Working backwards from an average to a missing value

What to assume they know

• Mean = total ÷ count, and reading a dot plot or a list as a total — The mean as the point where the distances balance
• Naming an unknown with a letter, forming a one- or two-step equation and solving it; carried in from earlier classes, since this chapter writes w and z without re-teaching the move
• Multiplying a decimal by a whole number, e.g. 39.2 × 10 and 25.6 × 15
• Dividing a whole number by a whole number to a decimal answer, e.g. 381 ÷ 15
• The median of an even-sized collection as the average of the two middle values — Whether adding a value raises or lowers the median
• That a magic-square-style constraint means every named line has the same total

What they should be able to do

• Convert a stated average and count into the total of the data
• Set up an equation with one unknown from a list of values and their stated mean, and solve it
• Check a recovered value by recomputing the average with it in place
• Reconstruct a total that was never listed, from an average and a count alone, and say what that total does and does not tell you about individual values
• Correct an average when one recorded value is known to be wrong by a stated amount, without re-collecting the data
• Build a collection of a stated size with a stated mean, and explain why there are infinitely many
• Build a collection of a stated size with a stated median, including one that is not a value in the collection
• Build collections in which the mean equals the median, and in which it exceeds it
• Fill blanks in a partly given collection to hit a stated mean or median, and count the possibilities under a stated restriction

Where it usually goes wrong

• "If a value is missing, the data is unusable." The chapter's opening image is a smudge, and the whole point is that the average is a record of the total, so one hole can be filled exactly. Two holes cannot.
• "You cannot find a total without the individual values." Venkayya's fifteen trees are never listed and the total is still 384. The average is not a summary that has forgotten the total; it is the total, divided.
• "So the average tells me the individual values." It tells you the total and nothing else. Two harvests of fifteen trees can share an average and have nothing else in common.
• "An error of 3 in one tree changes the average by 3." It changes the average by 3 ÷ 15. This is the same slip as adding a fixed number to every value and expecting the mean to move by that number times the count, and it is worth showing the two side by side.
• "You have to re-collect the data to fix the average." Not when the error is a known amount in a known direction. Fix the total, divide again.
• "A median has to be one of the data values." Item 2 sets a median of 15.5 over four values; the answer never needs 15.5 to appear at all.
• "Every filling-in question has finitely many answers." Item 4 has three; item 3 needs a convention before "how many" even means something. Being able to say which kind of question you have been handed is part of the skill.
• "The middle cell of a mean grid is free." It is the one cell that is not. Every filling of the grid has 10 in the centre.

Questions to check understanding

• Given a list with one entry missing and the mean, find the entry
• Given a mean and a count, state the total
• Given an average, a count and a known recording error, compute the corrected average
• Given a corrected average and the original, work back to the size of the error
• Construct a collection of stated size with a stated mean; with a stated median; with mean equal to median; with mean greater than median
• Fill blanks in a partly given collection to hit a stated mean, and list every filling allowed by a stated restriction
• Explain why a stated average does not determine the individual values, with two collections as evidence
• Complete a mean grid and explain why its centre entry is forced

Examples worth working on the board

Values marked printed appear on the page. Values marked not in the book are worked out here on the chapter's stated inputs. The chapter prints no answers to any exercise item and Part II has no answer appendix.

