Shifting and scaling the observations to make the arithmetic small

What to assume they know

• Carrying the frequencies through, for discrete and for grouped data — the variance of a frequency distribution and the formula in two column totals
• The same procedure once the data arrive already grouped — assumed mean, common factor and step-deviations for the mean
• Substituting a linear change of variable into a summation
• Taking a constant square outside a sum
• That the class width of a continuous distribution is a positive number

What they should be able to do

• Write the step-deviation substitution and its inverse
• Prove that adding a constant to every observation leaves the variance unchanged
• Prove that multiplying every observation by a positive constant multiplies the standard deviation by that constant and the variance by its square
• Recover the mean of the original data from the mean of the transformed data
• Recover the standard deviation of the original data from that of the transformed data, and say why only one of the two operations has to be undone
• Apply the assembled shortcut formula to a continuous distribution and check it against the direct computation
• State what the transformation results assume about the sign of the scale factor

Where it usually goes wrong

• "If the mean shifts by the assumed value, so does the spread." It does not. This is the single error the section exists to prevent, and the cancellation in line (4) is the place to show why.
• "Both the shift and the scale have to be undone in the standard deviation." Only the scale. Undoing the shift as well — adding the assumed value back into the spread — is the commonest wrong answer on this material.
• "Dividing by the class width makes the answer smaller, so it is an approximation." It is exact. The division is undone by the multiplication in line (4).
• "The variance is multiplied by the scale factor." By its square. The standard deviation gets the plain factor; the variance gets the square. Example 12 has a factor of 10 and a factor of 100 in front of its bracket.
• "The assumed value must be near the mean for the answer to be right." It must be near for the arithmetic to be small. The result is correct for any choice; a badly chosen one simply loses the benefit.
• "The relation between the two standard deviations holds for any scale factor." As printed it is stated with the factor multiplying directly, which reads correctly because a class width is positive. A negative scale factor would reverse the sign of the transformed observations and the relation would need the factor's size rather than the factor itself, since a standard deviation is never negative. The chapter never uses a negative factor; say the condition rather than leaving it implicit.
• "Example 12 is a new distribution." It is Example 10's distribution recomputed. The agreement of the two answers is the point.

Questions to check understanding

• Compute the mean and standard deviation of a continuous distribution by the shortcut method
• Given the variance of a data set, state the variance after every observation is increased by a constant
• Given the standard deviation of a data set, state it after every observation is multiplied by a constant
• Prove that the variance is unchanged by adding a constant to every observation
• Prove that multiplying every observation by a constant multiplies the variance by that constant's square
• Choose an assumed value and a scale factor for a given table and justify the choice

