Quartiles and Inter Quartile Range
Quartiles split ordered data into four equal parts. The median is the middle score, the lower quartile \(Q_1\) is the middle of the lower half and the upper quartile \(Q_3\) is the middle of the upper half. The interquartile range, \(Q_3-Q_1\), measures the spread of the middle 50% of the scores. This is NSW Year 10 Advanced maths (NESA outcome MA5‑DAT‑C‑01).
You will find five-number summaries and the IQR from lists, stem-and-leaf plots, column graphs and frequency tables, with odd and even numbers of scores. You will compare range and IQR when there is an extreme value, see what a 10% rise does to the IQR, and compare two food trucks and two classes. Harder questions find missing scores from a given summary, mean or IQR, and find the quartiles of 1 to \(n\). It is part of Year 10 Advanced Single Variable and Bivariant Statistics.
Every question with a fully worked solution.
- Quartiles and Inter Quartile Range - Video - Median, Quartiles and Interquartile Range Watch
Theory
The quartiles split ordered data into four roughly equal groups. The interquartile range, \(\text{IQR} = Q_3 - Q_1\), is the spread of the middle half of the data, so unlike the range it is not affected by a few extreme scores. This Year 10 Advanced maths page covers the five-number summary from lists, stem-and-leaf plots and frequency tables, what the quartiles say about percentages, and problems with unknown scores.
For data written in order: the median \(Q_2\) is the middle score. The lower quartile \(Q_1\) is the median of the lower half and the upper quartile \(Q_3\) is the median of the upper half. With an odd number of scores, leave the median out of both halves.
The five-number summary is: minimum, \(Q_1\), median, \(Q_3\), maximum.
The range \(=\) maximum \(-\) minimum measures the spread of all the data. The interquartile range \(= Q_3 - Q_1\) measures the spread of the middle \(50\%\). A small IQR means the middle scores are bunched together (consistent).
About \(25\%\) of scores lie below \(Q_1\), \(50\%\) below the median and \(75\%\) below \(Q_3\).
| Number of scores | Median | \(Q_1\) | \(Q_3\) |
|---|---|---|---|
| \(n = 15\) | 8th | 4th | 12th |
| \(n = 12\) | 6.5th (6th and 7th) | 3.5th (3rd and 4th) | 9.5th (9th and 10th) |
| \(n = 21\) | 11th | 5.5th (5th and 6th) | 16.5th (16th and 17th) |
Finding the quartiles
- Order the scores from smallest to largest (negative numbers first) and count them.
- Median: the middle score, or the average of the two middle scores.
- Split into lower and upper halves. If \(n\) is odd, the median is in neither half.
- \(Q_1\) and \(Q_3\): the median of each half.
- IQR \(= Q_3 - Q_1\). Give units.
From a table or graph
- Running totals: write the cumulative frequencies, then find the class or value that contains each required position.
- Combine data sets by listing all the scores together in one order before finding new quartiles.
- Range or IQR? If there is an outlier, the IQR is the better measure of spread; the range changes a lot when the outlier is removed.
- Unknown scores: write \(Q_1\), \(Q_3\) or the median in terms of the unknowns, use the mean for a total, and solve the equations.
Order the data:
The median is the \(\dfrac{15 + 1}{2} = 8\)th score, \(27\). Leave it out: each half has \(7\) scores and its middle is the \(4\)th.
| \(Q_1\) | \(=\) | \(22 \quad (\text{4th of } 18, \ldots, 26)\) |
| \(Q_3\) | \(=\) | \(33 \quad (\text{4th of } 28, \ldots, 45)\) |
Five-number summary: \(\boldsymbol{18,\ 22,\ 27,\ 33,\ 45}\).
| \(\text{range}\) | \(=\) | \(45 - 18 = \boldsymbol{27}\) |
| \(\text{IQR}\) | \(=\) | \(33 - 22 = \boldsymbol{11}\) |
| Stem | Leaf |
|---|---|
| \(4\) | \(1\ \ 3\ \ 4\ \ 6\ \ 6\ \ 8\) |
| \(5\) | \(0\ \ 1\ \ 3\ \ 5\ \ 8\) |
| \(6\) | \(2\) |
Key: \(4\,|\,1 = 4.1\) m
Find the median, \(Q_1\), \(Q_3\) and the IQR.With \(12\) scores the median is halfway between the \(6\)th and \(7\)th. Each half then has \(6\) scores, so each quartile is halfway between the \(3\)rd and \(4\)th of its half.
