Histograms · Study Notes · Edexcel IGCSE 4MA1 Higher

Histograms with UNEQUAL class widths use frequency density on the y-axis. AREA = FREQUENCY. The most-confused chart type for IGCSE students.

At a glance

Histogram = bar chart for CONTINUOUS data; bars TOUCH.
y-axis = frequency density, NOT frequency.
Frequency density = $\dfrac{\text{frequency}}{\text{class width}}$ .
Area of bar = frequency.
Class width = upper − lower of the interval.
Used when classes have UNEQUAL widths.
Read off: frequency = density × width.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

6.3 — Construct histograms for grouped continuous data with unequal class widths.
6.3 — Calculate frequency density.
6.3 — Read information (frequencies, totals) from a given histogram.
6.3 — Estimate frequencies for partial intervals.

Why frequency density?

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When classes have different widths, raw frequency misleads.

The problem. Suppose a class $0$ – $10$ has 5 students and a class $20$ – $80$ has 60 students. If you draw bars proportional to frequency, the second bar is 12 times taller — but the second class is also 6 times wider. The bar gives a false impression.

The fix. Plot frequency DENSITY: $\text{Frequency density} = \dfrac{\text{frequency}}{\text{class width}}$

Dividing frequency by class width corrects the unfair height a wide class would otherwise get.

This standardises by width — bar height represents "students per unit interval".

Worked example. Class $20 \le x < 30$ , frequency $25$ .

Class width = $30 - 20 = 10$ .
Frequency density = $25 / 10 = 2.5$ .

Worked example. Times (mins): $0$ – $10$ (5), $10$ – $20$ (15), $20$ – $40$ (12), $40$ – $80$ (8).

Class widths: 10, 10, 20, 40.
Densities: 0.5, 1.5, 0.6, 0.2.

Notice the second class has the LARGEST density even though the third has more students — because the third class is wider.

Edexcel tip. Always set up a table with columns: lower bound, upper bound, frequency, class width, frequency density. Show all working.

FD = freq / width.
Width = upper − lower.
Adjusts for unequal class widths.
Use a workings table.

See the full worked example for histogram →

Area = Frequency

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The big idea: area of each bar tells you how many.

On a histogram, the AREA of a bar equals the frequency for that class.

$\text{Area} = \text{height} \times \text{width} = \text{FD} \times \text{width} = \text{frequency}$

Why this matters.

A tall narrow bar can have the same frequency as a short wide bar.
Total frequency = total area under all bars.

Frequency is the bar's area — multiply class width by frequency density.

Worked example. Bar from $20$ to $35$ with FD $= 2.4$ .

Width = $15$ .
Frequency = area = $2.4 \times 15 = 36$ .

Estimating frequency for part of a class. Assume uniform distribution within the class.

Worked example. Class $30 \le x < 50$ with FD $= 1.5$ . Estimate number with $35 \le x < 50$ :

Sub-width = $15$ .
Estimated frequency = $1.5 \times 15 = 22.5 \approx 23$ .

Reverse direction. From a histogram, find the total number of data points: sum the areas of ALL bars.

Edexcel tip. "Use the histogram to find" → multiply density × width to get frequency. The most common 3-mark Higher question.

Area = frequency.
Multiply density × width.
Estimate partial bars by uniform assumption.
Sum all areas = total.

See the full worked example for histogram →

Drawing a histogram from a table

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Compute densities, label axes carefully, bars touching.

Steps:

Add columns for class width and frequency density to the frequency table.
Compute density for each class.
Choose suitable scales for both axes.
Plot bars with: width = class width, height = frequency density.
Bars TOUCH (no gaps — continuous data).
Label x-axis with units (e.g. "Time (minutes)") and y-axis "Frequency density".

Worked example table.

Time $t$	$f$	Width	FD
$0 \le t < 10$	5	10	0.5
$10 \le t < 20$	15	10	1.5
$20 \le t < 40$	12	20	0.6
$40 \le t < 80$	8	40	0.2

Then plot: 4 bars at the right widths and heights.

Bars touch, widths match the class intervals, heights are the frequency densities.