Examples worth working on the board

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

• The substitution (§13.5.4, p. 279). Each observation is replaced by its deviation from an assumed value, divided by a common factor, which for a continuous distribution is taken as the class width. The chapter labels the substitution and its inverse as line (1). Show both directions; the inverse is what section 6 needs.
• What happens to the mean (§13.5.4, p. 279). Substituting into the definition of the mean and splitting the sum gives the mean of the original data as the assumed value plus the scale factor times the mean of the transformed data. The chapter labels this line (3). Both operations appear in it — the shift as an addition, the scale as a multiplication.
• What happens to the variance (§13.5.4, pp. 279–280). Substituting into the definition of the variance, the assumed value cancels between the observation and the mean, the scale factor comes out of the square, and what is left is the scale factor squared times the variance of the transformed data. Lines (4) and (5) follow. The cancellation of the assumed value is the moment the explanation is built around: it is visible, it is one line, and it is the reason the shift is free.
• The gap argument, said in words. A variance is a weighted mean of squared differences from the mean. A shift moves every observation and the mean by the same amount, so every difference is unchanged. A scaling multiplies every observation and the mean by the same factor, so every difference is multiplied by that factor and every squared difference by its square. That is the whole proof in a sentence, and it should be said before the algebra rather than after.
• Example 12 (§13.5.4, pp. 280–281, Table 13.11). Classes 30–40 through 90–100 with frequencies 3, 7, 12, 15, 8, 3, 2; assumed value 65, common factor 10. Verified: N = 50; mid-points 35, 45, 55, 65, 75, 85, 95; the transformed values are −3, −2, −1, 0, 1, 2, 3; weighted they total −15 and their squares weighted total 105. Verified: the mean is 65 + (−15 ÷ 50) × 10 = 62; the variance is (100 ÷ 2500) × (50 × 105 − 225) = 5025 ÷ 25 = 201; the standard deviation is the root of 201, about 14.18. Every number entering the table is a single-digit integer, against mid-points in the tens and squared deviations in the hundreds and thousands on the direct route.
• The check that makes the section worth teaching. Verified: this is the same distribution as Example 10 of §13.5.3 (pp. 277–278) — the same seven classes with the same seven frequencies — and it returns the same mean, 62, the same variance, 201, and the same standard deviation. Put the two tables side by side; the agreement is the evidence that the transformation was undone correctly.
• A printed cross-reference to handle with care (p. 280). The sentence introducing Example 12 sends the reader to Example 11 for the data. The data actually used is Example 10's, and the answers reproduced are Example 10's; Example 11 is the discrete distribution on values 3, 8, 13, 18 and 23 with a standard deviation of about 6.12. Cite the table caption, not the sentence.
• The label on Example 12 (p. 280) is set in the plural where a singular is meant — one example carries a heading reading as though several followed. Confirmed on the page image. Harmless, but anyone transcribing labels should know it is the book's and not a typing slip of their own.
• Miscellaneous Example 15 (p. 284, in full, with its Note). Every observation is increased by a constant, and the variance is shown to be unchanged. The chapter proves it in full: the new mean is the old mean plus the constant, so the constant cancels inside every squared deviation. The Note beneath restates it for subtraction as well. This is the shift half of §13.5.4 standing on its own.
• Miscellaneous Example 13 (pp. 282–283). Twenty observations with variance 5 are each doubled. Verified: the total of the squared deviations is 100 before and 400 after, so the new variance is 20, which is four times the old. The Note beneath generalises: multiplying by a constant multiplies the variance by that constant's square. This is the scale half of §13.5.4 standing on its own.
• Miscellaneous Exercise items (p. 286). Item 3: six observations with mean 8 and standard deviation 4 are each multiplied by 3; the new mean and standard deviation are asked for. Verified: 24 and 12. Item 4 asks for a proof that multiplying every observation by a constant multiplies the mean by that constant and the variance by its square, and the printed statement carries the restriction that the constant is not zero.
• Exercise 13.2, the items that name the shortcut (pp. 281–282). Item 6: mean and standard deviation of the values 60 through 68 with frequencies 2, 1, 12, 29, 25, 12, 10, 4, 5. Verified, taking 64 as the assumed value with a unit scale: N = 100, the transformed weighted total is 0 so the mean is exactly 64, the weighted total of the transformed squares is 286, the variance is 2.86 and the standard deviation about 1.69. Item 9: mean, variance and standard deviation of heights in centimetres in classes 70–75 through 110–115 with 3, 4, 7, 7, 15, 9, 6, 6, 3 children. Verified, taking 92.5 as the assumed value with a scale of 5: N = 60, the transformed weighted total is 6 and the weighted total of their squares is 254, so the mean is 93, the variance is 15204 × 25 ÷ 3600, about 105.58, and the standard deviation about 10.28.

Figures to have open

• A two-panel movement of a dot row on an axis: first translated, then compressed, with a measured gap carried through both. An added figure and the core of the argument; §13.5.4 prints no diagram of its own.
• Fig 13.3 and Fig 13.4 from p. 267 may be reused here as the picture of the two operations — the chapter drew them for the mean and never redrew them for the variance. Say that they are being borrowed forward.
• Table 13.11 and Table 13.9 side by side, matched row for row, so the two routes to 201 can be compared. The chapter's own tables (pp. 280, 277).
• A units-and-factors card: the factor 10 on the standard deviation, 100 on the variance. Standard schematic.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 13 "Statistics", §13.5.4, whose printed heading names the shortcut route to a variance and a standard deviation, runs pp. 279–281, with the displayed lines labelled (1) to (5) and Table 13.11 on p. 280.
• Example 10, pp. 277–278, with Table 13.9 on p. 277, is the distribution Example 12 recomputes; that is the comparison section 9 rests on.
• Miscellaneous Example 13 and its Note, pp. 282–283; Miscellaneous Example 15 and its Note, p. 284.
• Exercise 13.2 items 6 and 9, pp. 281–282; Miscellaneous Exercise items 3 and 4, p. 286.
• The chapter's Summary, p. 287, carries the shortcut formula and the substitution in one block.