 # Understanding Averages

## Averages

In your work as Stubble Development Officer for the RSPB (Royal Society for the Protection of Beards), you are asked to determine just exactly how long are the member’s beards.

You decide to measure the length of 11 individual beards.

It’s taxing and itchy work and takes ages but eventually, you succeed. Tired, but triumphant you present your data to The Grand Wizard Beard-Master of the RSPB. “Hah!” she says. “This is all very well, but I’m too busy to look at 11 numbers”. “Can’t you give me one that represents them all?”

YES you can. You need a measure of the central tendency of the 11 beard lengths that comprise your data set. You need to calculate an AVERAGE.

There are 3 kinds of average that concern us:

The arithmetic mean, the median and the mode.

Most people will have heard of the arithmetic mean (commonly called simply the mean or in normal conversation the average)

If you add up all your beard lengths and divide by 11 you will have calculated the arithmetic mean of those 11 measurements:

Beard lengths/m: 0.2, 0.1, 0.5, 1.8, 0.1, 0.1, 0.2, 0.9, 0.2, 0.2, 0.6

0.2 + 0.1 + 0.5 + 1.8 + 0.1 + 0.1 + 0.2 + 0.9 + 0.2 + 0.2 + 0.6 = 0.45

= The arithmetic mean of your sample of 11 measurements.

The symbol for the arithmetic mean of your sample is: . This is an X with a line over the top (a bar) and so it’s referred to as X-bar.

Your is only based on 11 samples, it’s likely that if you measured more than 11 beards you would get a different result. If you could somehow swing an invitation to the annual RSPB Awards Ceremony (which ALL members MUST attend on pain of shaving), you might measure the beard-length of every single member.

If you managed this unlikely feat, you would have the true mean of the whole population of beards. This value is given the symbol: μ

For most things you might measure (e.g. Tree girths, shell-lengths, people-heights, seed-weights and so on) you will not be able to measure them all, so you will end up with sample arithmetic means.

Our value of (0.45m) is all very well, but did you notice that there was one measurement that was way different to the others? Obviously that was the chin-wobblingly astonishing 1.8m example (fourth from the left in the list). Including this observation makes our average much larger than it otherwise would have been. This makes our arithmetic mean not very representative of the sample as a whole, which (if you remember back to the beginning) is sort of, the whole point.

DO NOT DESPAIR...... We can get around this catastrophic situation by using a different kind of mean: THE MEDIAN.

To obtain the median all you do is put the data in order (from lowest to highest) and find the middle value. So for our beard-length data:

0.1, 0.1, 0.1, 0.2, 0.2, 0.2, 0.2, 0.5, 0.6 , 0.9, 1.8,

The figure in red is the middle value

Our median is 0.2. Most reasonable RSPB members would accept that as a more representative value for beard-length than the arithmetic mean.

Astute readers will have noticed that your author has cunningly used 11 pieces of data in order to avoid the problem of what to do if there is an even number of items of data, where you would have no middle value. All you do in that case is take the mean of the value of the middle pair. For example look at the row of 11 numbers below:

1 2 2 2 3 4 4 4 5 6

The middle pair are 3 and 4, so the median = 3 + 4/2 = 3.5

The alternative to both the arithmetic mean and the median is THE MODE.

The mode is a measure of central tendency based on the numbers of observations. In our beard-length example the data were as follows:

0.1, 0.1, 0.1, 0.2, 0.2, 0.2, 0.2, 0.5, 0.6 , 0.9, 1.8,

3 values of 0.1 (red)

4 values of 0.2 (blue)

just one of each of the other values (green)

It’s clear that the most common value was 0.2, there are four items of data which have that value, so the mode is 0.2.