Squaring removes the sign and punishes the far-out values

What to assume they know

• Where this measure breaks down, and why another was needed — why the chapter leaves the mean deviation behind
• That a square of a real number is never negative, and is zero only for zero
• Expanding a square of a difference
• The sum of the squares of the first n natural numbers
• Reading a dot diagram on a number line

What they should be able to do

• State why squaring removes the sign problem without introducing a modulus
• Show that the total of the squared deviations about the mean is zero exactly when every observation equals the mean
• Compute the total of the squared deviations for each of the chapter's two sets
• Use the sum-of-squares formula for the first n natural numbers to evaluate the larger of those totals without adding thirty-one terms
• Explain why that total ranks the two sets in the opposite order to their actual spread
• Divide by the number of observations and show that the ranking corrects
• Define the variance and give its symbol
• Quantify how much more weight squaring gives to a distant observation than averaging distances does

Where it usually goes wrong

• "Squaring is just another way to drop the minus sign, so it must give the same answer." It gives a different answer and a different ranking of magnitudes, because it stretches large deviations more than small ones. Set A's outermost pair carries 56 per cent of the mean deviation and 71 per cent of the variance.
• "A bigger total of squares means more scatter." Not across data sets of different sizes. That is exactly what §13.5's two sets are printed to refute.
• "So the total of squares is useless." It is not — it is zero precisely when there is no scatter at all, and it is the numerator of everything that follows. It is unusable as a comparator, which is a narrower complaint.
• "Dividing by six and by thirty-one is unfair to set B." It is the only way the two are comparable at all. Dividing by the count is what turns a total into a per-observation figure.
• "The variance of set A is bigger, so set A's observations are bigger." Both sets have mean 30. Variance says nothing about where the data sits; it was built from deviations, which are indifferent to the location of the centre.
• "291.67 is exact." It is 1750/6 rounded to two places. Keep the fraction in the working.
• "Set B is more spread because it covers thirty-one values." It covers a narrower interval, 15 to 45, against set A's 5 to 55. The number of observations is not the width.

Questions to check understanding

• Compute the total of the squared deviations for a short list
• Use the sum-of-squares formula for the first n natural numbers to evaluate the total for a run of consecutive integers
• Given two data sets of different sizes, explain why their totals of squares cannot be compared directly
• Compute the variance of a small data set from its definition
• State the condition under which the variance of a data set is zero
• Explain in one sentence why squaring weights distant observations more heavily than averaging distances does

Examples worth working on the board

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

• Set A (§13.5, p. 272): 5, 15, 25, 35, 45, 55 — six observations. Verified: the total is 180 and the mean is 30. The deviations are −25, −15, −5, 5, 15, 25; their squares are 625, 225, 25, 25, 225, 625; the total is 1750. The chapter prints this addition in full.
• Set B (§13.5, p. 272): every whole number from 15 to 45 — thirty-one observations. Verified: the mean is 30, by symmetry, and the deviations run from −15 through 0 to 15, each non-zero size occurring twice. So the total of the squares is twice the sum of the squares of the first fifteen natural numbers. Using n(n + 1)(2n + 1)/6 with n = 15 gives 15 × 16 × 31 ÷ 6 = 1240, and twice that is 2480. The chapter takes exactly this route and prints the formula in a parenthesis.
• The comparison that fails. Verified: set A's deviations reach 25 either side of the mean and set B's reach only 15, yet 1750 is less than 2480. The total ranks B as the more dispersed. The reason is arithmetic and not subtle: every additional observation adds a non-negative term, and B has twenty-five more observations than A. The total is partly a measure of spread and partly a count.
• When the total vanishes (§13.5, p. 272). Each term is a square, so the total can only be zero if every term is zero, which means every observation equals the mean. So the quantity does at least detect the complete absence of scatter, and that is what makes squares a candidate in the first place. Show the one-line argument; it is the only thing in the section that is proved rather than computed.
• Fig 13.5 and Fig 13.6 (§13.5, p. 273). Two number lines, each ticked and labelled 0, 5, 10, …, 60, with an arrow marked at 30 for the mean. Fig 13.5 carries set A's six dots, evenly spaced at intervals of 10 across the width; Fig 13.6 carries set B's thirty-one dots as a dense unbroken block between 15 and 45. Both share the same scale and the same marked mean. Redraw them stacked.
• The repair (§13.5, p. 273). Verified: 1750 ÷ 6 is 291.67 to two places, and 2480 ÷ 31 is exactly 80. The ranking now agrees with the pictures, and with the deviation ranges the chapter quoted. That quotient is the variance, and the chapter writes it with a squared sigma.
• How much weight squaring adds (not in the book). Verified: for set A the two outermost observations contribute 1250 of the total of 1750 — about 71 per cent from two of the six observations. If distances are averaged instead of squares, the same two contribute 50 of a total of 90, about 56 per cent. Squaring did not merely remove the sign; it moved a substantial share of the measure onto the observations furthest out. That is a design choice.
• The two measures side by side for these sets (not in the book). Verified: set A has mean deviation about the mean 15 and variance 291.67; set B has mean deviation 240/31, about 7.74, and variance 80. Both measures rank A above B; they disagree about by how much, A being about twice B on the first and about 3.6 times B on the second.

Figures to have open

• Fig 13.5 and Fig 13.6 redrawn stacked on one 0-to-60 axis with ticks every 5 and the mean arrow at 30. These are the chapter's own figures (p. 273) and the comparison only reads because the two share a scale; redraw as a schematic.
• A stacked-bar breakdown of set A's total, one band per observation, shown twice — once for distances and once for squares — so the shift of weight onto the outer pair is visible. An added figure.
• A pair of parabolic mappings for section 2: the same signed deviation on a number line and its square on a second axis, for a positive and a negative case. Standard schematic.
• No figure is required for the sum-of-squares formula; it is a single displayed line in the chapter (p. 272).

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 13 "Statistics", §13.5 "Variance and Standard Deviation", printed pp. 271–274 up to the displayed definition of the variance at the top of p. 274.
• Set A and set B, their totals and the sum-of-squares formula are on p. 272; Fig 13.5 and Fig 13.6, the two quotients and the naming of the variance are on p. 273.
• The reason the chapter is here at all is §13.4.3, p. 271, which belongs to Where this measure breaks down, and why another was needed.