Dividing through by the class width to shrink the arithmetic further

What to assume they know

• Shifting the origin: why guessing a centre cannot change the answer — shifting the origin, and the proof that the choice of assumed mean cannot move the answer
• The direct method, and where it becomes unwieldy — the direct method and why its cost is arithmetic
• Dividing a signed number by a positive number, including cases giving a decimal
• Factoring a constant out of a sum, and that dividing every term of a sum by h divides the sum by h
• Common factors and the greatest common divisor of a small set of numbers
• Class size, and the fact that classes need not all share one

What they should be able to do

• Build the u column as (x − a) ÷ h for a given assumed mean and step
• Compute a grouped mean as a + h × (Σf·u ÷ Σf)
• Derive that formula from the definition of u, naming the property used at each step
• Choose h as a common divisor of the deviations rather than assuming it must be the class width
• Apply the method to a distribution whose classes have unequal widths, and justify why this is legitimate
• Verify a step-deviation answer against the direct method on the same data
• Decide which of the three methods a given table calls for, and defend the choice

Where it usually goes wrong

• "h must equal the class size." It must divide the deviations usefully. Example 3 uses 20 against widths of 40, 50 and 100, and the answer is right. The class size is merely the choice that works when all classes share one.
• "Unequal classes rule the method out." They do not rule anything out. What the chapter says twice — in the mode formula's symbol list on p. 184 and the median's on p. 193 — is that those printed formulas are written assuming the classes are equal in size. It then says twice more, in the Remark on p. 186 and Remark 2 on p. 197, that both measures can be found for unequal classes and that it will not go into how. The median formula in fact runs unchanged with h read as the median class's own width. The mean has no such caveat at all, and Example 3 is the chapter demonstrating it.
• "−3.75 in the u column means I picked the wrong h." It means 75 is not a multiple of 20. The arithmetic proceeds unharmed; only the tidiness suffers. Choosing h = 5 removes the fraction and enlarges everything else.
• "The three methods are three formulas to memorise." They are one formula at three settings of a and h. A student who sees that has two fewer things to remember and a check they can run for free.
• "Multiply by h at the start, when building u." You divide by h to build u and multiply by h at the end. Reversing this is the single commonest slip, and it shows up as an answer wrong by a factor of h².
• "Forgetting h just gives a slightly wrong answer." Dropping the final multiplication on Example 3 gives 200 − 2.36, about 197.6, against 152.89 — not slightly wrong. The check against the direct method catches it instantly.
• "The mean must land inside the busiest class." Example 3's mean, 152.89, falls in 150–250, while the busiest class is 100–150. The mean is not a location of frequency.

Questions to check understanding

• Compute a grouped mean by the step-deviation method, showing the u and f·u columns
• State the assumed mean and the step used, and justify both
• Apply the method to a distribution with unequal class widths and explain why it remains valid
• Verify a step-deviation result against the direct method on the same table
• Given a distribution, decide which method is most appropriate and give a reason referring to the sizes of the class marks and frequencies
• Interpret the resulting mean in the context of the data, which Example 3 explicitly asks for
• Find a missing frequency using the step-deviation form of the mean equation

