A-level Mathematics

11.5 Measures of central tendency and variation, including standard deviation — Learn

Prerequisites: the mean, median and mode from GCSE; using a formula, and squaring and square-rooting with a calculator.

By the end of this section you can: choose and calculate an average; find quartiles and the interquartile range; calculate the standard deviation from a list and from summary statistics; and use the mean and standard deviation to compare two distributions.

1. Measures of central tendency

An average is a single value that stands for the centre of a data set. The three averages are the mean, the median and the mode.

Use the mean for symmetric data with no outliers. Use the median when the data are skewed or contain an outlier. Use the mode for the most common category.

Example 1. Find the mean, median and mode of 3,4,4,6,8,9,113, 4, 4, 6, 8, 9, 11.

mean=3+4+4+6+8+9+117=457=6.43.\text{mean} = \frac{3 + 4 + 4 + 6 + 8 + 9 + 11}{7} = \frac{45}{7} = 6.43.

The middle value is the fourth, so the median is 6. The value 4 occurs twice, and no other value occurs more than once, so the mode is 4.

2. Measures of variation

A measure of variation says how spread out the data are. Two data sets can have the same mean and very different spreads.

3. Quartiles and the interquartile range

Put the data in order. A quartile position uses

position of Q1=n+14,position of Q3=3(n+1)4.\text{position of } Q_1 = \frac{n + 1}{4}, \qquad \text{position of } Q_3 = \frac{3(n + 1)}{4}.

If the position is not a whole number, interpolate between the two values on either side. The interquartile range is

IQR=Q3−Q1.\text{IQR} = Q_3 - Q_1.

Example 2. Find Q1Q_1, Q3Q_3 and the IQR of 3,5,8,10,12,143, 5, 8, 10, 12, 14.

There are n=6n = 6 values. The position of Q1Q_1 is

6+14=1.75,\frac{6 + 1}{4} = 1.75,

which is three-quarters of the way from the 1st value to the 2nd:

Q1=3+0.75(5−3)=4.5.Q_1 = 3 + 0.75(5 - 3) = 4.5.

The position of Q3Q_3 is

3(6+1)4=5.25,\frac{3(6 + 1)}{4} = 5.25,

so

Q3=12+0.25(14−12)=12.5.Q_3 = 12 + 0.25(14 - 12) = 12.5.

The interquartile range is

IQR=12.5−4.5=8.\text{IQR} = 12.5 - 4.5 = 8.

4. The standard deviation

The standard deviation is the root mean square deviation from the mean. It measures how far, on average, the values sit from the mean.

For a data set x1,x2,…,xnx_1, x_2, \ldots, x_n with mean xˉ\bar{x},

σ=∑(x−xˉ)2n.\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}.

A useful calculating form is

σ=∑x2n−xˉ2.\sigma = \sqrt{\frac{\sum x^2}{n} - \bar{x}^2}.

The standard deviation has the same units as the data. A large standard deviation means the values are spread widely around the mean. A small standard deviation means they cluster near the mean.

Some calculators and spreadsheets divide by n−1n - 1 instead of nn, and call the result the sample standard deviation. Both are accepted at A-level, but you must say which one you used. The method above divides by nn.

Example 3. Find the standard deviation of 2,4,6,8,102, 4, 6, 8, 10.

The mean is

xˉ=2+4+6+8+105=6.\bar{x} = \frac{2 + 4 + 6 + 8 + 10}{5} = 6.

The squared deviations from the mean are

16, 4, 0, 4, 16.16,\ 4,\ 0,\ 4,\ 16.

Their sum is 40, so

σ=405=8=2.83 (3 s.f.).\sigma = \sqrt{\frac{40}{5}} = \sqrt{8} = 2.83 \ (3 \text{ s.f.}).

5. Standard deviation from summary statistics

Sometimes you are given the totals rather than the list. Use the calculating form.

Example 4. A data set has n=10n = 10, ∑x=60\sum x = 60 and ∑x2=800\sum x^2 = 800. Find the mean and the standard deviation.

The mean is

xˉ=6010=6.\bar{x} = \frac{60}{10} = 6.

Then

σ=80010−62=80−36=44=6.63 (3 s.f.).\sigma = \sqrt{\frac{800}{10} - 6^2} = \sqrt{80 - 36} = \sqrt{44} = 6.63 \ (3 \text{ s.f.}).

6. Using the mean and standard deviation to compare

Two distributions are often compared with the mean and the standard deviation together. The mean describes the centre, and the standard deviation describes the spread.

Example 5. Two classes sit the same test. Class A has mean 60 and standard deviation 4. Class B has mean 60 and standard deviation 12. Compare the classes.

Both classes have the same mean, so their typical scores agree. Class B has the larger standard deviation, so its scores are more spread out. Class B contains more very high and very low scores, and Class A is more consistent.

Example 6. Every value in a data set is increased by 10. Explain what happens to the mean and the standard deviation.

Adding 10 moves every value up by 10, so the mean rises by 10. The distances from the mean do not change, so the standard deviation is unchanged.

7. Try it: what moves when the data change

Predict first. The figure shows five readings, their mean and a bar one standard deviation wide either side of the mean. Before you touch the controls, predict which of the two changes more when you move the fifth reading a long way from the others: the mean or the standard deviation. Write your prediction down.

Test. Move the slider for the fifth reading and watch the marker for the mean and the bar for the standard deviation. Then move the slider that adds the same amount to every reading, and watch both again.

Explain. Write two sentences. The first explains why the mean moves by only one-fifth of the change in one reading. The second explains why adding the same amount to every reading leaves the standard deviation unchanged.

Watch out. Three mistakes are common.

Forgetting to square the deviations. The formula uses ∑(x−xˉ)2\sum(x - \bar{x})^2, not ∑(x−xˉ)\sum(x - \bar{x}). The unsquared deviations add to zero for any data set, so the squares are essential.

Dividing by n−1n - 1 without saying so. If your calculator uses n−1n - 1, state it. The two values are close for a large sample, but they differ for a small one.

Comparing means without comparing spreads. Two data sets can share a mean and have very different spreads. Always report a measure of variation as well.

Re-check: the standard deviation cannot be negative, and it is zero only when every value is the same.

You should now be able to

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