1MA1 · S3StatisticsHigherGrades 7–8

Histograms

Q: When do I need frequency density?

Whenever classes have unequal widths. With equal widths, bar heights can be frequencies, but density is still correct.

Q: How is a histogram different from a bar chart?

Histograms show continuous data with touching bars whose area represents frequency; bar charts show categories with gaps.

Q: What does a tall, thin bar mean?

Many values packed into a narrow range — high density.

A histogram looks like a bar chart for continuous data, but with one crucial difference: when the classes have different widths, the height of each bar is not the frequency. It is the $\textbf{frequency density}$ , chosen so that the $\textbf{area}$ of each bar represents the frequency. Getting that idea right is the whole of this topic.

45 min Statistics

The big picture

If a class twice as wide were drawn at its frequency, it would look twice as important simply because it is wide. Dividing frequency by class width fixes this: frequency density is frequency per unit, so the bar’s area — width × density — equals the frequency.

Edexcel histogram questions work in both directions: complete a table and draw the histogram, or read frequencies from a drawn histogram, often for part of a class, such as "estimate the number of values between 12 and 25".

What you'll learn

Calculate frequency density for classes of unequal width
Draw histograms with frequency density on the vertical axis
Find frequencies from a histogram using area
Estimate frequencies within part of a class and estimate the median

Frequency density

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

Frequency $=$ frequency density × class width — the area of the bar.

Class width is the difference between the class boundaries: $15 \le t < 20$ has width 5.

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

Bar height.

\text{frequency} = \text{density} \times \text{width}

Bar area.

Example 1Complete frequency densities for:

0 \le t < 10

: 12;

10 \le t < 15

: 20;

15 \le t < 20

: 18;

20 \le t < 40

: 30.

12 \div 10 = 1.2

20 \div 5 = 4

18 \div 5 = 3.6

30 \div 20 = 1.5

AnswerFrequency densities 1.2, 4, 3.6, 1.5

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

A class of width 15 has frequency 45. What is its frequency density?

Your answer

Show worked solution

Solution

1 $45 \div 15 = 3$ .

Answer: $3$

Drawing histograms

Use a continuous horizontal scale with class boundaries; bars touch, with no gaps.

The vertical axis is labelled "frequency density" with an even scale.

Each bar spans its class exactly, with height equal to its frequency density.

Tip — Label the vertical axis "frequency density", not "frequency". Using frequency heights with unequal widths loses most of the marks.

Reading histograms

To find the frequency of a class, multiply its height (frequency density) by its width.

If the vertical scale is not labelled with numbers, use a class whose frequency is known to work out what one unit of area represents.

Example 1Using the histogram above, estimate how many values lie between 12 and 25.

112 to 15 is in the class with density 4:

3 \times 4 = 12

215 to 20 is a whole class: 18.

320 to 25 is in the class with density 1.5:

5 \times 1.5 = 7.5

4Total

= 12 + 18 + 7.5

AnswerAbout 37.5, so roughly 37 or 38 values

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

The bar for $30 \le t < 50$ has frequency density 0.8. What is the frequency?

Your answer

Show worked solution

Solution

1Frequency $= 0.8 \times 20 = 16$ .

Answer: $16$

Why this works

Taking part of a class assumes values are spread evenly across it — the same assumption as reading from a cumulative frequency curve. That is why the answer is an estimate.

Estimating the median

Find the total frequency and the median position $\tfrac{n}{2}$ .

Add class frequencies until you reach the class containing the median, then work out how far into that class you need to go.

Example 1For the same data (total 80), estimate the median.

1Median position: 40.

2Cumulative: 12 by 10, 32 by 15. Need 8 more from the class

15 \le t < 20

, which has density 3.6.

3Width needed:

8 \div 3.6 = 2.22

, so median

\approx 15 + 2.22

AnswerMedian

\approx 17.2

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

The class $10 \le t < 20$ has frequency 30. Estimate how many values lie between 10 and 18.

Your answer

Show worked solution

Solution

118 is $\frac{8}{10}$ of the way through the class.
2 $\frac{8}{10} \times 30 = 24$ .

Answer: $24$

Think like an examiner

Label the vertical axis "frequency density".
Show width × density calculations when reading frequencies.
For part of a class, show the fraction of the width used.

Histograms

\text{f.d.} = \tfrac{f}{\text{width}}

Frequency density.

f = \text{f.d.} \times \text{width}

Frequency from a bar.

Watch out for these

Drawing bar heights as frequencies with unequal widths.

Heights must be frequency densities.

Using class midpoints as class widths.

Width is upper boundary minus lower boundary.

Reading the frequency straight off the vertical axis.

Multiply the height by the width.

Leaving gaps between bars.

Histogram bars touch because the data are continuous.

Stretch yourself

On a histogram with no numbers on the vertical axis, the bar for $10 \le x < 20$ is 6 cm tall and represents 42 people. The bar for $20 \le x < 25$ is 8 cm tall. How many people does it represent?

Hint — Work out how many people 1 cm² of bar represents, using horizontal distance in class units.

Reveal solution

SolutionThe first bar has frequency density

\tfrac{42}{10} = 4.2

and is drawn 6 cm tall, so each centimetre of height represents a density of

0.7

. The second bar is 8 cm tall, so its density is

8 \times 0.7 = 5.6

, and its frequency is

5.6 \times 5 = 28

people. Check by area:

10 \times 6 = 60

units for 42 people is 0.7 people per unit, and

5 \times 8 = 40

units gives 28 ✓.

\blacksquare

Questions students ask

When do I need frequency density?

Whenever classes have unequal widths. With equal widths, bar heights can be frequencies, but density is still correct.

How is a histogram different from a bar chart?

Histograms show continuous data with touching bars whose area represents frequency; bar charts show categories with gaps.

What does a tall, thin bar mean?

Many values packed into a narrow range — high density.

Key takeaways

Frequency density $=$ frequency ÷ class width.
In a histogram, bar area represents frequency.
Bars touch and the vertical axis is frequency density.
Read frequencies with width × density, including parts of classes.
Estimate the median by finding how far into the median class the middle value lies.

Exam-style questions

Practise it the way it is examined

2 original questions · 7 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

Heights: $0 < h \le 20$ : 10; $20 < h \le 30$ : 24; $30 < h \le 35$ : 15; $35 < h \le 50$ : 18.

(a)Work out the frequency densities.[2]
Show mark scheme (a)
- B1Two correct from 0.5, 2.4, 3, 1.2
- B1All four correct
(b)Estimate the number of heights greater than 32.[2]
Show mark scheme (b)
- M1 $\frac{3}{5} \times 15 = 9$
- A1 $9 + 18 = 27$
Answer: $27$

Question 2Grade 83 marks

Use the same data (total 67).

Estimate the median height.[3]
Show mark scheme
- M1Median is the 33.5th value, in the $20$ – $30$ class
- M1 $20 + \frac{23.5}{24} \times 10$
- A1 $29.8$
Answer: $29.8$

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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