Examples worth working on the board

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

• The worked case (Table 13.5, p. 176). The Example 1 marks again: classes 10–25 through 85–100, counts 2, 3, 7, 6, 6, 6, class marks 17.5 to 92.5, a = 47.5, h = 15. Verified: deviations −30, −15, 0, 15, 30, 45 all divide by 15, giving u values −2, −1, 0, 1, 2, 3; products −4, −3, 0, 6, 12, 18; total 29; mean 47.5 + 15 × (29 ÷ 30) = 47.5 + 14.5 = 62. The whole product column now fits in two digits.
• Example 2 (pp. 178–179). Classes 15–25 through 75–85, counts 6, 11, 7, 4, 4, 2, 1, a = 50, h = 10. Verified: u values −3, −2, −1, 0, 1, 2, 3; products −18, −22, −7, 0, 4, 4, 3; total −36; mean 50 + 10 × (−36 ÷ 35) = 50 − 10.2857… = 39.71 to two places, matching the direct and assumed-mean routes on the same page.
• Example 3, the load-bearing one (p. 180). How many wickets each of 45 bowlers has claimed across a one-day career. Classes 20–60, 60–100, 100–150, 150–250, 250–350, 350–450 with counts 7, 5, 16, 12, 2, 3. Verified: the six widths run 40, 40, 50, 100, 100, 100 — three distinct values, and h is taken as 20, which matches none of them. Class marks 40, 80, 125, 200, 300, 400; deviations from a = 200 are −160, −120, −75, 0, 100, 200; dividing by 20 gives −8, −6, −3.75, 0, 5, 10; products −56, −30, −60, 0, 10, 30; total −106; mean 200 + 20 × (−106 ÷ 45) = 200 − 47.111… = 152.89. The chapter reads this back as the average number of wickets these bowlers took.
• The independent check on Example 3 — this is the moment. Verified by the direct method on the same table: products 7 × 40 = 280, 5 × 80 = 400, 16 × 125 = 2000, 12 × 200 = 2400, 2 × 300 = 600, 3 × 400 = 1200, totalling 6880; and 6880 ÷ 45 = 152.888…, the same 152.89. The unequal widths and the fractional u did no damage whatever.
• h is free too. Verified: rerun Example 3 with h = 5, which divides every one of the six deviations exactly. The u column becomes −32, −24, −15, 0, 20, 40; the products −224, −120, −240, 0, 40, 120; the total −424; and the mean 200 + 5 × (−424 ÷ 45) = 200 − 47.111… = 152.89 again. Two different h's, two different tables, one answer. Show both columns side by side: h = 5 keeps every entry whole but makes them large, h = 20 keeps them small at the cost of one fraction. That is the real trade, and it is a choice, not a rule.
• Collapsing the family. Verified on Table 13.5's data: with a = 47.5 and h = 1 the u column is just the deviation column, the total is 435, and the mean is 47.5 + 1 × (435 ÷ 30) = 62 — the assumed mean method. With a = 0 and h = 1 the u column is the class mark column, the total is 1860, and the mean is 0 + 1 × (1860 ÷ 30) = 62 — the direct method. One formula, three settings.
• Where the method has nothing to offer — Exercise 13.1 Q8 (p. 183). Days absent, classes 0–6, 6–10, 10–14, 14–20, 20–28, 28–38, 38–40 with counts 11, 10, 7, 4, 4, 3, 1 over 40 students. Verified: widths 6, 4, 4, 6, 8, 10, 2; class marks 3, 8, 12, 17, 24, 33, 39; deviations about a = 17 are −14, −9, −5, 0, 7, 16, 22, whose greatest common divisor is 1. There is no useful h, so the chapter's own advice runs out and the direct method is the honest choice. Verified: products 33, 80, 84, 68, 96, 99, 39, totalling 499; mean 499 ÷ 40 = 12.475 days.
• Exercise 13.1 questions where the step is clean. Q2, h = 20; Q4, heartbeats, h = 3 with class marks 66.5, 69.5, 72.5, 75.5, 78.5, 81.5, 84.5 and a = 75.5 — verified u values −3, −2, −1, 0, 1, 2, 3, products −6, −8, −3, 0, 7, 8, 6, total 4, mean 75.5 + 3 × (4 ÷ 30) = 75.9; Q6, h = 50 — verified u values −2, −1, 0, 1, 2, products −8, −5, 0, 2, 4, total −7, mean 225 + 50 × (−7 ÷ 25) = ₹211; Q9, h = 10 — verified u values −2, −1, 0, 1, 2, products −6, −10, 0, 8, 6, total −2, mean 70 + 10 × (−2 ÷ 35) = 69.43%.

Figures to have open

• A step-ruler figure: the original scale along the top, a second axis beneath it ticked every h units and numbered by u, with each class mark dropping onto its u value. This is what makes "step" mean something. Standard schematic.
• Example 3's six classes drawn to true relative width on one axis — 40, 40, 50, 100, 100, 100 — with the h = 20 ruler laid beneath. The visible mismatch is the argument of section 8. Standard schematic; the chapter draws nothing.
• A two-column comparison of Example 3 at h = 20 and h = 5, ending on the same mean. Standard schematic.
• A dial or slider graphic for the (a, h) family collapsing to the three named methods. Standard schematic.
• Nothing needs to come from the printed page; this chapter has no diagrams.

Where this sits in the book

• NCERT Class 10 Mathematics, Chapter 13 "Statistics", §13.2, p. 176 — the common factor of 15 spotted, u defined, and Table 13.5.
• The derivation, the substitution giving 62, the method named, and the four bullet points that follow it, p. 177. The last of those bullets is what licenses h to be something other than the class size.
• Example 3 and Table 13.8, p. 180.
• The Remark on choosing between the methods, including the advice to take h as a divisor of the deviations when class sizes are unequal, p. 179.
• The three formulas collected in §13.5, pp. 200–201.
• Exercise 13.1, questions 2, 4, 6, 8 and 9, pp. 181–183.