Signed deviations cancel to nothing, so the sign has to be discarded

What to assume they know

• The cheapest measure of spread, and everything it fails to notice — that the range ignores the interior and never mentions a centre
• The definition of the mean as total divided by count, in sigma notation
• Splitting a sum: that a sum of differences is the difference of the sums
• The modulus of a real number, and that |a − b| is the distance between a and b on the number line
• That the mean of a data set lies somewhere between the smallest and the largest observation, and strictly between them unless every observation is the same

What they should be able to do

• Prove that the deviations of a data set about its own mean add to zero
• Explain why that identity kills the average-of-deviations proposal for every data set at once, rather than for particular ones
• State the modulus of a difference as the distance between two points on the number line
• Define the mean deviation about a general central value a, and write it in sigma notation
• Show that the identity is about the mean specifically, by exhibiting a data set whose deviations about its median do not add to zero
• Explain why averaging only the positive deviations recovers no new information
• State the chapter's Remark: that mean deviation may be taken about any measure of central tendency, and which two are used in practice

Where it usually goes wrong

• "The deviations happened to cancel for that example." They cancel for every example. Show the proof, not a second example — a second example is what makes students believe it is a coincidence.
• "Then use the median as the centre and the problem goes away." It does not go away and it does not stay either: about the median the signed deviations need not be zero, as the Example 3 data shows, but they still mix signs and still under-report the spread. The modulus is applied regardless of which centre is chosen.
• "Taking absolute values is a trick to make the answer come out nice." It is a change of question. Signed deviation records which side and how far; distance records how far. Dispersion is a question about how far.
• "|x − a| is x − a with a minus sign stuck on if needed." That is what it computes to, but the reason it is the right object is geometric: it is the length of the segment between the two points, and length has no sign.
• "Averaging just the positive deviations would work as well." It gives exactly half the mean deviation — but only if the positive total is divided by the full count of observations. Divide it instead by how many deviations were positive, the natural reading of "averaging", and the halving disappears: on this brief's own first worked example the four positive deviations average to 2.75, which is the mean deviation itself, not half of it. Name the divisor or the claim is wrong.
• "Mean deviation is defined only about the mean." The chapter defines it about an arbitrary central value and then narrows to two by convention. The notation M.D. (a) carries the general case.

Questions to check understanding

• Prove that the deviations of a data set about its mean add to zero
• Given a short list, compute the signed deviations about the mean and about the median and comment on the two totals
• State the definition of mean deviation about a general central value in sigma notation
• Explain in one sentence why absolute values are introduced
• Given the sum of the positive deviations about the mean, write down the mean deviation
• Short-answer: about which centre are the signed deviations guaranteed to cancel

Examples worth working on the board

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

• The failing construction, on the chapter's own eight numbers (the data of Example 1, §13.4.1, p. 261): 6, 7, 10, 12, 13, 4, 8, 12. Verified: the total is 72 and the count is 8, so the mean is 9, and the signed deviations are −3, −2, 1, 3, 4, −5, −1, 3. Verified: those eight add to 0.
• Why the zero was inevitable. Adding the deviations gives the total of the observations less the count times the mean. But the mean is by construction the total divided by the count, so the second term is the total again and the difference is zero. Nothing about the particular eight numbers entered the argument, which is the point: every data set does this.
• What the average of the deviations therefore is. Zero divided by the count, which is zero for every data set, scattered or not. The chapter writes this out on p. 259. A quantity that is the same for all data can report nothing about any of them.
• Positive deviations alone. Verified: for the eight numbers above, the positive deviations are 1, 3, 4 and 3, adding to 11; the negative ones are −3, −2, −5 and −1, of size 3, 2, 5 and 1, also adding to 11. The two halves are always equal, because they must cancel — so keeping only one half is the unsigned total halved, and carries no information the unsigned total did not.
• The modulus, applied. Verified: the eight distances are 3, 2, 1, 3, 4, 5, 1, 3, adding to 22, and dividing by 8 gives 2.75. That number is computed properly in the next topic; here it is the payoff of the repair.
• The definition (§13.4, p. 260). The mean deviation about a value a is the total of the distances of the observations from a, divided by how many observations there are, and the chapter writes it M.D. (a). It is defined about any central value; the Remark on p. 260 records that mean and median are the two used in practice.
• A data set whose deviations about the median do not cancel (the data of Example 3, §13.4.1, p. 262): 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. Verified: sorted, this is 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21, so with eleven observations the median is the sixth entry, 9. Verified: the signed deviations about 9 are −6, −6, −5, −4, −2, 0, 1, 3, 9, 10, 12, and they add to 12, not to 0. Section 10 needs this: the cancellation theorem is about the mean, and an explanation that states it about "the centre" has overstated it.
• The bound the chapter reasons from (§13.4, p. 259). For the mean of a data set that is not constant, the value sits strictly inside the range, so some deviations must come out negative and some positive. Do not extend that to any central value: on the four readings 1, 1, 1 and 5 the median is 1, which is the smallest observation, and every deviation about it is zero or positive. That is the informal version of the argument; the identity in section 4 is the exact one, and only for the mean.

Figures to have open

• A number line carrying the eight observations of Example 1 with the mean marked at 9, arrows drawn from 9 to each observation, first as signed arrows and then as plain lengths. The chapter does not print this figure; it is added here and it is what makes the modulus feel inevitable.
• A balance schematic: the same eight points as weights on a beam pivoted at the mean, level. Standard schematic, and the visual form of the section 4 identity.
• Two stacked panels for section 10, one for the mean and one for the median of the Example 3 data, with the running totals shown.
• §13.4 itself prints no figure on pp. 259–260, and p. 260 carries only the four numbered steps and the displayed formulas.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 13 "Statistics", §13.4 "Mean Deviation", printed pp. 259–260, including the Remark on p. 260 about which central values are used.
• The eight numbers used throughout are the data of Example 1, §13.4.1, p. 261; the eleven numbers in section 10 are the data of Example 3, §13.4.1, p. 262. Both are computed properly in the next topic.
• The requirement this topic starts from is the closing paragraph of §13.3, p. 259.