• The wrestlers' weights (Part II p.109, printed; the list is set as handwritten artwork down the right-hand side of the page, with a circled note reading Avg 39.2 beside it and a dark smudge where the last entry should be). Read off the printed page, the nine legible weights in kilograms are 42, 40, 39, 33, 48, 38, 42, 35 and 32; the tenth is the smudged one, and the number of players is ten.
• The worked solution (Part II p.109, printed in full). With the unknown called w, the equation is (42 + 40 + 39 + 33 + 48 + 38 + 42 + 35 + 32 + w) ÷ 10 = 39.2, which simplifies to 349 + w = 392, so w = 43. The missing weight is 43 kg. An added check to show: 392 ÷ 10 really is 39.2, and 43 sits inside the range of the other nine, which is a sanity test worth teaching.
• Venkayya's coconut harvest (Part II p.109, printed). The average harvest per tree is 25.6, there are 15 trees, and the individual per-tree counts are not given at all. One tree's count was written down as 3 more than the true number. The page's solution names the recorded total z, gets z = 25.6 × 15 = 384, subtracts the error to reach 381, and reports the corrected average as 381 ÷ 15 = 25.4. The page states in as many words that the per-tree data is not available and asks whether the total can still be found — that question is the pivot of the whole topic.
• What the reconstructed total cannot do (not in the book). 384 is recoverable; the fifteen individual counts are not. Any fifteen numbers totalling 384 fit the same average, so a class should be asked to invent two very different harvests with the same mean. This is the honest limit of working backwards, and it is also the reason the correction works: the error was in the total, so it can be repaired in the total.
• The shortcut worth naming (not in the book). Removing 3 from a total shared by 15 trees moves the average by 3 ÷ 15 = 0.2, which takes 25.6 to 25.4 without computing either total. Both routes should be shown; the second is the same reasoning as the insertion rule in Which added values move the mean, and in which direction.
• Item 7 (Part II p.115, printed): the mean of 8, 13, 10, 4, 5, 20, y and 10 is 10.375, and y is wanted. Not in the book: eight values averaging 10.375 total 83; the seven given values total 70; so y = 13. A useful aside when explaining it — 10.375 is 83 ÷ 8 exactly, which is why the stated mean has three decimal places, and a class can be asked why a mean of eight whole numbers had to end in .375 or .25 or .5 or .125.
• Item 8 (Part II p.115, printed): fifteen values have mean 134; find the total. Not in the book: 2010. Two lines long, and the point is that no information about the individual values is needed or obtainable.
• Item 2 (Part II p.127, printed as five construction tasks, two examples wanted for each). Give three numbers with mean 8; four with median 15.5; five with mean 13.6; six with mean equal to the median; six with mean greater than the median. Not in the book: the targets convert to totals of 24 and 68 for the first and third; 15.5 as a median of four values needs the two middle ones to total 31, for instance 10, 15, 16, 40 or 2, 15.5, 15.5, 100 — a good moment to note that a median need not be one of the values. For the last two, six values symmetric about their centre give mean equal to median, and dragging the largest value far to the right makes the mean exceed the median while leaving the median untouched — the same contrast as Whether adding a value raises or lowers the median section 9.
• Item 1, Mean Grids (Part II p.127, printed; the three-by-three grid is drawn as empty blue cells to the right of the text). Fill nine distinct numbers so that every row, every column and each diagonal averages 10. Not in the book: averaging 10 over three cells means each of those eight lines totals 30, so this is a magic square with constant 30. Adding 5 to every entry of the classical three-by-three square built from 1 to 9 gives 7, 12, 11 / 14, 10, 6 / 9, 8, 13, which satisfies all eight lines and uses nine distinct numbers. The centre cell is forced to be 10 — the four lines through the centre together give a relation that pins it — which is the one piece of real mathematics in the item and the answer to its part (ii): yes, other fillings exist, but none of them can move the middle.
• Item 4 (Part II p.127, printed): fill two blanks so that 3, 11, ___, ___, 15 and 6 has mean 6.5, and count the possibilities if only counting numbers are allowed. Not in the book: six values averaging 6.5 total 39, the four given values total 35, so the two blanks total 4. With counting numbers that is 1 and 3, 2 and 2, or 3 and 1 — three ordered fills, two if the pair is treated as unordered. This is the cleanest finite-count item in the chapter.
• Item 3 (Part II p.127, printed): fill three blanks so that 5, 21, 14, ___, ___, ___ has median 13, and count the possibilities under the same restriction. Not in the book: with six values the two middle ones must total 26. Two fills that work: 3, 12 and 40, which sorts to 3, 5, 12, 14, 21, 40 with 12 and 14 in the middle; and 12, 13, 13, which sorts to 5, 12, 13, 13, 14, 21. The count is the item's real work and it depends on a convention the item does not state — see §Notes.

Figures to have open

• The notepad figure from Part II p.109 — a column of handwritten weights, a circled average, and a smudge where one entry should be. This is the chapter's own artwork and it is what makes the problem feel like a problem. Redraw it; do not reproduce it.
• A total-and-shares diagram: one long bar labelled with the total, cut into equal shares, then one share replaced by a labelled unknown. Not in the chapter, and it carries sections 2 and 6.
• The three-by-three grid with all eight lines drawn as translucent bands, each carrying 30. Standard schematic.
• A pair of very different fifteen-tree harvests with the same total, drawn as two rows of coconut counts. Standard schematic; this is the figure that stops the average being over-read.
• The coconut-farm illustration on Part II p.109 is decorative and can be dropped.
• No photograph is needed.

Where this sits in the book

• NCERT Ganita Prakash Class 8, Part II, printed Chapter 5, "Tales by Dots and Lines", §5.1 "The Balancing Act", the unnumbered subsection "Finding the Unknown", Part II p.109. The wrestlers' case occupies the upper half of the page and the coconut case the lower half; the subsection begins immediately below the median argument and runs to the foot of the page, with "Mean and Median with Frequencies" starting on Part II p.110.
• Exercise items owned by this topic: Part II p.115 items 7 and 8; Part II p.127 items 1, 2, 3 and 4. The Part II p.115 set is shared with three sibling topics and the Part II p.127 set with two.
• Sibling topics: the direction-of-change reasoning this topic inverts is Which added values move the mean, and in which direction (mean) and Whether adding a value raises or lowers the median (median); the missing dot on a plot with a stated mean, Part II p.113 item 2, is filed under Which added values move the mean, and in which direction.
• Forward pointer: computing the same two statistics from a frequency table, and then in a spreadsheet, is Mean and median from a frequency table, by hand and in a spreadsheet, Part II pp.110–113.