Shifting the origin: why guessing a centre cannot change the answer

What to assume they know

• The direct method, and where it becomes unwieldy — the direct method, and why its only problem is the size of the intermediate products
• Grouping loses the raw values, so we stand the class mark in for them — class marks as stand-ins for whole classes
• Signed arithmetic: adding and subtracting negative numbers, and totalling a column containing both signs
• That Σf is the number of observations, so Σf·a ÷ Σf is a ÷ 1
• Splitting a sum over a subtraction: Σ(p − q) equals Σp − Σq
• Factoring a constant out of a sum: Σ(f·a) equals a·Σf

What they should be able to do

• Choose an assumed mean from a table and justify the choice on arithmetic grounds
• Build the deviation column d = x − a and the product column f·d for a given distribution
• Compute a grouped mean as a + Σf·d ÷ Σf
• Derive that the mean of the deviations equals the true mean less a, naming the property of sums used at each step
• Predict, before computing, that changing the assumed mean will not change the final answer, and verify it on a second choice
• Explain what a badly chosen assumed mean actually costs
• Recognise the direct method as the special case in which nothing is subtracted

Where it usually goes wrong

• "The assumed mean is a guess at the answer, so a good guess gives a better answer." It is not a guess at anything. Every choice returns the identical mean; the derivation leaves no room for the choice to matter.
• "a must be one of the class marks." It is convenient — one deviation becomes 0 and one product vanishes — but it is not required. A centre of 100 works on this data and lies above every class mark.
• "a must lie inside the range of the data." No. See the a = 100 case.
• "You add a back because it seems fair." You add it back because the algebra produces exactly a and nothing else.
• "If the deviations total zero the method has failed." A total of zero means the chosen centre happened to be the mean. That is the method succeeding perfectly, not failing — try a = 62 on this data.
• "Negative deviations should be dropped or made positive." The signs are the whole mechanism. Making them positive computes a different quantity entirely and destroys the cancellation.
• "Different a, different answer — I got 62 and my neighbour got 62.5." One of you made an arithmetic slip. Because the answer cannot depend on a, disagreement between two choices is a reliable error detector — a genuinely useful exam habit.

Questions to check understanding

• Compute a grouped mean by the assumed mean method, showing the d and f·d columns
• State the assumed mean used and justify the choice
• Recompute the same mean with a different a and confirm the answer is unchanged
• Complete a derivation of x̄ = a + Σf·d ÷ Σf with steps left blank
• Given a table and a stated mean, find a missing frequency by the deviation route, which usually gives a tidier equation than the direct route
• Explain why the choice of assumed mean cannot affect the result

Examples worth working on the board

Values marked verified are worked out here on the chapter's printed data. The chapter itself computes only the first case and asserts the rest.

• The worked case, a = 47.5 (Table 13.4, p. 175). The Example 1 marks regrouped: classes 10–25, 25–40, 40–55, 55–70, 70–85, 85–100 with counts 2, 3, 7, 6, 6, 6 and class marks 17.5, 32.5, 47.5, 62.5, 77.5, 92.5. Verified: deviations −30, −15, 0, 15, 30, 45; products −60, −45, 0, 90, 180, 270; total 435; mean 47.5 + 435 ÷ 30 = 47.5 + 14.5 = 62.
• Activity 1, all six guesses (p. 176). The chapter asks the student to redo the table with each class mark in turn as the centre, states that 62 comes out every time, and leaves the reason as a question. Verified:
  • a = 17.5 → deviations 0, 15, 30, 45, 60, 75; products 0, 45, 210, 270, 360, 450; total 1335; 1335 ÷ 30 = 44.5; and 17.5 + 44.5 = 62.
  • a = 32.5 → products −30, 0, 105, 180, 270, 360; total 885; 885 ÷ 30 = 29.5; 32.5 + 29.5 = 62.
  • a = 47.5 → total 435; 14.5; 62. (the chapter's own case)
  • a = 62.5 → products −90, −90, −105, 0, 90, 180; total −15; −15 ÷ 30 = −0.5; 62.5 − 0.5 = 62.
  • a = 77.5 → products −120, −135, −210, −90, 0, 90; total −465; −15.5; 77.5 − 15.5 = 62.
  • a = 92.5 → products −150, −180, −315, −180, −90, 0; total −915; −30.5; 92.5 − 30.5 = 62. Notice the pattern: the six column-totals run 1335, 885, 435, −15, −465, −915, falling by exactly 450 each time. That is 30 observations × 15 of shift — the shortfall tracks the shift precisely, which is the theorem visible as a number sequence.
• A centre outside the data. Verified: take a = 100, above every class mark. Deviations −82.5, −67.5, −52.5, −37.5, −22.5, −7.5; products −165, −202.5, −367.5, −225, −135, −45; total −1140; −1140 ÷ 30 = −38; and 100 − 38 = 62. Worth showing, because students believe a has to sit inside the data. It does not. It only has to be a number.
• A centre of zero. Verified: with a = 0 every deviation is the class mark itself, the products total 1860, and the mean is 0 + 1860 ÷ 30 = 62 — which is the direct method exactly. So the two methods are one method with a chosen at different places. Present this as the unification it is, but see the note below about the chapter's own wording on non-zero choices.
• Example 2 by the same route (pp. 178–179). Classes 15–25 through 75–85, counts 6, 11, 7, 4, 4, 2, 1, class marks 20 to 80 in tens, and the chapter takes a = 50. Verified: deviations −30, −20, −10, 0, 10, 20, 30; products −180, −220, −70, 0, 40, 40, 30; total −360; mean 50 + (−360 ÷ 35) = 50 − 10.2857… = 39.71 to two places, agreeing with the direct method's 1390 ÷ 35.
• The derivation to show (p. 175). Start from the average of the d column. Replace each d by x − a. Split the sum into a part over the x's and a part over the a's. The first part is the true mean. The second is a·Σf ÷ Σf, and the Σf's cancel, leaving a. So the deviation-average is the true mean minus a, and adding a back recovers it. Four lines, and only two facts: sums split over subtraction, and constants factor out.

Figures to have open

• A number line carrying the six class marks, with a draggable origin flag. As the flag moves, the six deviation arrows re-length and the running total changes, while a separate readout showing a + (total ÷ 30) stays fixed at 62. This single interactive-feeling image is the topic, and the chapter has nothing like it. Standard schematic, must be built.
• The four-line derivation set as annotated equations with the cancelling Σf's struck through. Standard schematic.
• A six-column comparison panel for Activity 1. 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", §13.2, p. 174 — the assumed mean introduced, and the deviation defined.
• Table 13.4 and the four-line derivation, p. 175.
• The substitution giving 62, the method named, and Activity 1, p. 176.
• Example 2's assumed-mean column, Table 13.7 and the Remark on method choice, pp. 178–179.
• The chapter's summary of the three formulas, §13.5, pp. 200–201.