Where this measure breaks down, and why another was needed

What to assume they know

• Working it out about the mean and about the median, for a plain list — mean deviation about the mean and about the median for a plain list
• The same procedure once the data arrive already grouped — the grouped versions, and how much bookkeeping they cost
• Expanding a square of a difference
• That |a| has no expansion into terms in a
• The batting records of §13.1 and their equal medians

What they should be able to do

• State the three objections §13.4.3 raises against the mean deviation
• Give a data set on which the median is an unrepresentative centre, and say why
• Verify on the chapter's own data that the mean deviation about the mean is not smaller than the one about the median, and identify when the two are equal
• Explain why that inequality holds, using the counting argument for the median
• Explain what "carrying a measure into further algebra" would require, and why a modulus prevents it
• Expand a squared deviation and point to the term structure a modulus has no counterpart for
• State the replacement §13.5 proposes and what it preserves from mean deviation

Where it usually goes wrong

• "The mean deviation is wrong." It is not wrong; it is limited. It is still the intuitive measure — the average distance from a centre — and it is still examinable, with an exercise spanning parts of two pages — twelve items, about a page of type all told. The chapter moves on for tractability, not for correctness.
• "The mean deviation about the mean is always bigger than about the median." Not always — Examples 1 and 2 make them equal, because in both the two centres coincide. Say "not smaller".
• "Absolute values are just harder to compute with." They are not harder to compute with at all; a distance is easy. What they resist is being expanded, rearranged and recombined after the computation.
• "Squaring is chosen because it gives bigger numbers." It is chosen because it is a polynomial operation that removes the sign. That it also weights distant observations more heavily is a consequence, and the next module treats it as one.
• "The median is a bad centre." §13.4.3's complaint is narrow: when variability is very high, a single middle value stands for little. For tightly clustered data the median is an excellent centre and the chapter has just used it in three worked examples — the ungrouped one, the discrete one and the continuous one. The other worked examples in that stretch are all about the mean.
• "§13.5 abandons everything from §13.4." It keeps the whole plan — deviations about a centre, averaged — and changes one step: how the sign is removed.

Questions to check understanding

• State the objections the chapter raises to the mean deviation
• Given a data set, decide whether its median is a representative centre and justify the verdict
• Compute both mean deviations for a small list and comment on which is larger
• Give a data set for which the two mean deviations are equal, and say what makes them equal
• Explain why a measure built from absolute values cannot be expanded
• Short-answer: what single change does §13.5 make to the plan of §13.4

Examples worth working on the board

Values marked verified are worked out here on data printed in this chapter.

• The paragraph itself (§13.4.3, p. 271). Three claims in order: a median is not a representative centre when variability is very high, so a mean deviation computed about it cannot be leaned on; the total of the unsigned deviations about the mean exceeds the total about the median, which the chapter reads as a mark against the mean version; and a measure built from absolute values cannot be manipulated further, which is what forces the chapter onward.
• An unrepresentative median (the batting record of §13.1, p. 257). Batsman A's ten innings sorted are 0, 5, 30, 30, 42, 64, 71, 80, 91, 117. Verified: his median is 53, and 53 is not a score he made; five of his ten innings ended at 42 or below, and two of the other five exceeded 80. A single value at 53 describes none of the innings and stands between two quite different modes of behaviour. That is what §13.4.3's first sentence is pointing at.
• The inequality on the chapter's own data (Example 3, §13.4.1, p. 262): 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. Verified: the median is 9 and the mean deviation about it is 58/11, about 5.27; the mean is 111/11, about 10.09, and the mean deviation about it is 652/121, about 5.39. The mean version is the larger, by about 0.12.
• The case of equality (Examples 1 and 2, §13.4.1, pp. 261–262). Verified: the eight numbers 6, 7, 10, 12, 13, 4, 8, 12 have mean 9 and median 9, and the twenty numbers of Example 2 have mean 10 and median 10. When the two centres coincide the two mean deviations are the same number, 2.75 and 6.2 respectively. The inequality is not strict.
• The reason for the inequality. The total distance from a reference point falls as the point moves toward the middle of the data and stops falling when the number of observations on each side balances — which is what a median is. So no reference point beats a median, and the mean is one reference point among many. Move the point one small step right and the total changes by that step times the difference between the counts on the two sides; the change is negative exactly while more observations lie ahead.
• Why that is not, by itself, a fault. Being larger than the smallest possible value is not a defect in a measure — it is what any measure taken about a centre other than the minimising one must do. The mean deviation about the mean is a perfectly well-defined quantity, and the chapter goes on to use the mean, not the median, as the centre for the whole of §13.5. Present §13.4.3's second sentence faithfully and then say why it does not carry the weight the paragraph puts on it.
• What the modulus blocks, concretely. A squared deviation expands: the square of an observation less the mean becomes the square of the observation, less twice the product, plus the square of the mean. That expansion is what lets §13.5.3 turn the variance into a formula in the totals of the observations and of their squares (pp. 276–277), with no deviation column at all. There is no corresponding expansion of a modulus, so no such formula exists for mean deviation: you must know the centre before you can compute a single term, and you cannot recover the mean deviation of a combined data set from the mean deviations of its parts.
• The opening of §13.5 (p. 271). Squaring is offered as the other way of killing the sign. Every square is non-negative, so the cancellation problem of §13.4 does not return, and the square is a polynomial, so everything the modulus forbade becomes available.

Figures to have open

• Batsman A's innings on a number line with the median marked and the two behaviour clusters shaded. Derived from the data of p. 257 and Fig 13.1 on p. 258; redraw as a schematic.
• A graph of total distance against the position of the reference point for the eleven numbers of Example 3 — a piecewise straight line with its lowest stretch over the median. An added figure and the clearest single picture of why the median minimises; the chapter prints nothing like it.
• Side-by-side expansion panels: the square of a difference opening into three terms, and a modulus with no expansion beneath it. Standard schematic.
• §13.4.3 prints no figure — printed p. 271 carries the last four items of Exercise 13.1, the limitations paragraph and the opening of §13.5.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 13 "Statistics", §13.4.3 "Limitations of mean deviation" and the opening of §13.5 "Variance and Standard Deviation", both printed on p. 271.
• The data used for the inequality is Example 3 of §13.4.1, p. 262; the equality cases are Examples 1 and 2, pp. 261–262; the batting record is §13.1, p. 257.
• The expansion the modulus lacks is carried out for the square in §13.5.3, pp. 276–277. That is a deliberate forward reference and it belongs to Carrying the frequencies through, for discrete and for grouped data.