Carrying the frequencies through, for discrete and for grouped data

What to assume they know

• Taking the root to get the spread back into the units of the data — variance and standard deviation for a plain list
• The same procedure once the data arrive already grouped — frequencies as weights, mid-points, and the total N
• Expanding a square of a difference
• Splitting a sum of three terms into three sums, and taking a constant outside a summation
• That the weighted total of the observations equals N times the mean

What they should be able to do

• Write, in terms of the frequencies, the variance of a discrete frequency distribution and its standard deviation
• Compute both for a discrete distribution using a five-column working table
• Replace the classes of a continuous distribution by mid-points and compute the variance
• Derive the second form of the variance by expanding the square and using the weighted total of the observations
• State the identity that the variance is the mean of the squares less the square of the mean, and read off the inequality it forces between those two quantities
• Compute a standard deviation from the two column totals alone, without a deviation column
• Decide, for a given distribution, which of the two routes will be less work

Where it usually goes wrong

• "The two formulas give different variances." They are the same quantity, and Example 10 and Example 12 work the same distribution by two different routes and reach 201 both times. Be exact about which two routes those are: Example 10 goes by the definition and Example 12 by the step-deviation shortcut of §13.5.4, so the pair evidences definition-against-shortcut. The one worked example that applies the alternative formula itself is Example 11. If a student's two routes disagree, the arithmetic is wrong, not the theory.
• "The second formula is an approximation." It is an identity, derived in four algebraic steps with nothing dropped.
• "You still need the mean for formula (3)." You do not. The mean was substituted out; the formula needs only the two column totals and N. That is precisely why it exists.
• "Squaring the observations is the same as squaring the deviations." It is not, and the identity says by exactly how much they differ — the square of the mean.
• "N is the number of rows in the table." N is the total of the frequency column: 30 in Example 9, 50 in Example 10, 48 in Example 11.
• "Since the mid-points stand in for the classes, the answer is approximate." The computation is exact for the reconstructed data; the approximation lives in the reconstruction, not in the arithmetic. Say which is which.
• "The observation with the largest frequency contributes most." Not necessarily. In Example 9 the value 32 occurs once and contributes 324 of the 1374, more than the value 11 contributes with a frequency of nine.

Questions to check understanding

• Compute the mean, the variance and the standard deviation of a discrete frequency distribution
• Do the same for a continuous distribution using mid-points
• Derive the second form of the variance from the definition
• Compute a standard deviation from the totals of the observations and of their squares, without forming a deviation column
• Find the variance of the first n natural numbers in terms of n
• Convert a distribution with gaps into a continuous one and then find its standard deviation
• Explain why the mean of the squares of a data set cannot be less than the square of its mean