Edexcel tip. Mark scheme allocations:

M1: at least one density correct.
M1: more than one density correct.
A1: all densities correct AND drawn with correct widths AND correct heights.

Set up table with FD column.
Width × density → bar.
Bars TOUCH (continuous).
Label axes properly.

Quick recap

Histogram: continuous data, bars touch, y-axis = frequency density.
FD = frequency / class width.
Area of bar = frequency.
From histogram: frequency = density × width.
Unequal class widths → must use density (not raw frequency).
Estimate partial intervals by assuming uniform distribution within the class.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Frequency density = $\dfrac{\text{frequency}}{\text{class width}}$ .
Area = frequency.
Bars TOUCH (continuous data).
Class width = upper − lower.
y-axis label: 'Frequency density'.

How it’s examined

Histograms appear once on every Higher Tier paper as a 4-6 mark question. Common formats: (a) given table, draw histogram; (b) given histogram, find frequencies. Examiner reports flag: (1) plotting frequency on y-axis (not density), (2) gaps between bars, (3) wrong class widths.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024, 4MA1/1H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on histogram, written exactly the way a tutor would explain them at the board.

1Calculate frequency density
Higher• frequency density
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Question
Class $20 \le x < 30$ has frequency $25$ . Find the frequency density.
Step-by-step solution
1. Step 1
  Class width = $30 - 20 = 10$ .
2. Step 2
  Frequency density = frequency / class width = $25 / 10 = 2.5$ .
Answer
$2.5$
2Build histogram from a frequency table
Higher• Adapted from 4MA1/2H May/Jun 2024 Q22• histogram
▼
Question
Times (mins): $0 \le t < 10$ (5), $10 \le t < 20$ (15), $20 \le t < 40$ (12), $40 \le t < 80$ (8). Find the frequency densities.
Step-by-step solution
1. Step 1
  Class widths: $10, 10, 20, 40$ .
2. Step 2
  Frequency densities: $5/10 = 0.5$ ; $15/10 = 1.5$ ; $12/20 = 0.6$ ; $8/40 = 0.2$ .
Answer
Densities: $0.5, 1.5, 0.6, 0.2$ (in order).
Examiner tip
Always lay out a 'frequency density' column in your table. It's the y-axis, NOT the frequency.
3Find frequency from histogram
Higher• Adapted from 4MA1/1H Jan 2024 Q23• histogram
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Question
From a histogram: a bar covers $20 \le x < 35$ with frequency density $2.4$ . Find the frequency.
Step-by-step solution
1. Step 1
  Class width = $35 - 20 = 15$ .
2. Step 2
  Frequency = density × width = $2.4 \times 15 = 36$ .
Answer
Frequency = $36$
4Estimate a partial-bar frequency
Higher• histogram, partial
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Question
Bar covers $30 \le x < 50$ with frequency density $1.5$ . Estimate the number with $35 \le x < 50$ .
Step-by-step solution
1. Step 1
  Width of $35 \le x < 50$ is $15$ . Density is $1.5$ .
2. Step 2
  Estimate = $1.5 \times 15 = 22.5 \approx 23$ (round appropriately).
Answer
Approximately $22$ or $23$
Examiner tip
Assume uniform distribution within the class. Mark scheme allows reasonable rounding.
5Two bars of the same height [1 mark]
Getting started• Paper 1H/2H style — 1 mark• histograms, area
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Question
On a histogram, the bar for $40 \leqslant x < 60$ and the bar for $60 \leqslant x < 65$ are the same height. The bar for $40 \leqslant x < 60$ represents $36$ people.

Work out the number of people represented by the bar for $60 \leqslant x < 65$ . [1]
Step-by-step solution
1. Step 1
  The bars are the same height, so the frequency is proportional to the class width. The widths are $20$ and $5$ .
2. Step 2
  B1 The second bar is a quarter as wide, so it holds a quarter as many people.
  $36 \times \frac{5}{20} = 9$
Answer
$9$ people
Examiner tip
Examiner tip: Equal heights mean equal frequency density: $36 \div 20 = 1.8$ for both bars, so the second bar holds $1.8 \times 5 = 9$ .

