Analysing Dot Plots and Stem and Leaf
Dot plots and stem-and-leaf plots show every value, so you can read the mode, range, median and mean straight from the display. This subtopic rounds off data analysis in NSW Year 7 maths (NESA outcome MA4‑DAT‑C‑02).
You will find the median by counting to the middle position, calculate the mean from stacks of dots or from stems and leaves, read the range and mode, describe clusters, gaps and skew, compare back-to-back plots, and find a missing value from a known mean. A class experiment growing bean plants with and without fertiliser is one example. This subtopic is part of Year 7 Analysing Data.
Theory
Dot plots and stem-and-leaf plots show every value, so you can read the mean, median, mode and range straight from them. This Year 7 maths page shows how, and how to describe the shape of the data and compare two groups.
Dot plot. Each dot is one value. Mode: the tallest stack. Range: the last dot minus the first dot. Median: count the dots from the left to the middle position. Mean: multiply each value by its number of dots, add, then divide by the number of dots.
Stem-and-leaf plot. Read the stem and leaf together using the key (4 | 2 \(= 42\)). The leaves are in order, so the values are already sorted. For the mean, add every value, then divide by the number of leaves.
Shape. A cluster is a group of values close together; a gap is an empty stretch; an outlier is a value far from the rest. Data are symmetrical if both sides match, positively skewed if the tail stretches right, and negatively skewed if it stretches left. A tail pulls the mean towards it.
Reading statistics from a display:
| Statistic | Dot plot | Stem-and-leaf plot |
|---|---|---|
| mode | tallest stack | leaf repeated most in a row |
| range | last dot − first dot | last leaf − first leaf (with stems) |
| median | count dots to the middle | count leaves to the middle |
| mean | value × dots, add, ÷ dots | add all values, ÷ leaves |
How to find the median from a display
- Count all the dots or leaves (\(n\)).
- Find the middle position \(\dfrac{n + 1}{2}\).
- Count from the smallest value to that position (two middle values if \(n\) is even).
(i) the mode
(ii) the range
(iii) the median
(iv) the mean.
| Stem | Leaf |
|---|---|
| 5 | 1 4 7 |
| 6 | 3 3 5 6 |
| 7 | 7 8 8 9 9 |
| 8 | 5 6 6 6 8 |
| 9 | 1 2 4 |
Key: 6 | 3 = 63
| \(94 - 51\) | \(=\) | \(43\) |
| \(\dfrac{78 + 79}{2}\) | \(=\) | \(78.5\) |
| \(1518 \div 20\) | \(=\) | \(75.9\) |
(i) \(86\) (three times)
(ii) \(43\)
(iii) \(20\) values, so the median is halfway between the two middle values, \(78\) and \(79\): \(78.5\).
(iv) The sum is \(1518\), so the mean is \(75.9\).
(i) the mode
(ii) the range
(iii) the median
(iv) the mean (to \(1\) decimal place).
Sum: \(1 + 2 + 3 + 2 \times 4\) \(+\, 3 \times 5 + 4 \times 6\) \(+\, 4 \times 7 + 3 \times 8\) \(+\, 9 + 10\), which is \(124\), from \(21\) dots.
| \(10 - 1\) | \(=\) | \(9\) |
| \(124 \div 21\) | \(=\) | \(5.90\ldots\) |
(i) \(6\) and \(7\) (four dots each)
(ii) \(9\)
(iii) The middle dot of the \(21\) is above \(6\): median \(6\).
(iv) \(\approx 5.9\)
| Stem | Leaf |
|---|---|
| 1 | 5 8 |
| 2 | 0 3 3 7 |
| 3 | 1 4 9 |
| 4 | 2 6 |
Key: 2 | 3 = 23
| \(46 - 15\) | \(=\) | \(31\) |
| \(318 \div 11\) | \(=\) | \(28.90\ldots\) |
Median \(27\) (the \(6\)th value), mode \(23\), range \(31\), mean \(\approx 28.9\) years.
(i) Find the mean and the median.
(ii) Leave out the outlier and find the new mean.
Value \(\times\) dots: \(1 \times 2 + 2 \times 4\) \(+\, 3 \times 2 + 4 \times 1\) \(+\, 10 \times 1\), which is \(30\), from \(10\) dots. The two middle dots are both above \(2\).
| \(30 \div 10\) | \(=\) | \(3\) |
| \((30 - 10) \div 9\) | \(=\) | \(2.22\ldots\) |
(i) Mean \(3\), median \(2\)
(ii) \(\approx 2.2\). The data are positively skewed: the outlier pulled the mean up, but the median stays \(2\).
| 7A | Stem | 7B |
|---|---|---|
| 8 2 | 1 | 9 |
| 7 5 4 1 | 2 | 2 6 8 9 |
| 3 0 | 3 | 1 4 5 8 |
| 1 | 4 |
Key: 4 | 2 | 6 means 24 for 7A and 26 for 7B
| \(41 - 12\) | \(=\) | \(29\) |
| \(38 - 19\) | \(=\) | \(19\) |
Nine values each, so the median is the \(5\)th. 7A: median \(25\), range \(29\). 7B: median \(29\), range \(19\). 7B has the higher median and the smaller range, so it did better and more consistently.
Common pitfalls
Frequently asked questions
How do you find the median from a dot plot?
Count the dots, find the middle position, then count dots from the left to that position.
How do you find the mean from a stem-and-leaf plot?
Add every value (stem and leaf together) and divide by the number of leaves.
How do you find the range from a stem-and-leaf plot?
Last leaf on the last stem minus the first leaf on the first stem, read with their stems.
What does an outlier do to the mean and median?
It pulls the mean towards it; the median hardly changes.
How do you compare two groups in a back-to-back plot?
Compare their medians (centre) and ranges (spread).