Summarising Data - Measures of Centre
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- Summarising Data - Measures of Centre - Video - Mean, Median, Mode, and Range - How To Find It! Watch
Theory
The mean, median and mode measure the centre of a set of data, and the range measures its spread. This Year 9 maths page finds all four from a list and from a frequency table, works with negative scores, works backwards from a mean to a missing score, and shows why the median is the better choice when the data have an outlier.
The mean is the total of the scores shared out evenly: add the scores and divide by how many there are. People often call it the average.
The median is the middle score once the data are in order. With an even number of scores there are two middle scores, so average them.
The mode is the score that appears most often. It can be any score, even \(0\): “the mode is \(0\)” means \(0\) appears more often than any other score, and says nothing about the mean or the total. Data with one mode are unimodal, with two modes bimodal, and with more than two multimodal. If every score appears equally often there is no mode, and the data are uniform.
The mean, median and mode are measures of centre. The range is a measure of spread: the highest score minus the lowest. It is one number, not two.
An outlier is a score a long way from the rest. It drags the mean and stretches the range, but it barely moves the median.
The median of \(n\) sorted scores is in position \(\dfrac{n + 1}{2}\). For \(7\) scores that is the \(4\)th. For \(8\) scores it is \(4.5\), so average the \(4\)th and \(5\)th.
From a frequency table, add an \(f \times x\) column. Then:
| Score \(x\) | Frequency \(f\) | \(f \times x\) |
|---|---|---|
| \(2\) | \(6\) | \(12\) |
| \(3\) | \(4\) | \(12\) |
| \(4\) | \(6\) | \(24\) |
| \(5\) | \(4\) | \(20\) |
| Total | \(\sum f = 20\) | \(\sum fx = 68\) |
Here the mean is \(68 \div 20 = 3.4\). Counting down the frequency column, scores \(1\) to \(6\) are \(2\)s and scores \(7\) to \(10\) are \(3\)s, so the \(10\)th score is \(3\) and the \(11\)th is \(4\). The median is \(3.5\). Two scores tie for the highest frequency, so the data are bimodal, with modes \(2\) and \(4\). The range is \(5 - 2 = 3\).
How to find the mean, median, mode and range
- Sort the data from smallest to largest. Count the scores.
- Mean: add the scores and divide by the count. From a frequency table, divide \(\sum fx\) by \(\sum f\).
- Median: find position \(\dfrac{n+1}{2}\). From a frequency table, count down the frequency column to that position.
- Mode: find the score with the highest count. Look for a tie.
- Range: subtract the lowest score from the highest.
- Unknown score: write the total as mean \(\times\) count, or put the unknown in the median position, and solve.
Add the scores with their signs, then divide by how many there are.
Sum: \(-6 + 2 - 1 + 4 - 1 - 4 = -6\)
| \(\text{mean}\) | \(=\) | \(\dfrac{-6}{6}\) |
| \(\) | \(=\) | \(\mathbf{-1}\) |
\(-1\) appears twice and every other score once, so the mode is \(-1\).
Sort the data first:
There are \(6\) scores, so the middle is between the \(3\)rd and \(4\)th. Average them. The range is the highest minus the lowest.
| \(\text{median}\) | \(=\) | \(\dfrac{11 + 14}{2}\) |
| \(\) | \(=\) | \(\mathbf{12.5}\) |
| \(\text{range}\) | \(=\) | \(20 - 5\) |
| \(\) | \(=\) | \(\mathbf{15}\) |
The median does not have to be one of the scores.
| Score \(x\) | \(10\) | \(11\) | \(12\) | \(13\) | \(14\) |
|---|---|---|---|---|---|
| Frequency \(f\) | \(4\) | \(7\) | \(9\) | \(6\) | \(3\) |
Multiply each score by its frequency and add. Then divide by the total frequency.
\(\sum f = 4 + 7 + 9 + 6 + 3 = 29\)
\(\sum fx = 40 + 77 + 108\) \(+ 78 + 42\) \(= 345\)
| \(\text{mean}\) | \(=\) | \(\dfrac{345}{29}\) |
| \(\) | \(\approx\) | \(\mathbf{11.90}\) |
With \(29\) scores the median is the \(15\)th. Count down the frequencies: the \(10\)s fill places \(1\) to \(4\), the \(11\)s places \(5\) to \(11\), and the \(12\)s places \(12\) to \(20\). So the median is \(12\).
The highest frequency is \(9\), for the score \(12\), so the mode is \(12\).
Turn each mean back into a total: \(\text{sum} = \text{mean} \times \text{number of scores}\).
| \(\text{total of four}\) | \(=\) | \(17 \times 4 = 68\) |
| \(\text{total of five}\) | \(=\) | \(18 \times 5 = 90\) |
| \(\text{fifth score}\) | \(=\) | \(90 - 68\) |
| \(\) | \(=\) | \(\mathbf{22}\) |
Check: \(68 + 22 = 90\) and \(90 \div 5 = 18\). Removing a score works the same way: find the old total, subtract the score, and divide by the new count.
Common pitfalls
Frequently asked questions
How do you find the mean, median, mode and range?
Mean: add the scores and divide by how many there are. Median: sort the scores and take the middle one, or the average of the middle two. Mode: the most common score. Range: highest minus lowest.
How do you find the median with an even number of scores?
Sort the scores and find the two in the middle. Add them and divide by \(2\). For \(5\), \(9\), \(11\), \(14\), \(16\), \(20\) the middle two are \(11\) and \(14\), so the median is \(12.5\).
Can a set of data have two modes?
Yes. If two scores tie for the most common, the data are bimodal and both are modes. If every score appears equally often, there is no mode.
How do you find a missing number when you know the mean?
Multiply the mean by the number of scores to get the total. Subtract the scores you know. What is left is the missing score.
Which is better, the mean or the median?
When the data have an outlier, the median is better because the outlier drags the mean towards it. When the data are spread evenly with no outliers, the mean uses every score and works well.