Working it out about the mean and about the median, for a plain list

What to assume they know

• Signed deviations cancel to nothing, so the sign has to be discarded — why the modulus is taken, and the definition of M.D. (a)
• Computing a mean, and the median rules for odd and even counts
• Sorting a short list by hand
• Reading and writing a sum in sigma notation
• Division producing a recurring decimal, and rounding to two places

What they should be able to do

• Carry out the four steps of §13.4.1 for the mean deviation about the mean of a given list
• Carry out the same four steps about the median, including the sorting step the mean does not require
• Apply the odd-count and even-count median rules correctly
• Compute the mean deviation for each of the four ungrouped data sets set in Exercise 13.1
• Explain, by a counting argument, why shifting the reference point away from a median cannot decrease the total distance
• Recognise data sets where the mean and the median coincide, and say why those cannot illustrate the difference between the two mean deviations
• Round a recurring quotient to two decimal places as the chapter's examples do

Where it usually goes wrong

• "The mean deviation about the mean and about the median are the same thing computed twice." They are different numbers for most data. On the Example 3 list they are about 5.39 and about 5.27.
• "The median version is bigger because the median is smaller." The direction has nothing to do with which centre is numerically smaller. The median version is never the larger of the two, whichever way round the two centres sit.
• "You can find the median without sorting." Every median step in these examples begins by arranging the data in order, and the chapter's Example 5 points out when the given data is already ordered so that step can be skipped.
• "The median of a twelve-value list is the sixth value." It is the average of the sixth and the seventh. Exercise 13.1 items 3 and 4 are set precisely to catch this.
• "5.27 is the exact answer." It is 58/11 rounded. Keep the fraction in view and round only at the end, or a student who rounds early will report 5.3 and believe the difference from 5.27 is the book's error.
• "Dropping the sign is the same as ignoring the negative observations." The distances of the below-centre observations are kept in full; only the direction is discarded. In Example 1 the largest single distance, 5, comes from an observation below the mean.

Questions to check understanding

• Compute the mean deviation about the mean of a list of eight to twelve values
• Compute the mean deviation about the median of a list of even length, showing the sorting step
• Given a list, state whether its mean and median coincide and what follows for the two mean deviations
• Given the total of the distances and the count, write down the mean deviation
• Explain why the mean deviation about the median is not larger than the one about the mean
• Report a recurring quotient to two decimal places

Examples worth working on the board

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

• The four steps (§13.4.1, p. 260). Fix the centre; subtract it from each observation; drop the signs; average. Step 4 is the definition of M.D. (a) from the previous topic, so the procedure adds nothing to the definition — it just orders the work. The chapter's Note on p. 261 says the steps need not be written out separately once they are understood.
• Example 1 (§13.4.1, p. 261). Data: 6, 7, 10, 12, 13, 4, 8, 12. Verified: total 72, count 8, mean 9; signed deviations −3, −2, 1, 3, 4, −5, −1, 3; distances 3, 2, 1, 3, 4, 5, 1, 3, totalling 22; mean deviation 2.75.
• Example 2 (§13.4.1, pp. 261–262). Data: 12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5 — twenty observations. Verified: total 200, so mean 10; the twenty distances total 124; mean deviation 6.2. This example is here for the bookkeeping, not the idea: twenty distances is where a student's arithmetic starts failing, and the round mean of 10 is what keeps it survivable.
• Example 3 (§13.4.1, p. 262). Data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21. Verified: sorted, 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21; eleven observations, so the median is the sixth entry, 9. Verified: the distances from 9 are 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12, totalling 58; the mean deviation is 58 ÷ 11, which the chapter reports to two places as 5.27.
• The same eleven numbers about their mean (not in the book). Verified: the total is 111, so the mean is 111/11, which is 10 and 1/11 — about 10.09, not a whole number. Verified: the eleven distances then total 652/11, and the mean deviation is 652/121, about 5.39. So for this data the mean deviation about the mean is the larger of the two, by about 0.12. This is the comparison §13.4.3 will lean on, and Example 3 is the only one of the three where it can be made.
• Why Examples 1 and 2 cannot make it. Verified: Example 1 sorted is 4, 6, 7, 8, 10, 12, 12, 13, whose two middle entries are 8 and 10, so its median is 9 — the same as its mean. Verified: Example 2 sorted has 9 and 11 in the tenth and eleventh places, so its median is 10 — again the same as its mean. In both, the two mean deviations are identical, 2.75 and 6.2 respectively. An explanation that claims the two centres differ and then demonstrates on Example 1 has demonstrated nothing.
• Why the median minimises. Slide the reference point a small step to the right. Every observation still to the right of it gets a step closer, every observation still to the left gets a step further, and the total distance changes by the step times the difference of those two counts. The total keeps falling as long as more observations lie ahead than behind, and stops falling exactly when the counts balance — which is what "median" means. The chapter states the resulting inequality in §13.4.3 (p. 271) without giving this reason.
• Exercise 13.1, the ungrouped items (pp. 270). About the mean: item 1 is 4, 7, 8, 9, 10, 12, 13, 17; item 2 is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44. About the median: item 3 is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17; item 4 is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49. Verified, by working added here: item 1 has mean 10 and mean deviation 3; item 2 has mean 50 and mean deviation 8.4; item 3 has twelve observations, median 13.5 and mean deviation 28/12, about 2.33; item 4 has ten observations, median 47.5 and mean deviation 7. Items 3 and 4 are the ones that teach the even-count rule, and both produce a median that is not a whole number and is not one of the observations.

Figures to have open

• A four-column working table — observation, deviation, distance, running total — that fills row by row. The chapter shows this only as printed lists in §13.4.1; the tabular form arrives in §13.4.2. Standard schematic.
• A number line for section 7 carrying the eleven sorted observations of Example 3, with a movable reference marker and live counts of observations to its left and right. An added figure; the chapter prints no diagram in §13.4.1.
• Two stacked distance-fans for section 6: the same eleven points with spokes to 9 and with spokes to 111/11, the two totals shown. An added figure.
• §13.4.1 prints no figure between pp. 260 and 262 — the pages carry the four steps, the displayed formulas, the Note and the three examples.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 13 "Statistics", §13.4.1 "Mean deviation for ungrouped data", printed pp. 260–262, with the Note on p. 261 fixing M as the symbol for the median in this chapter.
• Exercise 13.1 items 1 to 4, printed p. 270. Items 5 onward belong to the next topic.
• The inequality between the two mean deviations is stated in §13.4.3, p. 271, and is developed there rather than here.
• The median rules for odd and even counts are restated in §13.1, p. 258.