S3StatisticsHigherGrades 7–8

Histograms

Q: Why not just use a bar chart?

With unequal class widths, bar charts exaggerate wide classes. Area keeps the comparison fair.

Q: Can a histogram have equal class widths?

Yes; then heights are proportional to frequency, but AQA questions usually use unequal widths.

Q: What is the modal class in a histogram?

The class with the tallest bar — the highest frequency density.

Suppose 16 people are aged 0–20 and 15 are aged 30–35. A bar chart would make the groups look about the same, even though the second is squeezed into a class a quarter as wide. Histograms fix this by making the $\textbf{area}$ of each bar represent frequency. The height becomes $\textbf{frequency density}$ .

45 min AQA GCSE Maths

The big picture

Frequency density $= \dfrac{\text{frequency}}{\text{class width}}$ . A tall bar means data are crowded closely together, not simply that there are many values.

Every histogram question is either "draw it" (work out the densities) or "read it" (area gives frequency). Estimating how many values lie in part of a class uses the same idea: take the matching fraction of the bar’s area.

What you'll learn

Calculate frequency density and draw histograms
Read frequencies from a histogram using area
Estimate frequencies within part of a class
Estimate the median and mean from a histogram

Drawing a histogram

Ages of 73 people: $0 < a \le 20$ : 16; $20 < a \le 30$ : 24; $30 < a \le 35$ : 15; $35 < a \le 50$ : 18.

Frequency densities: $\dfrac{16}{20} = 0.8$ , $\dfrac{24}{10} = 2.4$ , $\dfrac{15}{5} = 3$ , $\dfrac{18}{15} = 1.2$ .

Draw bars with no gaps, each as wide as its class and as tall as its frequency density.

Your turnSame method, new numbers — try it before moving on.

A class has frequency 36 and class width 12. What is its frequency density?

Your answer

Show worked solution

Solution

1Frequency density $= \frac{\text{frequency}}{\text{class width}} = \frac{36}{12} = 3$ .

Answer: $3$

Tip — Add a frequency density column to the table before drawing anything — it keeps the scale decisions simple.

Reading a histogram

Frequency $= \text{frequency density} \times \text{class width}$ .

A bar from 10 to 25 with height 1.6 represents $1.6 \times 15 = 24$ values.

Example 1Using the age histogram, estimate how many people are aged between 25 and 40.

25

30

5 \times 2.4 = 12

30

35

5 \times 3 = 15

35

40

5 \times 1.2 = 6

AnswerAbout 33 people

Your turnSame method, new numbers — try it before moving on.

On a histogram, the bar for $20 < a \le 50$ has frequency density 1.4. How many people are in this class?

Your answer

Show worked solution

Solution

1Frequency $=$ frequency density $\times$ class width.
2 $1.4 \times 30 = 42$ .

Answer: $42$

Why this works

This assumes values are spread evenly within each class. That is why the answer is an estimate — the raw data might be bunched at one end of a class.

Estimating the median

Find the position of the median, then count through the areas.

For 73 people, the median is at $\dfrac{73}{2} = 36.5$ . The first class contains 16, so we need 20.5 more from the 20–30 class of 24.

Median $\approx 20 + \dfrac{20.5}{24} \times 10 = 28.5$ years.

Estimating the mean

Find each frequency, multiply by the class midpoint, add, and divide by the total — exactly as for a grouped frequency table.

Tip — If the histogram has no scale on the vertical axis, use a known frequency to work out what one square represents.

Think like an examiner

Label the vertical axis "frequency density".
Show frequency density calculations in a table.
State that part-class estimates assume an even spread.

Remember these

\text{FD} = \dfrac{\text{frequency}}{\text{class width}}

Frequency density.

\text{frequency} = \text{FD} \times \text{class width}

Area of a bar.

\bar{x} \approx \dfrac{\sum fx}{\sum f}

Estimated mean from midpoints.

Watch out for these

Plotting frequency on the vertical axis.

Plot frequency density.

Reading frequency as the height of a bar.

Frequency is the area.

Leaving gaps between bars.

Histogram bars touch because the data are continuous.

Using a whole class when only part is needed.

Use the fraction of the class width required.

Stretch yourself

A histogram shows these frequency densities: $0 < h \le 10$ : 1.5; $10 < h \le 15$ : 4; $15 < h \le 25$ : 2.2; $25 < h \le 40$ : 0.6. Find the total frequency, estimate how many values are greater than 20, and estimate the mean.

Hint — Convert each bar to a frequency first.

Reveal solution

SolutionFrequencies:

1.5 \times 10 = 15

4 \times 5 = 20

2.2 \times 10 = 22

0.6 \times 15 = 9

. Total

= 66

. Greater than 20: half of the 15–25 class gives 11, plus 9, so about 20 values. Mean

\approx \dfrac{15(5) + 20(12.5) + 22(20) + 9(32.5)}{66} = \dfrac{1057.5}{66} \approx 16.0

\blacksquare

Questions students ask

Why not just use a bar chart?

With unequal class widths, bar charts exaggerate wide classes. Area keeps the comparison fair.

Can a histogram have equal class widths?

Yes; then heights are proportional to frequency, but AQA questions usually use unequal widths.

What is the modal class in a histogram?

The class with the tallest bar — the highest frequency density.

Key takeaways

Frequency density $=$ frequency $\div$ class width.
Bar area represents frequency.
Bars touch and have widths equal to the classes.
Estimate part-class frequencies by taking a fraction of the area.
Median and mean can be estimated from bar areas.

Exam-style questions

Practise it the way it is examined

2 original questions · 6 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 74 marks

Times taken: $\begin{array}{c|c} \text{Time } t \text{ (s)} & \text{Frequency} \\ \hline 0 < t \le 10 & 12 \\ 10 < t \le 15 & 18 \\ 15 < t \le 30 & 24 \\ 30 < t \le 50 & 16 \end{array}$

(a)Work out the frequency density for each class.[2]
Show mark scheme (a)
- B1Two correct from 1.2, 3.6, 1.6, 0.8
- B1All four correct
(b)Estimate the number of times greater than 25 seconds.[2]
Show mark scheme (b)
- M1 $\frac{5}{15} \times 24 = 8$ from the $15$ – $30$ class
- A1 $8 + 16 = 24$
Answer: $24$

Question 2Grade 82 marks

On a histogram, the bar for $0 < x \le 20$ has height 0.8 and represents 16 people. The bar for $20 < x \le 30$ has height 1.5. How many people does it represent?[2]
Show mark scheme
- M1Frequency $= 1.5 \times 10$
- A1 $15$
Answer: $15$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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