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This lesson is Higher tier. A histogram looks like a bar chart, but it is built for continuous data grouped into unequal class widths — and that one difference changes everything. In a histogram it is the area of each bar, not its height, that represents frequency. The height is the frequency density. Mastering frequency density, and the reverse skill of finding a frequency from an area, is a reliable source of Higher-tier marks in the OCR GCSE Mathematics (J560) Statistics strand, and it is a topic where careless candidates routinely lose marks by treating a histogram like an ordinary bar chart.
The topic is mostly AO1 (calculating frequency density and drawing the histogram accurately) with a strong AO2/AO3 strand (reading frequencies back from areas and solving problems where a frequency or class width is unknown). OCR command words include "Draw", "Work out", "Estimate" and "Use the histogram".
| Term | Meaning |
|---|---|
| Histogram | A diagram for continuous data where bar area represents frequency |
| Frequency density | Frequency ÷ class width; the height of each bar |
| Class width | Upper boundary − lower boundary of a class |
| Continuous data | Measured data that can take any value in a range |
| Unequal class widths | Classes that are not all the same size — the reason histograms are needed |
If class widths were all equal, a bar chart would do. But continuous data is often grouped into unequal widths — narrow classes where data is dense, wide classes where it is sparse. If you simply plotted frequency as height, the wide classes would look far bigger than they should. The fix is to plot frequency density so that area = frequency:
frequency density=class widthfrequency,frequency=frequency density×class width.
Because area = frequency, the total area of all the bars equals the total frequency.
A class is 20<w≤40 and contains 32 items. Work out its class width and frequency density.
Solution: Class width =40−20=20. Frequency density =2032=1.6.
Common error: using the midpoint (30) or the frequency (32) as the height. The height must be the frequency density, here 1.6.
The masses, w grams, of 84 letters are grouped below. Complete the frequency density column.
| Mass w (g) | Frequency | Class width | Frequency density |
|---|---|---|---|
| 0<w≤10 | 8 | 10 | 0.8 |
| 10<w≤20 | 20 | 10 | 2.0 |
| 20<w≤40 | 32 | 20 | 1.6 |
| 40<w≤70 | 24 | 30 | 0.8 |
Solution: Divide each frequency by its class width: 108=0.8; 1020=2.0; 2032=1.6; 3024=0.8. Notice the widest class (30) has the same frequency density as the narrowest, even though it holds three times as many letters as the first class — that is exactly why height alone would mislead. The histogram is shown below.
To read a frequency back from a histogram, calculate the area of the bar: area = frequency density × class width.
From the letter histogram, a bar has frequency density 1.6 over the class 20<w≤40. Show that it represents 32 letters.
Solution: Frequency =frequency density×class width=1.6×(40−20)=1.6×20=32 letters. ✓
A histogram bar covers 5<x≤15 with a frequency density of 3.4. How many items does it represent?
Solution: Class width =15−5=10. Frequency =3.4×10=34 items.
Common error: reading the frequency directly off the vertical axis (3.4). The axis shows frequency density; the frequency is the area, found by multiplying by the class width.
In a histogram, a class 0<t≤4 has frequency density 6, and another class 4<t≤10 has frequency density 5. How many items are there in total across these two classes?
Solution: First class: 6×4=24. Second class: 5×6=30. Total =24+30=54 items.
A histogram bar for the class 100≤x<160 has a frequency density of 0.45. How many items does it represent?
Solution: Class width =160−100=60. Frequency =0.45×60=27 items. Notice how a low frequency density over a wide class can still represent a sizeable frequency — another reason height alone is misleading.
Two classes of a histogram have the same frequency of 24. The first is 0<x≤8; the second is 8<x≤32. Work out the frequency density of each and comment on the bar heights.
Solution: First class width =8, so frequency density =824=3. Second class width =24, so frequency density =2424=1. Although both classes contain 24 items, the first bar is three times as tall as the second, because the same frequency is squeezed into a third of the width. This is exactly the visual correction that frequency density provides.
Some questions do not give the frequency-density axis directly; instead they tell you the frequency of one bar and ask you to use it to scale the rest. The trick is to find how many items one unit of area represents.
On a histogram, the bar for the class 10<x≤20 has a frequency density read as 4, and you are told this class contains 40 items. A second bar covers 20<x≤50 with frequency density 2. How many items are in the second class?
Solution: Check the key bar: 4×10=40 items ✓, confirming area = frequency directly. Second class: 2×(50−20)=2×30=60 items.
Some OCR histograms do not label the vertical axis with frequency density at all. Instead they tell you the frequency of one class and expect you to use the area of that bar to work out what a unit of area is worth, then apply it to the other bars. The method is:
This works because, in any histogram, area is proportional to frequency.
A histogram has no frequency-density scale. The bar for 0<x≤10 has height 3 (in grid units) and width 10, and is known to represent 60 items. The bar for 10<x≤30 has height 4.5 and width 20. How many items does the second bar represent?
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