Trying to find math inside everything else

In both geometry and calculus this year the opportunity has come up to ask, “What is a mean average?” Students can usually pull together an answer that gives me a procedure (“add them up and then divide by how many there are”) but not one that really explains what that procedure shows.

With my tutoring students, I usually get to show them the method that my mother taught me when I was young, that I’ve never seen elsewhere. The procedure works by answering the question from the previous paragraph: a mean average is the value you’d have if the quantity were distributed evenly. (If we have 20 total cookies, how many to give each person so they are the same. If we have a certain budget for salaries, how to adjust them so everyone gets paid the same.)

So the method my mother taught me works that way – not adding and dividing, but redistribution. Let’s see an example.

Let’s take these five numbers I got from rolling a d100 five times. I want to average them. First, I’m gonna take a guess of around what the average is. The 10 is gonna pull it down a lot, so let’s guess the average is 70. So first I’ll redistribute the numbers from the ones larger than 70 to the ones lower than 70, like so.

Okay, now 4 of the numbers are the same, but the last one is too low. At least now when I redistribute I can take the same amount from all of them. Let’s do 5.

Pretty close! We have that 1 spare, so we’ll break that into a fraction, giving us a mean average of 65.2.

This idea of redistribution helps clarify the average value of a function or the average rate of change in calculus. Average value of a function is the y-value we’d have if we had the same total value (area) but redistributed so that all the y-values are the same (a constant).

The function in black, and its average value in purple.

The average rate of change of a function is the rate we would be going if we were going at a constant rate (aka draw a straight line, the secant).

You may be thinking that’s great, but often adding and dividing will be a cleaner and faster algorithm, and you’re not wrong. However, this method really shines when you have a question like one of these.

So let’s think. We have four tests, but we don’t know the fourth one.

We do know that we want an 89 average, which means we want all of the tests to be equal to 89. So let’s do that.

So we need a total of 10 extra points to get to that 89 average. Those points can’t come from nowhere – they have to come from the 4th test.

Therefore, that last test must be 99. Perfectly balanced, as all things should be.

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