Histograms and Stem-and-Leaf Plots
Histograms show numerical data as touching bars over a number line, while stem-and-leaf plots keep every score visible in order. This NSW Year 9 Advanced maths subtopic revises both Stage 4 displays (NESA outcome MA4‑DAT‑C‑02) and analyses the data in them.
You will read frequencies, find the median, mean, mode and range from each display, and choose a suitable display for survey data. You will compare groups with back-to-back and split stem-and-leaf plots, estimate a mean from grouped data using class centres, find missing leaves from a mean and range, and find the least and greatest possible mean of grouped data. It is part of Year 9 Advanced Investigating Data.
Theory
A histogram shows how often each score, or each group of scores, occurs. A stem-and-leaf plot shows every score, already in order. This Year 9 Advanced maths page shows how to read both displays and get the mean, median, mode and range from them.
A histogram is a graph for numerical data. The horizontal axis is a number line, so the bars touch. A column graph is for categories (colours, sports) and its bars have gaps.
The height of a bar is the frequency: how many scores it stands for. On a grouped histogram each bar covers a class interval such as \(20\)–\(29\). The class with the tallest bar is the modal class, and the middle of a class is its class centre.
In a stem-and-leaf plot each score is split into a stem (usually the tens digit) and a leaf (the units digit). A key says how to read it: \(4 \mid 7\) might mean \(47\), or with a decimal key \(8 \mid 4\) means \(8.4\). An ordered plot has its leaves in order, so you can count to the median.
A back-to-back plot shares one stem between two groups. The left-hand leaves are read outwards from the stem: \(7\;3 \mid 4\) is \(43\) and \(47\). In a split stem plot each stem appears twice: the first row holds leaves \(0\)–\(4\), the second leaves \(5\)–\(9\).
An outlier is a score that sits well apart from the rest.
Here \(x\) is a score (or a class centre) and \(f\) is its frequency, the bar height.
For an even \(n\) this lands halfway between two scores, so average them. Range \(=\) largest score \(-\) smallest score.
| Display | Mode | Median | Mean |
|---|---|---|---|
| stem-and-leaf | most common value | count along the ordered leaves | add every score, divide by \(n\) |
| histogram (single scores) | tallest bar | running totals of the bars | \(\sum fx \div \sum f\) |
| grouped histogram | modal class | class containing it | estimate with class centres |
Reading a stem-and-leaf plot
- Read the key so you know what a stem and a leaf stand for.
- Count the leaves to get \(n\).
- Find the median position, \(\dfrac{n + 1}{2}\), and count along the rows to it.
- Back-to-back? Read the left side from the stem outwards and treat each group on its own.
Reading a histogram
- Read every bar height and add them to get the total.
- Counts and percentages: add the bars that fit the condition ('less than \(20\)', 'at least \(3\)'), then divide by the total.
- Median: keep a running total of the bars until you pass the median position.
- Mean: multiply each score or class centre by its frequency, add, and divide by the total. For grouped data say it is an estimate.
- Check a claim (such as 'most students scored above the mean') by counting the bars on each side.
| 9P leaves | Stem | 9Q leaves |
|---|---|---|
| \(8\;\;5\;\;2\) | \(1\) | \(6\;\;9\) |
| \(7\;\;3\;\;3\;\;1\) | \(2\) | \(4\;\;5\;\;8\;\;9\) |
| \(6\;\;4\;\;0\) | \(3\) | \(2\;\;3\;\;5\;\;8\) |
| \(1\) | \(4\) | \(4\;\;7\) |
Key: \(2 \mid 1 \mid 4\) means \(12\) s for 9P and \(14\) s for 9Q
Find(i) the median time for each class
(ii) the range for each class
(iii) which class was faster overall.
Write out each list in order. For 9P, read each left-hand row from the stem outwards: \(8\;5\;2 \mid 1\) is \(12, 15, 18\).
(i) 9P has \(11\) times, so the median is the \(\dfrac{11 + 1}{2} = 6\)th. 9Q has \(12\), so the median is halfway between the \(6\)th and \(7\)th.
| \(\text{median}_{\text{9P}}\) | \(=\) | \(\mathbf{23}~\text{s}\) |
| \(\text{median}_{\text{9Q}}\) | \(=\) | \(\dfrac{29 + 32}{2} = \mathbf{30.5}~\text{s}\) |
(ii) Range \(=\) largest \(-\) smallest.
| \(\text{9P}\) | \(:\) | \(41 - 12 = \mathbf{29}~\text{s}\) |
| \(\text{9Q}\) | \(:\) | \(47 - 16 = \mathbf{31}~\text{s}\) |
(iii) A shorter time is better. 9P's median is \(7.5\) s lower and its spread is a little smaller, so 9P was faster overall.
