Median and Range
Every question with a fully worked solution.
- Median and Range - Video - Median and Range 1 Watch
Theory
The median is the value in the middle position of ordered data, and the range is the highest value minus the lowest. For \(2, 3, 3, 4, 4, 4, 5, 5, 6\) the median is the 5th value, \(4\), and the range is \(6 - 2 = 4\). This Year 8 maths page shows how.
Order the data and count it. Call the number of values \(n\).
An odd number of values: the median is the single middle value, in position \(\dfrac{n + 1}{2}\).
An even number of values: there are two middle values, in positions \(\dfrac{n}{2}\) and \(\dfrac{n}{2} + 1\). The median is halfway between them: add them and divide by \(2\). It need not be a data value.
Ordered displays. A frequency table, a stem-and-leaf plot and a dot plot already hold the values in order. Keep a running total of the frequencies, leaves or dots, then read off the value in the middle position.
Range \(=\) highest value \(-\) lowest value. It is a measure of spread: a small range means consistent data. If the lowest value is negative, subtracting it adds: \(7 - (-4) = 11\). The median is a measure of centre.
Outliers change the range a lot but the median very little.
Working backwards. If the median is given, the value in the middle position must equal it. If the range is given, highest \(=\) lowest \(+\) range, and lowest \(=\) highest \(-\) range.
Middle positions:
| Number of values | Middle position(s) |
|---|---|
| \(9\) (odd) | 5th |
| \(10\) (even) | 5th and 6th |
| \(14\) (even) | 7th and 8th |
| \(23\) (odd) | 12th |
Finding the median
- Order the values (or use an ordered display).
- Count them: \(n\).
- Odd \(n\): take the value in position \(\dfrac{n + 1}{2}\). Even \(n\): average the values in positions \(\dfrac{n}{2}\) and \(\dfrac{n}{2} + 1\).
Finding the range
- Find the highest and the lowest values.
- Subtract: highest \(-\) lowest (subtracting a negative adds).
(i) \(2, 3, 3, 4, 4, 4, 5, 5, 6\)
(ii) \(2, 3, 3, 4, 4, 5, 6, 7, 7, 8\)
(iii) Find the range of the data in (i).
(i) There are \(9\) values, so the 5th is the middle.
(ii) There are \(10\) values, so the 5th and 6th are the middle.
| \(\text{Median}\) | \(=\) | \(4\) |
| \(\text{Median}\) | \(=\) | \(\dfrac{4 + 5}{2} = 4.5\) |
| \(\text{Range}\) | \(=\) | \(6 - 2 = 4\) |
(i) \(4\)
(ii) \(4.5\)
(iii) \(4\)
In order: \(3, 5, 7, 9, 13, 16, 19, 24\). With \(n = 8\), use the 4th and 5th values.
| \(\text{Median}\) | \(=\) | \(\dfrac{9 + 13}{2} = 11\) |
| \(\text{Range}\) | \(=\) | \(24 - 3 = 21\) |
Median \(11\) mm, range \(21\) mm.
(i) Find the median and the range.
(ii) A \(17\)th player joins and scores \(33\). Find the new median.
| Stem | Leaf |
|---|---|
| 1 | 2 5 8 |
| 2 | 0 1 4 6 7 9 |
| 3 | 0 3 3 5 8 |
| 4 | 2 6 |
Count the leaves: \(3 + 6 + 5 + 2 = 16\), so the middle values are the 8th and 9th: running totals put the 1st to 3rd on stem \(1\) and the 4th to 9th on stem \(2\).
| \(\text{8th and 9th}\) | \(=\) | \(27 \text{ and } 29\) |
| \(\text{Median}\) | \(=\) | \(\dfrac{27 + 29}{2} = 28\) |
| \(\text{Range}\) | \(=\) | \(46 - 12 = 34\) |
(i) Median \(28\) goals, range \(34\) goals.
(ii) With \(17\) values the median is the 9th alone: \(29\) goals.
(i) Find the median and the range.
(ii) Find both again without the \(\$240\).
In order: \(45, 46, 48, 50, 52, 55, 58, 60, 240\).
| \(\text{Median}\) | \(=\) | \(52\) |
| \(\text{Range}\) | \(=\) | \(240 - 45 = 195\) |
| \(\text{New median}\) | \(=\) | \(\dfrac{50 + 52}{2} = 51\) |
| \(\text{New range}\) | \(=\) | \(60 - 45 = 15\) |
The outlier moves the range by \(180\) but the median by only \(1\).
(ii) Overnight temperatures ranged from \(-4^\circ\)C to \(7^\circ\)C. Find the range.
(i) \(20\) students, so the 10th and 11th values. Running totals: \(0\) pets holds the 1st to 4th, \(1\) pet the 5th to 10th, \(2\) pets the 11th to 15th.
| \(\text{Median}\) | \(=\) | \(\dfrac{1 + 2}{2} = 1.5\) |
| \(\text{Range}\) | \(=\) | \(7 - (-4) = 11\) |
(i) \(1.5\) pets
(ii) \(11^\circ\)C
(i) Find \(m\).
(ii) One more game is played; the new score is the lowest of the eight, and the range becomes \(29\). Find it.
(i) The known scores in order are \(22, 27, 31, 35, 40, 44\). With \(7\) scores the median is the 4th; since \(31 < 33 < 35\), \(m\) is the 4th.
| \(m\) | \(=\) | \(33\) |
| \(\text{Lowest}\) | \(=\) | \(44 - 29 = 15\) |
(i) \(m = 33\)
(ii) \(15\). Check: \(44 - 15 = 29\).
Common pitfalls
Frequently asked questions
How do you find the median?
Put the values in order and take the middle one. With an even number of values, take the two middle values and halve their sum.
What is the range?
The highest value minus the lowest value: for \(3\), \(8\), \(14\) the range is \(11\).
Can the median be a decimal?
Yes. With an even number of values the median is halfway between two values, such as \(4.5\).
Why use the median instead of the mean?
An outlier pulls the mean a long way but hardly moves the median, so the median can better describe a typical value.
How do I find the median from a frequency table?
Keep a running total of the frequencies until you reach the middle position, then read off that value.