On Average(s)

One of the concepts that standardised tests use to trouble students is averages, ironically enough. Ironic, because calculating averages is one of the few things from maths classes we tend to retain over time – how could averages ever give us trouble?

While calculating averages is easy – simply add everything up and divide by how many things there are – understanding what they are often isn’t so straight forward. In fact, we tend to think of averages as being precisely what they aren’t – and the test is counting on that.

First off: averages aren’t the middle number in the distribution: that’s the median. And they’re also not the most likely number: that’s the mode.

So, what exactly are averages, then?

An average is best thought of as being the balancing point in the data: the place where all the deviations to one side are perfectly balanced by all the deviations to the other! Consider the following distribution:

1 2 3 4 5

It’s pretty obvious that the average here is 3 – and that’s in the middle, right? It is – but only because of where all the other numbers are compared to 3! Let’s look at the distance from ‘3’ to each of the numbers:

-2 -1 0 1 2
1 2 3 4 5

Those distances balance each other out around 3, so 3 is the average. (Also, (1+2+3+4+5)/5 totally equals 3, so the old method works too.) What’s interesting though is that, if we chose different numbers, we would still see the same thing! Consider:

1 5 6 9 19

Here, the average is 8: (1+5+6+9+19 = 40 and 40/5 = 8). So what happens with those distances?

-7 -3 -2 +1 +11
1 5 6 9 19

As you can see, the distances add up to zero! The numbers balance around the average, because that’s what an average does!

So How Does This Help?

By looking at the distances from the average instead of the numbers themselves, it makes it a lot easier to calculate an answer on the types of questions you are likely to get on a standardised exam. Basically – smaller numbers are simpler and easier to deal with, they take less time and you are less likely to make a mistake! Check out this example question:

A group of children had their heights measured for a project. The 18 boys had an average height of 47 inches, while the girls had an average height of 44. If the group as a whole had an average height of 46 inches, then how many girls are in the class?
a) 7
b) 8
c) 9
d) 10
e) 11

Answer and Explanation – The Hard Way

Attacking this question in the traditional way would involve a LOT of annoying calculation. You could try stating that the group’s average height would equal the sum of the kids individual heights divided by the number of kids – so:

46 = (47(18) + 44g)/(18+g)

where g is the number of girls in the group. We could then move that denominator to the other side to get

46(18+g) = 47(18) + 44g
828 + 46g = 846 + 44g
2g = 18
g = 9

All of that is true! And ‘9’ is the right answer! But, seriously – who has time for this? Not you – not on a standardised exam. So, what’s the easier way to do it? Let’s check out that balance.

Answer and Explanation – the Easy Way

First off: how far from the average height are the boys? Well, on average, they’re one inch taller and there are 18 boys – so, that’s 18 inches altogether! To balance that, the girls would have to be 18 inches shorter altogether – and since it’s 2 inches per girl, that gives us 9 girls.

The maths here are literally as simple as multiplying 18 times 1 and then dividing by 2: smaller numbers, easier calculations and it takes almost no time at all.

The Upshot

The point of this all is that maths questions on standardised tests are not simply about memorising formulae and working really quickly: they’re about testing your ability to think critically and use the proper tool for the proper job. Understanding what averages really are allows you to use methods that are simpler, easier and quicker – so you can get more questions right, feel less stress and achieve your high score!

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