Read the bar heights: \(3, 8, 12, 9, 5, 3\).
| \(n\) | \(=\) | \(3 + 8 + 12 + 9 + 5 + 3 = \mathbf{40}\) |
(ii) 'At least \(3\)' means \(3\), \(4\) or \(5\) books.
| \(\dfrac{9 + 5 + 3}{40} \times 100\%\) | \(=\) | \(\dfrac{17}{40} \times 100\% = \mathbf{42.5\%}\) |
(iii) \(40\) values, so the median is halfway between the \(20\)th and \(21\)st. Keep a running total: \(3\), then \(3 + 8 = 11\), then \(11 + 12 = 23\). Both the \(20\)th and \(21\)st are in the bar for \(2\) books.
| \(\text{median}\) | \(=\) | \(\mathbf{2}~\text{books}\) |
(iv) Each bar stands for \(f\) students who each read \(x\) books, so add up \(f \times x\).
| \(\sum fx\) | \(=\) | \(0 + 8 + 24 + 27 + 20 + 15 = 94\) |
| \(\bar{x}\) | \(=\) | \(\dfrac{94}{40} = \mathbf{2.35}~\text{books}\) |
(i) The tallest bar: the modal class is \(100\)–\(119\) minutes.
(ii) The median is halfway between the \(16\)th and \(17\)th. Running totals: \(4\), \(13\), \(24\). Both lie in \(100\)–\(119\) minutes.
(iii) A class centre is the middle of its class: \(\dfrac{60 + 79}{2} = 69.5\), then \(89.5, 109.5, 129.5, 149.5\).
| \(\sum fx\) | \(=\) | \(4(69.5) + 9(89.5) + 11(109.5)\) |
| \(\) | \(\) | \(+\ 6(129.5) + 2(149.5) = 3364\) |
| \(\bar{x}\) | \(\approx\) | \(\dfrac{3364}{32} = 105.125 \approx \mathbf{105.1}~\text{min}\) |
(iv) Smallest: every student at the bottom of their class, at \(60, 80, 100, \ldots\) minutes. Largest: every student at the top, which is \(19\) minutes more each.
| \(\text{smallest}\) | \(=\) | \(\dfrac{3060}{32} = \mathbf{95.625}~\text{min}\) |
| \(\text{largest}\) | \(=\) | \(\dfrac{3060 + 32 \times 19}{32} = \mathbf{114.625}~\text{min}\) |
The mean is only an estimate: the histogram hides the actual times inside each class.
| Stem | Leaf |
|---|---|
| \(5\) | \(2\;\;6\;\;8\) |
| \(6\) | \(1\;\;4\;\;k\;\;9\) |
| \(7\) | \(0\;\;3\;\;7\) |
Key: \(5 \mid 2\) means \(52\)
The mean is \(64.7\).(i) Find \(k\).
(ii) Find the median.
(iii) One more score is added, above all the others, and the range becomes \(40\). Find the new score.
(i) Mean \(\times\) number of scores gives the total.
| \(\text{total}\) | \(=\) | \(64.7 \times 10 = 647\) |
| \(\text{known nine}\) | \(=\) | \(580\) |
| \(60 + k\) | \(=\) | \(647 - 580 = 67\) |
| \(k\) | \(=\) | \(\mathbf{7}\) |
Check the order: \(64 \le 67 \le 69\ \checkmark\)
(ii) \(10\) scores: halfway between the \(5\)th and \(6\)th, which are \(64\) and \(67\).
| \(\text{median}\) | \(=\) | \(\dfrac{64 + 67}{2} = \mathbf{65.5}\) |
(iii) The smallest score is still \(52\).
| \(\text{new score} - 52\) | \(=\) | \(40\) |
| \(\text{new score}\) | \(=\) | \(\mathbf{92}\) |
Common pitfalls
Frequently asked questions
What is the difference between a histogram and a column graph?
A histogram is for numerical data and its bars touch, because the horizontal axis is a number line. A column graph is for categories and its bars have gaps.
How do you find the median from a stem-and-leaf plot?
Count the leaves to get \(n\), find the position \(\dfrac{n + 1}{2}\), then count along the ordered leaves to that position. For an even \(n\), average the two middle scores.
How do you find the mean from a histogram?
Multiply each score by the height of its bar, add these products, and divide by the total of the bar heights: \(\bar{x} = \dfrac{\sum fx}{\sum f}\).
Why is the mean from a grouped histogram only an estimate?
The histogram shows how many scores are in each class but not their actual values, so the class centre is used in place of each score.
What is a split stem-and-leaf plot?
Each stem is written twice. The first row holds the leaves \(0\) to \(4\) and the second row holds the leaves \(5\) to \(9\), which spreads out data that would otherwise crowd onto a few stems.
How do you read a back-to-back stem-and-leaf plot?
The two groups share the stem. Leaves on the right are read normally; leaves on the left are read from the stem outwards. Treat each group as its own data set.