The Mean, Median, Mode and Range
The mean, median and mode describe the centre of a set of data, and the range describes its spread. This NSW Year 9 Core maths subtopic revises all four from Stage 4 (NESA outcome MA4‑DAT‑C‑02), with several questions that work backwards to a missing score.
You will find the mean as the total divided by the count, the median after sorting, for an odd or even number of scores, the mode (including bimodal data) and the range. You will use frequency tables to find the mean and median, find a missing score from a given mean, as for a golfer's fifth round, see how adding or removing a score changes the mean, and choose the median when one house price of \(\$4\,200\,000\) would drag the mean far above the rest. It is part of Year 9 Core Investigating Data.
Every question with a fully worked solution.
- The Mean, Median, Mode and Range - 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 revises all four from earlier years, finds them from a frequency table, 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. 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, then divide by how many there are.
| \(\text{sum}\) | \(=\) | \(6 + 8 + 11 + 15\) |
| \(\) | \(=\) | \(40\) |
| \(\text{number of scores}\) | \(=\) | \(4\) |
| \(\text{mean}\) | \(=\) | \(\dfrac{40}{4}\) |
| \(\) | \(=\) | \(\mathbf{10}\) |
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.
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 \(90 \div 5 = 18\).
Work out both measures and compare them with the data.
| \(\text{mean}\) | \(=\) | \(\dfrac{8830}{5}\) |
| \(\) | \(=\) | \(\$1766\) |
| \(\text{median}\) | \(=\) | \(\text{3rd of 5 sorted scores}\) |
| \(\) | \(=\) | \(\$980\) |
Four of the five workers earn less than the mean, because the \(\$5000\) outlier drags it up. The median, \(\$980\), is the better summary.
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.