Examples worth working on the board

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

• The discrete formula (§13.5.2, p. 275). The standard deviation is the root of the weighted total of the squared deviations divided by N, where N is the total of the frequencies. The chapter labels the displayed line (2). It is the §13.5.1 definition with each squared deviation counted as many times as its value occurs.
• Example 9 (§13.5.2, pp. 275–276, Table 13.8). Values 4, 8, 11, 17, 20, 24, 32 with frequencies 3, 5, 9, 5, 4, 3, 1. Verified: N = 30; the products total 420, so the mean is 14; the deviations are −10, −6, −3, 3, 6, 10, 18; their squares are 100, 36, 9, 9, 36, 100, 324; weighted, those total 1374; the variance is 45.8 and the standard deviation is the root of 45.8, reported as 6.77. Note that the single observation at 32 contributes 324 of the 1374 — nearly a quarter of the whole measure from one of the thirty observations.
• The continuous case (§13.5.3, p. 276). Each class is replaced by its mid-point and the discrete method is then applied unchanged. No new idea; the mid-point assumption is the same one §13.4.2 made and carries the same cost.
• Example 10 (§13.5.3, pp. 277–278, Table 13.9 on p. 277). Classes 30–40 through 90–100 with frequencies 3, 7, 12, 15, 8, 3, 2. Verified: N = 50; mid-points 35, 45, 55, 65, 75, 85, 95; the products total 3100, so the mean is 62; the squared deviations are 729, 289, 49, 9, 169, 529, 1089; weighted they total 10050; the variance is 201 and the standard deviation is the root of 201, reported as 14.18.
• The derivation (§13.5.3, pp. 276–277). Four moves. First, expand the squared deviation into the square of the observation, plus the square of the mean, less twice their product. Second, split the weighted sum into three weighted sums. Third, note that the frequencies alone total N and that the observations weighted by their frequencies total N times the mean — this is what empties the third term and turns the second into the square of the mean times N. Fourth, divide by N. What is left is the weighted mean of the squares of the observations, less the square of the mean.
• The consequence worth stating. A variance is never negative, so that identity forces the mean of the squares to be at least the square of the mean, for any data whatever, with equality only when every observation is the same. The chapter does not draw this out; it is one line and it is the sort of thing a student is asked to justify.
• Formula (3) (§13.5.3, p. 277). Substituting the mean itself as the weighted total divided by N and clearing denominators gives the standard deviation as one over N, times the root of N times the weighted total of the squared observations, less the square of the weighted total of the observations. Two column totals and N — nothing else.
• Example 11 (§13.5.3, pp. 278–279, Table 13.10 on p. 278). Values 3, 8, 13, 18, 23 with frequencies 7, 10, 15, 10, 6. Verified: N = 48; the products total 614; the squares of the values are 9, 64, 169, 324, 529 and, weighted, total 9652. Formula (3) gives one forty-eighth of the root of 48 × 9652 − 614², which is one forty-eighth of the root of 463296 − 376996 = 86300; the root is about 293.77 and the standard deviation is about 6.12. Verified, and this is the point of the example: the mean here is 614 ÷ 48, which is 12.7916… and does not terminate. A deviation column built on it would be five decimal places wide in every row. The chapter chose this data so that the new formula would be visibly cheaper.
• Exercise 13.2, the items belonging here (pp. 281–282). Mean and variance for: item 1, the list 6, 7, 10, 12, 13, 4, 8, 12; item 2, the first n natural numbers; item 3, the first ten multiples of 3; item 4, values 6, 10, 14, 18, 24, 28, 30 with frequencies 2, 4, 7, 12, 8, 4, 3; item 5, values 92, 93, 97, 98, 102, 104, 109 with frequencies 3, 2, 3, 2, 6, 3, 3; item 7, classes 0–30 through 180–210 with frequencies 2, 3, 5, 10, 3, 5, 2; item 8, classes 0–10 through 40–50 with frequencies 5, 8, 15, 16, 6. Item 10 asks for the mean diameter of a set of circles measured in millimetres, and for their standard deviation; the classes are given as 33–36, 37–40, 41–44, 45–48 and 49–52 with 15, 17, 21, 22 and 25 circles, together with a printed hint to close the gaps by shifting every limit half a unit. Verified, by working added here: item 1 has mean 9 and variance 9.25; item 2 has mean (n + 1)/2 and variance (n² − 1)/12; item 3 has mean 16.5 and variance 74.25; item 4 has mean 19 and variance 43.4; item 5 has mean 100 and variance 640/22, about 29.09; item 7 has mean 107 and variance 2276; item 8 has mean 27 and variance 132; item 10, once the classes are closed, has mean diameter 43.5 mm, variance 30.84 and standard deviation about 5.55 mm.
• A connection worth flagging (not in the book). Verified: Exercise 13.2 item 1 is the same eight numbers as Example 1 of §13.4.1, and its mean and variance, 9 and 9.25, are exactly the mean and variance quoted in item 1 of the Miscellaneous Exercise on p. 286, which withholds two of the eight observations and asks for them back. So the chapter sets the same data three times in three different directions.

Figures to have open

• Working tables mirroring Tables 13.8, 13.9 and 13.10 in column order, so a student can follow in the book. The chapter's own tables (pp. 275, 277, 278).
• A step-by-step algebra panel for sections 6 to 10, carrying the derivation line by line with the cancelling term highlighted. The chapter prints this derivation as a block of displayed lines on pp. 276–277; the explanation's job is to slow it down.
• A two-column-totals card for section 10 — the only two numbers formula (3) needs, with N. Standard schematic.
• A bar of contributions for Example 9, one bar per value, heights proportional to the weighted squared deviations, so the single observation at 32 stands out. An added figure.

Where this sits in the book

• NCERT Class XI Mathematics, Chapter 13 "Statistics". §13.5.2, whose printed heading names the discrete frequency distribution, runs pp. 275–276; §13.5.3, whose printed heading names the continuous frequency distribution, runs pp. 276–279 and carries both the subheading on p. 276 that introduces the second formula and the displayed line labelled (3) on p. 277.
• Tables 13.8, 13.9 and 13.10, on pp. 275, 277 and 278.
• Exercise 13.2 items 1 to 5, 7, 8 and 10, printed pp. 281–282. Items 6 and 9 ask explicitly for the shortcut method and belong to Shifting and scaling the observations to make the arithmetic small.
• Item 1 of the Miscellaneous Exercise, p. 286, uses the same eight observations as Exercise 13.2 item 1; that is a deliberate cross-reference within the chapter.