Median and Range
The median is the middle value once the data are in order, and the range is the highest value minus the lowest. In NSW Year 7 maths they describe the centre and the spread of a dataset (NESA outcome MA4‑DAT‑C‑02).
You will order data and find the median for an odd or even number of values, calculate the range, even with temperatures below zero, see how an outlier changes the mean and range but hardly moves the median, find a missing value from the range, and compare two datasets. Median house prices and a swimmer’s lap times use these ideas. This subtopic is part of Year 7 Analysing Data.
Theory
The median is the middle value when the data are in order. The range is the highest value minus the lowest value. The median is a measure of centre; the range is a measure of spread. This Year 7 maths page shows both, for odd and even numbers of values, frequency tables and data with outliers.
The median is a measure of centre. Put the values in order and count them. If the number of values \(n\) is odd, the median is the middle value, in position \(\dfrac{n + 1}{2}\). If \(n\) is even, there are two middle values: add them and divide by \(2\).
The range is a measure of spread: range \(=\) highest value \(-\) lowest value. A small range means the values are close together (consistent).
An outlier changes the mean and the range a lot, but hardly moves the median.
Missing value from the range. Turn the formula around: highest \(=\) lowest \(+\) range, and lowest \(=\) highest \(-\) range.
The rules:
How to find the median
- Order the values and count them.
- Odd: the value in position \(\dfrac{n + 1}{2}\).
- Even: add the two middle values and divide by \(2\).
How to find the range
- Find the highest and lowest values.
- Subtract: highest \(-\) lowest.
(i) \(2, 3, 3, 4, 4, 4, 5, 5, 6\)
(ii) \(2, 3, 3, 4, 4, 5, 6, 7, 7, 8\)
Then find the range of the first set.
| \(\dfrac{4 + 5}{2}\) | \(=\) | \(4.5\) |
| \(6 - 2\) | \(=\) | \(4\) |
(i) \(9\) values, so the \(5\)th: median \(4\).
(ii) \(10\) values, so halfway between the \(5\)th and \(6\)th: \(4.5\).
Range \(= 6 - 2 = 4\).
In order: \(8, 12, 15, 19, 21, 27, 32\). The middle position is \(\dfrac{7 + 1}{2} = 4\).
| \(32 - 8\) | \(=\) | \(24\) |
Median \(19\) pages, range \(24\) pages.
In order: \(-4, -2, 0, 1, 3, 5\). Six values, so the median is halfway between \(0\) and \(1\).
| \(\dfrac{0 + 1}{2}\) | \(=\) | \(0.5\) |
| \(5 - (-4)\) | \(=\) | \(9\) |
Median \(0.5^\circ\)C, range \(9^\circ\)C.
| Days | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Students | 4 | 6 | 3 | 2 | 1 |
There are \(4 + 6 + 3 + 2 + 1 = 16\) values, so the median is halfway between the \(8\)th and \(9\)th. Values \(1\) to \(4\) are \(0\); values \(5\) to \(10\) are \(1\).
| \(4 - 0\) | \(=\) | \(4\) |
Median \(1\) day. Range \(4\) days (use the values, not the frequencies).
| \(80 + 95 + 100 + 110 + 615\) | \(=\) | \(1000\) |
| \(1000 \div 5\) | \(=\) | \(200\) |
Mean \(\$200\), median \(\$100\). The \(\$615\) outlier pulls the mean up, so the median better describes a typical price.
| \(x\) | \(=\) | \(6 + 12\) |
| \(\) | \(=\) | \(18\) |
The lowest value is \(6\), so \(x = 18\).
Common pitfalls
Frequently asked questions
How do you find the median?
Put the values in order and take the middle value, or the halfway point of the two middle values.
What if there is an even number of values?
Add the two middle values and divide by \(2\). The median of \(4\) and \(5\) is \(4.5\).
How do you find the range?
Highest value minus lowest value.
Which is affected more by an outlier, the mean or the median?
The mean. The median hardly moves.
What does a small range tell you?
The values are close together, so the data are consistent.