At a glance. One lesson of about 60 minutes. It assumes grouped data (Lesson 1) and area of a rectangle. Students find frequency density, read frequencies from areas, draw a histogram from a table, and estimate a median by area.
Calculator: no calculator. No calculator is needed: the divisions and products are chosen to be short, and the idea that area means frequency is the reasoning.
What students will learn
Know
- Frequency density = frequency divided by class width.
- The area of a bar is the frequency.
- Class width = upper boundary minus lower boundary.
Understand
- Height shows how crowded the values are, not how many there are.
- A wide, low bar can hold more values than a narrow, tall one.
- The modal class of unequal classes has the highest density.
Do
- Compute frequency density from a table.
- Read frequencies and totals from bars.
- Estimate a median by locating the middle value inside its class.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: density, area and class width (questions 1 to 8).
- Criterion B. Investigating patterns: the second figure. Ask students to predict how the height changes when the width is halved or doubled.
- Criterion C. Communicating: labelling the vertical axis frequency density and showing width × height for each bar (the worked example and question 5), or frequency ÷ width (the faded example).
- Criterion D. Applying mathematics in real-life contexts: age bands, journey times; judging whether a picture is fair.
The one thing to do with the figure
Make students commit before they press. In the first figure, ask “When the height becomes frequency density, which bar will be the tallest?” and take a class number from each pair before pressing the button.
In the second figure, ask what happens to the area if the width doubles and the frequency stays the same, then check with the sliders.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 7 min | Hook: the age bands. Collect predictions about the biggest bar and the busiest band. |
| 10 min | First figure and key idea: density and area. |
| 10 min | Second figure, the warning card and the worked example; pairs finish the faded example. |
| 5 min | The check. If many choose B or D, return to the rectangle: what is its area? |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 8. |
The answers
| Warm-up | 50; 24; 5. |
|---|---|
| Hook | With height as frequency a wide band can look largest because it covers more years. The busiest band per year of age is the one with the greatest density, which needs each count divided by its width. |
| Worked example | 30, 42 and 24, total 96. |
| Faded example | Heights 3.6, 3 and 1.3. |
| 1. Frequency from a bar | C. A adds; B divides width by height; D divides height by width. |
| 2. Frequency density | 4.5. 54 ÷ 12. |
| 3. Frequency from height and width | 18. 5 × 3.6. |
| 4. Density from class boundaries | 2.2. 33 ÷ 15. |
| 5. Total from a histogram | 84. 18 + 48 + 18. |
| 6. Same frequency, a different width | 1.2. 24 ÷ 20. |
| 7. A median from a histogram | 16.25. 15 + 10 × 5 ÷ 40. |
| 8. Frequency of one bar | D. A is the height; B uses the lower boundary as the width; C uses the upper boundary. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| A, B or D in question 1 | Adds, divides the wrong way, or divides height by width. | Sketch the rectangle and label its sides. |
| 648, 42 or 66 in question 2 | Multiplied, subtracted or added the frequency and the width. | Ask: is it values per unit of width? What does 'per' mean? |
| 0.72, 8.6 or 3.6 in question 3 | Divided, added, or gave the height as the frequency. | Ask for the area of the rectangle in square units. |
| 3.3, 1.32 or 48 in question 4 | Used a boundary as the width, or added. | Subtract the boundaries to find the width first. |
| 4.8, 48 or 66 in question 5 | Added the heights, used a common width, or left out a bar. | Write each class's frequency in a column, then add. |
| 4.8, 2.4 or 19.2 in question 6 | Kept the height, used a boundary as the width, or multiplied. | Find the frequency first, then redistribute. |
| 20, 55 or 25 in question 7 | Took the midpoint of the class, a position, or the class end. | Count 50 values, then ask how many more are needed to reach 55. |
| A, B or C in question 8 | Gave the height, or used a boundary as the width. | Subtract the boundaries to find the width. |
Questions worth asking
- Why must the vertical axis be labelled frequency density? Because the height is not a count: it is a count per unit of width.
- What would the histogram look like if all the classes were the same width? The heights would be proportional to the frequencies, so the tallest bar would be the largest class.
- Could a bar have a height less than 1 and still hold many values? Yes: if the class is wide enough, the area can be large.
- Why can we only estimate the median from a histogram? We do not know where inside a class the values lie.
Differentiation
A way in for anyone stuck
- Give a table with a width column and a density column ready to fill.
- Use a checklist: width? divide? label? area to check?
- Use squared paper so that area can be counted as squares.
Stretch
- Prove that if two adjacent classes have the same density, merging them gives a class with the same density.
- A histogram has a last class that is open-ended. Why can its bar not be drawn without an assumption?
Students learning in English as an additional language
“Density” here means how crowded something is. “Frequency density” is the number of values per unit. “Class width” is how wide a class is. “Area” is width times height of a rectangle.
| Word | Plain meaning | Example |
|---|---|---|
| histogram | a graph of continuous data in which the area of each bar is the frequency | bars with no gaps and a density axis |
| frequency density | frequency divided by class width | values per minute of waiting time |
| class width | upper boundary minus lower boundary | the width of the class 30 to under 45 |
| continuous data | data that can take any value on a scale | heights or times |
| area | width times height; in a histogram, the frequency | the size of a bar |
| modal class | the class with the highest frequency density | the tallest bar when the heights show density |
Sentence starters for explaining aloud or in writing:
- The width of this class is … because ….
- The frequency density is … because ….
- I know the frequency is … because the area is ….
Two things not to say
“The tallest bar has the most values.” Only if the classes have equal width; otherwise compare area.
“Leave gaps between the bars.” A histogram is for continuous data, so the bars touch.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- DensityFrequency 40, width 8. Find the frequency density.
5. - FrequencyA bar is 6 wide and 2.5 tall. Find the frequency.
15.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- ContextA class of 30 to under 50 minutes of daily screen time holds 45 students. What height is drawn for its bar?
45 ÷ 20 = 2.25. - Earlier work (Lesson 2)A curve for 80 values reads the median at which count?
40.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningTwo bars have equal area. One is three times as wide as the other. How do their heights compare?
The wider one is a third as tall. - Working backwardsA histogram has a total area of 150. One bar has height 3 and width 10. What fraction of the values are in it?
30 ÷ 150 = 1/5.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
Both figures work with the keyboard (tab to a slider or button, then the arrow keys or Enter) and by touch. Under each one, “Show me the steps” gives the same states in words.
No device? Use squared paper and let one square stand for 2 values: draw the bars and count squares to check the area.