Stem-and-Leaf Plots
Theory
A stem-and-leaf plot sorts data into rows while keeping every score. This Year 9 maths page shows how to build ordered and split-stem plots, read keys such as \(13 \mid 2 = 13.2\), find the median, mode, range and mean from a plot, describe its shape, and compare two groups on a back-to-back plot.
A stem-and-leaf plot splits each score in two. Usually the stem is the tens digit and the leaf is the units digit, so \(6 \mid 4\) means \(64\). Every leaf is one score, so a repeated leaf is a repeated score.
The key says what a stem and leaf stand for. With \(13 \mid 2 = 13.2\) the stem is the whole number and the leaf is the tenths digit. With \(10 \mid 0 = 100\) the stem is the hundreds and tens.
In an ordered plot the leaves in each row run from smallest to largest, so the whole plot reads in order from the top row to the bottom row. A stem with no scores still gets its own row, left blank, so gaps show up.
A split-stem plot uses each stem twice. The first row holds the leaves \(0\) to \(4\) and the second row, often marked with a \(*\), holds \(5\) to \(9\). It spreads out data that are bunched on only a few stems.
A back-to-back plot shows two groups on one stem. The right-hand leaves read left to right as usual. The left-hand leaves are read from the stem outwards, so the leaf next to the stem is the smallest.
The shape of a plot shows clusters (scores bunched together), gaps (empty rows) and outliers (scores far from the rest).
How to use a stem-and-leaf plot
- Read the key. Decide what one stem and leaf stand for, including any decimal point or split stem.
- Count the leaves to get \(n\).
- Range: last score minus first score.
- Median: count along the leaves to position \(\dfrac{n+1}{2}\). With an even \(n\), average the two middle scores.
- Mode: find the most repeated leaf. Mean: add every score and divide by \(n\).
- Shape: look for clusters, gaps, outliers and the direction of any tail.
- Back-to-back: read the left leaves outwards, work out each group separately, then compare centre and spread.
\(52\), \(47\), \(61\), \(58\), \(44\), \(55\), \(63\), \(49\), \(50\), \(57\), \(66\), \(41\)
Sort the scores first: \(41\), \(44\), \(47\), \(49\), \(50\), \(52\), \(55\), \(57\), \(58\), \(61\), \(63\), \(66\). Leaves \(0\)–\(4\) go on the first row of each stem and \(5\)–\(9\) on the starred row.
| Stem | Leaf |
|---|---|
| 4 | 1 4 |
| 4* | 7 9 |
| 5 | 0 2 |
| 5* | 5 7 8 |
| 6 | 1 3 |
| 6* | 6 |
Key: \(4 \mid 1 = 41\) \(4^{*} \mid 7 = 47\)
There are \(12\) leaves, so the median is the average of the \(6\)th and \(7\)th scores.
| \(\text{median}\) | \(=\) | \(\dfrac{52 + 55}{2} = \mathbf{53.5}\) |
| \(\text{range}\) | \(=\) | \(66 - 41 = \mathbf{25}\) |
(i) Find the median and the mean.
(ii) What percentage of the swimmers took less than \(13\) seconds?
| Stem | Leaf |
|---|---|
| 11 | 4 8 |
| 12 | 0 3 3 7 |
| 13 | 1 5 6 |
| 14 | 2 |
Key: \(11 \mid 4 = 11.4\) seconds
(i) The scores are \(11.4\), \(11.8\), \(12.0\), \(12.3\), \(12.3\), \(12.7\), \(13.1\), \(13.5\), \(13.6\), \(14.2\). With \(10\) scores, average the \(5\)th and \(6\)th.
| \(\text{median}\) | \(=\) | \(\dfrac{12.3 + 12.7}{2} = \mathbf{12.5}\) |
| \(\text{sum}\) | \(=\) | \(126.9\) |
| \(\text{mean}\) | \(=\) | \(\dfrac{126.9}{10} = \mathbf{12.69}\) |
(ii) Times below \(13\) are the leaves on stems \(11\) and \(12\): \(2 + 4 = 6\) swimmers.
| \(\dfrac{6}{10} \times 100\%\) | \(=\) | \(\mathbf{60\%}\) |
| Red | Stem | Blue |
|---|---|---|
| 8 | 1 | |
| 7 5 2 | 2 | 6 8 9 |
| 8 4 3 1 | 3 | 0 1 2 4 5 6 |
| 5 | 4 |
Key: \(5 \mid 2 \mid 6\) means \(25\) for Red and \(26\) for Blue
Read Red’s leaves outwards from the stem: \(18\), \(22\), \(25\), \(27\), \(31\), \(33\), \(34\), \(38\), \(45\). Blue: \(26\), \(28\), \(29\), \(30\), \(31\), \(32\), \(34\), \(35\), \(36\). Each team has \(9\) scores, so the median is the \(5\)th.
| \(\text{median Red}\) | \(=\) | \(31\) |
| \(\text{median Blue}\) | \(=\) | \(31\) |
| \(\text{range Red}\) | \(=\) | \(45 - 18 = 27\) |
| \(\text{range Blue}\) | \(=\) | \(36 - 26 = 10\) |
The medians are equal, so neither team scores more in a typical game. Blue’s range is much smaller, so Blue is more consistent.
(i) Describe the shape of the data.
(ii) Find the median.
(iii) Find the range with and without the outlier.
| Stem | Leaf |
|---|---|
| 1 | 2 5 8 |
| 2 | 0 1 3 4 4 6 7 9 |
| 3 | 0 2 5 |
| 4 | |
| 5 | |
| 6 | 8 |
Key: \(2 \mid 4 = 24\) minutes
(i) Most times form a cluster from \(20\) to \(29\) minutes. Stems \(4\) and \(5\) are empty, a gap, and \(68\) sits far above the rest: it is an outlier. Apart from the outlier the data are roughly symmetrical about the \(20\)s.
(ii) With \(15\) scores the median is the \(8\)th: \(12\), \(15\), \(18\), \(20\), \(21\), \(23\), \(24\), \(\mathbf{24}\), \(\ldots\)
| \(\text{median}\) | \(=\) | \(\mathbf{24}\) |
| \(\text{range with}\) | \(=\) | \(68 - 12 = \mathbf{56}\) |
| \(\text{range without}\) | \(=\) | \(35 - 12 = \mathbf{23}\) |
One outlier more than doubles the range.
Common pitfalls
Frequently asked questions
How do you read a stem-and-leaf plot?
Use the key to turn each stem and leaf back into a score. With the key \(5 \mid 2 = 52\), the row \(5 \mid 2 \ 6 \ 8\) holds the scores \(52\), \(56\) and \(58\).
How do you find the median of a stem-and-leaf plot?
Count the leaves to find how many scores there are. Find the middle position, \(\dfrac{n+1}{2}\). Count along the leaves, which are already in order, to that position. With an even number of scores, average the two middle ones.
What is a split-stem plot?
A stem-and-leaf plot that uses each stem twice. The first row holds the leaves \(0\) to \(4\) and the second, often marked with a star, holds \(5\) to \(9\). It is used when the data are bunched on a few stems.
How do you read a back-to-back stem-and-leaf plot?
The stem is in the middle. The right-hand leaves are read left to right as usual. The left-hand leaves are read from the stem outwards, so the leaf next to the stem is the smallest.
How can you tell the shape from a stem-and-leaf plot?
Turn the plot a quarter turn anticlockwise so the rows stand up like bars. A long tail of high scores is positively skewed, a long tail of low scores is negatively skewed, and matching halves are symmetrical.