹2662.50. A gap of over ₹800 between mode and mean, because the distribution has a long upper tail.
  • Q4, states by students per teacher, counts 3, 8, 9, 10, 3, 0, 0, 2 across 15–20 to 50–55. Verified: modal class 30–35; mode 30 + 5 × 1 ÷ 8 = 30.625. Mean, a = 32.5 and h = 5: u values −3, −2, −1, 0, 1, 2, 3, 4, products −9, −16, −9, 0, 3, 0, 0, 8, total −23 over 35, giving 32.5 − 3.286 = 29.21. This table is the mode's locality made visible: two classes are empty and two states sit far out at 50–55. Drop those two states and the mean falls to 917.5 ÷ 33 = 27.80, moving by 1.4 — while the mode does not move at all, because none of the three frequencies it uses has changed.
  • Q5, batsmen by career runs, counts 4, 18, 9, 7, 6, 3, 1, 1 across 3000–4000 to 10000–11000. Verified: modal class 4000–5000; mode 4000 + 1000 × 14 ÷ 23 = 4608.70 runs.
  • Q6, cars per three-minute period, counts 7, 14, 13, 12, 20, 11, 15, 8 across 0–10 to 70–80. Verified: modal class 40–50; mode 40 + 10 × 8 ÷ 17 = 44.71. Note: this table has visible secondary bumps at 10–20 and 60–70, so it is the one exercise where the chapter's restriction to single-peaked data is worth mentioning out loud.

Figures to have open

• A three-bar diagram — the class below, the modal class, the class above — with the two excesses drawn as vertical gaps and the mode's position on the horizontal axis tied to their ratio. As the neighbours' heights change, the mode slides. This is the topic's central image and the chapter draws nothing like it. Standard schematic, must be built.
• Examples 5 and 6 as a mirrored pair of that diagram, side by side, so the opposite leanings are seen at once. Standard schematic.
• A frequency-bar picture of Exercise 13.2 Q4 showing the two empty classes and the far-out pair of states, with the mean marker moving when they are removed and the mode marker staying put. Standard schematic.
• Nothing needs to come from the printed page; this chapter contains no diagrams at all.

Where this sits in the book

• NCERT Class 10 Mathematics, Chapter 13 "Statistics", §13.3 Mode of Grouped Data, pp. 183–186.
• The definition recalled and the note on multimodal data, p. 183; Example 4 and its frequency table, p. 183.
• The modal class named and the formula with its five symbols, p. 184.
• Example 5 worked, p. 185; Example 6 worked, with the mode-and-mean comparison and Remarks 1 and 2, p. 185.
• Activity 3 and the Remark declining unequal class sizes, p. 186.
• Exercise 13.2, questions 1–6, pp. 186–187.
• A Note to the Reader on continuity, p. 201, cited here as the condition the formula depends on.