The Mean, Median, Mode and Range
The mean, median and mode describe the centre of a dataset, and the range describes its spread. This NSW Year 9 Advanced maths subtopic revises these Stage 4 statistics (NESA outcome MA4‑DAT‑C‑02) and works backwards from them.
You will find all four from lists, frequency tables and dot plots, see how an outlier changes the mean, and choose the best measure for skewed house prices. You will find a missing score from a mean, decide whether a target average is possible, find scores from given statistics, see how adding and doubling change each measure, and correct a mean after a wrong mark. It is part of Year 9 Advanced 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 describe the centre of a dataset; the range describes its spread. This Year 9 Advanced maths page revises the earlier measures and pushes further: frequency tables, totals, outliers and finding unknown values.
The mean \(\bar{x}\) is the sum of the scores divided by the number of scores. The median is the middle score when the scores are in order (or the mean of the two middle scores). The mode is the most common score. The range is highest score \(-\) lowest score.
The mean, median and mode are measures of centre. The range is a measure of spread.
A dataset can have no mode (uniform: every score equally common), one mode (unimodal), two (bimodal) or more (multimodal).
An outlier is a score far from the rest. It drags the mean and stretches the range but barely moves the median, so the median is often the better "typical" value when there are outliers.
How to find the four statistics
- Order the data (or use a frequency table, dot plot or stem-and-leaf plot, which are already in order).
- Mean: add the scores and divide by how many. In a table, use \(\sum fx \div \sum f\).
- Median: find position \(\dfrac{n + 1}{2}\). In a table, count down a cumulative frequency column.
- Mode and range: the highest frequency; highest \(-\) lowest.
Harder questions
- Scores added, removed or wrong: change means into totals, adjust the total, then divide by the new count.
- Unknown score: write the mean as an equation and solve. A target mean above \(100\%\) of a test shows a goal is impossible.
- Unknowns in ordered data: the median position tells you which value equals the median; the mode tells you which values repeat.
- Choosing a measure: with an outlier, prefer the median; for the most popular category, use the mode.
Add an \(fx\) row and a cumulative frequency row.
| Siblings \(x\) | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| Frequency \(f\) | 4 | 9 | 7 | 3 | 1 | 1 |
| \(fx\) | 0 | 9 | 14 | 9 | 4 | 5 |
| Cumulative \(f\) | 4 | 13 | 20 | 23 | 24 | 25 |
| \(\text{mean}\) | \(=\) | \(\dfrac{\sum fx}{\sum f} = \dfrac{41}{25} = \mathbf{1.64}\) |
Median: \(25\) scores, so the middle one is the \(\dfrac{25 + 1}{2} = 13\)th. The cumulative frequency reaches \(13\) at \(x = 1\), so the median is \(\mathbf{1}\).
Mode: the highest frequency is \(9\), so the mode is \(\mathbf{1}\). Range: \(5 - 0 = \mathbf{5}\).
Turn each mean into a total: total \(=\) mean \(\times\) number of scores.
| \(\text{total of 8}\) | \(=\) | \(15 \times 8 = 120\) |
| \(\text{total of 10}\) | \(=\) | \(16 \times 10 = 160\) |
| \(\text{the two new scores}\) | \(=\) | \(160 - 120 = 40\) |
| \(\text{other score}\) | \(=\) | \(40 - 12 = \mathbf{28}\) |
Check: \(\dfrac{120 + 12 + 28}{10} = 16\) \(\checkmark\)
(i) Find the mean and median.
(ii) Which better describes a typical time?
(iii) Find the mean without the \(65\).
| \(\text{mean}\) | \(=\) | \(\dfrac{195}{9} = 21.666\ldots \approx \mathbf{21.7}\text{ min}\) |
| \(\text{median}\) | \(=\) | \(5\text{th score} = \mathbf{16}\text{ min}\) |
(ii) \(65\) is an outlier, far from the rest. It pulls the mean above all but one of the times, so the median is the better measure here.
| \(\text{(iii) mean}\) | \(=\) | \(\dfrac{195 - 65}{8} = \dfrac{130}{8} = \mathbf{16.25}\text{ min}\) |
Removing the outlier changes the mean a lot, the median only a little (to \(15.5\)), and the range from \(53\) to \(9\).
| Score | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Frequency | 5 | \(k\) | 4 | 6 |
There are \(15 + k\) scores. Scores of \(1\) fill positions \(1\) to \(5\); scores of \(2\) fill positions \(6\) to \(5 + k\). The middle position must land in the 2s.
Try values. If \(k = 5\): \(20\) scores, the middle two are the \(10\)th and \(11\)th, but the 2s only reach position \(10\), so the median is \(\dfrac{2 + 3}{2} = 2.5\).
| \(k = 6:\ n\) | \(=\) | \(21,\ \text{middle} = \dfrac{21 + 1}{2} = 11\text{th}\) |
| \(\text{2s fill positions}\) | \(=\) | \(6 \text{ to } 11 \ \checkmark\) |
So the smallest value is \(\mathbf{k = 6}\). (Any larger \(k\) also works, because the 2s then fill even more of the middle.)
Common pitfalls
Frequently asked questions
What is the difference between mean, median and mode?
The mean is the total divided by the number of scores. The median is the middle score in order. The mode is the most common score.
How do you find the median of an even number of scores?
Put the scores in order and find the two middle scores. The median is halfway between them, their mean.
How do you find the mean from a frequency table?
Multiply each score by its frequency, add these products, then divide by the total frequency: \(\bar{x} = \dfrac{\sum fx}{\sum f}\).
When is the median better than the mean?
When the data contain an outlier or are very lopsided. The outlier pulls the mean away from most scores but hardly changes the median.
Can a dataset have more than one mode?
Yes. Two scores tied for most common makes the data bimodal; three or more makes it multimodal. If every score is equally common, there is no mode.
How do you find a missing score when you know the mean?
Multiply the mean by the number of scores to get the total, then subtract the scores you know.