| \(\text{median}\) | \(=\) | \(\dfrac{4.8 + 5.0}{2} = \boldsymbol{4.9}~\text{m}\) |
| \(Q_1\) | \(=\) | \(\dfrac{4.4 + 4.6}{2} = \boldsymbol{4.5}~\text{m}\) |
| \(Q_3\) | \(=\) | \(\dfrac{5.3 + 5.5}{2} = \boldsymbol{5.4}~\text{m}\) |
| \(\text{IQR}\) | \(=\) | \(5.4 - 4.5 = \boldsymbol{0.9}~\text{m}\) |
Count leaves carefully: the \(3\)rd and \(4\)th leaves on stem \(4\) are \(4\) and \(6\).
| Siblings | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) |
|---|---|---|---|---|---|---|
| Frequency | \(7\) | \(12\) | \(10\) | \(6\) | \(3\) | \(2\) |
(i) Find \(Q_1\), the median, \(Q_3\) and the IQR.
(ii) What percentage of the students have more siblings than \(Q_3\)?
(i) Running totals: \(7,\ 19,\ 29,\ 35,\ 38,\ 40\). Each half has \(20\) students.
| \(Q_1\) | \(=\) | \(\text{average of 10th and 11th} = \boldsymbol{1} \quad (7 < 10,\ 11 \le 19)\) |
| \(\text{median}\) | \(=\) | \(\text{average of 20th and 21st} = \boldsymbol{2} \quad (19 < 20,\ 21 \le 29)\) |
| \(Q_3\) | \(=\) | \(\text{average of 30th and 31st} = \boldsymbol{3} \quad (29 < 30,\ 31 \le 35)\) |
| \(\text{IQR}\) | \(=\) | \(3 - 1 = \boldsymbol{2}\) |
(ii) More than \(3\) siblings: \(3 + 2 = 5\) students.
| \(\dfrac{5}{40} \times 100\%\) | \(=\) | \(\boldsymbol{12.5\%}\) |
This is not \(25\%\): with repeated values, the quarters are only approximately equal.
Each half has \(4\) scores, so \(Q_1\) is halfway between the \(2\)nd and \(3\)rd scores and \(Q_3\) between the \(6\)th and \(7\)th.
| \(Q_1\) | \(=\) | \(\dfrac{6 + a}{2},\quad Q_3 = \dfrac{b + 18}{2}\) |
| \(\dfrac{b + 18}{2} - \dfrac{6 + a}{2}\) | \(=\) | \(9\) |
| \(b + 12 - a\) | \(=\) | \(18\) |
| \(b - a\) | \(=\) | \(6 \quad (1)\) |
The mean gives the total: \(8 \times 12 = 96\). The known scores add to \(72\).
| \(72 + a + b\) | \(=\) | \(96\) |
| \(a + b\) | \(=\) | \(24 \quad (2)\) |
| \((1) + (2):\quad 2b\) | \(=\) | \(30\) |
| \(b\) | \(=\) | \(15,\quad a = 9\) |
Check: \(3, 6, 9, 10, 13, 15, 18, 22\) is in order; \(Q_1 = 7.5\), \(Q_3 = 16.5\), IQR \(= 9\ \checkmark\). So \(\boldsymbol{a = 9}\), \(\boldsymbol{b = 15}\).
Common pitfalls
Frequently asked questions
How do you find the lower and upper quartiles?
Put the scores in order and find the median. Split the data into a lower half and an upper half, leaving out the median if there is an odd number of scores. Q1 is the median of the lower half and Q3 is the median of the upper half.
What is the interquartile range?
It is the upper quartile minus the lower quartile. It measures the spread of the middle 50 percent of the data.
Why is the IQR better than the range?
The range uses only the two most extreme scores, so one outlier can make it very large. The IQR ignores the top and bottom quarters, so it is not affected by outliers.
What is a five-number summary?
The minimum, the lower quartile, the median, the upper quartile and the maximum. It is the information needed to draw a box plot.
How do you find quartiles from a frequency table?
Write the running totals of the frequencies. Work out the positions of the median and quartiles, then find which score contains each position.