Working backwards from an average to a missing value

0:00A coach writes down the weight of every player on the team. Ten players, ten numbers, and beside them a circled average: thirty-nine point two kilograms. Then something spills on the page. Nine of the weights are still legible. The tenth is a smudge. The instinct is that the data is ruined. One value gone, so the list is incomplete, and there is nothing to be done about it. That instinct is wrong, and the reason it is wrong is the most useful idea in this whole topic.

0:37An average is not a summary that has forgotten the numbers. It is a record of their total. Mean equals total divided by count. That gets taught as an instruction: add them up, divide by how many. But it is not an instruction. It is an equation, and an equation can be run in either direction. Multiply both sides by the count and the same statement says: total equals mean times count.

1:10Read forwards it turns data into a summary. Read backwards it turns a summary back into a fact about the data - the one fact that survives the dividing. So. Nine legible weights: forty-two, forty, thirty-nine, thirty-three, forty-eight, thirty-eight, forty-two, thirty-five, thirty-two. They come to three hundred and forty-nine. Call the smudged one w. Whatever w turns out to be, the ten weights together come to three hundred and forty-nine plus w, and dividing that by ten has to give thirty-nine point two.

1:52Now read the formula backwards. Ten values averaging thirty-nine point two come to three hundred and ninety-two. The total is known before the missing weight is. Three hundred and forty-nine plus w is three hundred and ninety-two, so w is forty-three. The missing weight is forty-three kilograms. Not estimated. Forced. Check it back into the average: put forty-three in, the total is three hundred and ninety-two, divide by ten, thirty-nine point two.

2:30And one more check worth making a habit. Forty-three sits between thirty-two and forty-eight, the lightest and the heaviest you could still read. A recovered value that lands outside the range of the others is not impossible, but it is worth a second look. One hole can be filled exactly. Two cannot - two unknowns and one equation is not a question with an answer. Now the same idea with nothing at all to lean on.

3:02A farmer has fifteen coconut trees, and the harvest averages twenty-five point six coconuts a tree. What did the fifteen trees produce altogether? Notice what you have not been given. The fifteen individual counts. Not one of them. There is no list to add up. And it does not matter. Fifteen values averaging twenty-five point six come to three hundred and eighty-four. That total was never written down anywhere, and it is still recoverable, because the average IS the total, divided by fifteen.

3:42Now be exact about what you have and have not just learned. Three hundred and eighty-four is a fact. The fifteen individual harvests are not. Here are fifteen trees where every one gave twenty-five or twenty-six. And here are fifteen trees where fourteen gave a single coconut each and one gave three hundred and seventy. Same total. Same average. The middle of the first is twenty-six; the middle of the second is one. The first spreads over a single coconut, the second over three hundred and sixty-nine.

4:21An average tells you the total. It tells you nothing else whatever. Which is exactly what makes a known mistake repairable. Suppose one tree's count was written down three too many. The true total is three hundred and eighty-four minus three: three hundred and eighty-one. Divide by fifteen and the corrected average is twenty-five point four. Nobody went back and re-counted a single tree. And here is the slip to avoid. The error was three, so the average drops by three - twenty-five point six down to twenty-two point six?

5:03No. The three was in the total, and the total is shared among fifteen. Three divided by fifteen is nought point two. The average drops by two tenths, to twenty-five point four, which is what repairing the total gave. Two routes, one answer. Two quick ones, now that the move is yours. Eight numbers - eight, thirteen, ten, four, five, twenty, y and ten - average ten point three seven five.

5:35Eight values at that average come to eighty-three. The seven you can see come to seventy. So y is thirteen. And notice where that awkward-looking average came from: eighty-three divided by eight is exactly ten point three seven five. Second: fifteen values average one hundred and thirty-four. What do they total? Two thousand and ten. There is nothing else to find here, and nothing else is findable. Turn the whole thing round now and build data to order.

6:12Give me three numbers with a mean of eight. Three numbers averaging eight come to twenty-four. So choose two of them however you like - five and four - and the third has to absorb whatever is left. Fifteen. Five numbers with a mean of thirteen point six? They come to sixty-eight; pick any four and the fifth is forced. There are infinitely many answers, and the total is the one thing every one of them has in common.

6:44Middles can be built to order too, and they behave quite differently. Four numbers with a middle of fifteen point five. The two middle ones have to average fifteen point five, so they have to come to thirty-one. Ten, fifteen, sixteen and forty does it - and notice that fifteen point five is not one of the four numbers. It never had to be. Now six numbers whose average equals their middle. Make them symmetric - four, six, nine, eleven, fourteen, sixteen - and both come out at ten.

7:21Then drag the largest one out to seventy-six. The middle does not move. It is still ten. The average goes to twenty. Here is a puzzle that looks like arithmetic and is really algebra. A three-by-three grid, nine different numbers, and every row, every column and both diagonals have to average ten. Eight lines in all. Three cells averaging ten means every one of those lines totals thirty. Seven, twelve, eleven. Fourteen, ten, six. Nine, eight, thirteen. All eight lines come to thirty.

8:03And now the real question: could some other filling put a different number in the middle? Four of the eight lines pass through the centre. Between them they cover every cell once, and the centre three extra times. Four lines of thirty is a hundred and twenty. All nine cells come to ninety. So three times the centre is thirty. The centre is ten. In every filling there is. Last, a question about questions.

8:36Fill two blanks so that three, eleven, blank, blank, fifteen and six average six point five. Six values at six point five come to thirty-nine. The four you have come to thirty-five. So the two blanks come to four. With counting numbers that is one and three, two and two, or three and one. Three fillings - or two, if you do not care which blank is which. Finite, and you can list them.

9:05Now a cousin of it. Fill three blanks so that five, twenty-one, fourteen and your three have a middle of thirteen. Three, twelve and forty works. Twelve, thirteen and thirteen works. How many are there? Counting up to twenty, ninety-seven. Up to forty, three hundred and thirty-seven. The count just keeps growing. So how many is not a question yet. Somebody has to say how large the numbers are allowed to be.

9:37An average lets you work backwards to one missing number exactly. Knowing when it does not is the same skill, pointed the other way.