Common mistake: Answering $36$ because the bars are the same height. On a histogram the area shows the frequency, so the narrower bar holds fewer people.
6Total frequency from every bar [2 marks]
Getting started• Paper 1H/2H style — 2 marks• histograms, frequency density
▼
Question
A histogram shows the times, $t$ minutes, that some people spent in a queue. The bars are:

$0 \leqslant t < 10$ : frequency density $1.4$

$10 \leqslant t < 15$ : frequency density $4$

$15 \leqslant t < 30$ : frequency density $2.2$

$30 \leqslant t < 50$ : frequency density $0.5$

Work out the total number of people. [2]
Step-by-step solution
1. Step 1
  M1 Work out the frequency of each bar: frequency density × class width.
  $1.4 \times 10 = 14, \quad 4 \times 5 = 20, \quad 2.2 \times 15 = 33, \quad 0.5 \times 20 = 10$
2. Step 2
  A1 Add the frequencies.
  $14 + 20 + 33 + 10 = 77$
Answer
$77$ people
Examiner tip
Examiner tip: The total frequency is the total area of the histogram. Write each frequency next to its bar so the method is clear.

Common mistake: Adding the heights, $1.4 + 4 + 2.2 + 0.5 = 8.1$ . The heights are frequency densities, not frequencies.
7Estimating over three bars [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• histograms, estimation
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Question
A histogram shows the times, $t$ seconds, taken by some children to run a race. Three of the bars are:

$10 \leqslant t < 20$ : frequency density $2.6$

$20 \leqslant t < 30$ : frequency density $3.5$

$30 \leqslant t < 50$ : frequency density $1.4$

Work out an estimate for the number of children with $18 \leqslant t < 42$ . [3]
Step-by-step solution
1. Step 1
  M1 Split the range at the class boundaries: $18$ to $20$ , $20$ to $30$ , $30$ to $42$ . Find the area of the histogram over each part.
  $2.6 \times 2 = 5.2, \quad 3.5 \times 10 = 35, \quad 1.4 \times 12 = 16.8$
2. Step 2
  M1 Add the three parts.
  $5.2 + 35 + 16.8 = 57$
3. Step 3
  A1
  $57 \text{ children}$
Answer
$57$ children
Examiner tip
Examiner tip: Estimating in part of a class assumes the values are spread evenly across it, so the area of the part of the bar you need is the estimate.

Common mistake: Using the width of the whole class for the end pieces, for example $1.4 \times 20 = 28$ instead of $1.4 \times 12$ . Work out $42 - 30 = 12$ first.
8Scale from a known bar measured in centimetres [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• histograms, no scale
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Question
A histogram shows the masses, $w$ grams, of some letters. The vertical axis has no scale.

The bar for $20 \leqslant w < 30$ is $8$ cm tall and the bar for $30 \leqslant w < 60$ is $3$ cm tall.

There are $32$ letters with $20 \leqslant w < 30$ .

Work out the number of letters with $30 \leqslant w < 60$ . [3]
Step-by-step solution
1. Step 1
  M1 The frequency density of the known bar is $32 \div 10 = 3.2$ , and it is $8$ cm tall.
  $3.2 \div 8 = 0.4 \text{ per cm}$
2. Step 2
  M1 The bar for $30 \leqslant w < 60$ is $3$ cm tall, so its frequency density is $3 \times 0.4 = 1.2$ .
  $1.2 \times 30$
3. Step 3
  A1
  $= 36$
Answer
$36$ letters
Examiner tip
Examiner tip: Build the missing scale from the bar you know: frequency ÷ width gives its frequency density, and its height tells you what $1$ cm stands for.

Common mistake: Scaling by height alone: $32 \times \frac{3}{8} = 12$ . The second bar is three times as wide, so it holds three times as many letters as that.
9Height of a missing bar from the total [4 marks]
Stretch• Paper 1H/2H style — 4 marks• histograms, completing a histogram
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Question
A histogram shows the masses, $m$ kg, of $200$ bags. The bar for $20 < m \leqslant 25$ has not been drawn. The other bars are:

$0 < m \leqslant 10$ : frequency density $2$

$10 < m \leqslant 20$ : frequency density $5.5$

$25 < m \leqslant 40$ : frequency density $3$

$40 < m \leqslant 60$ : frequency density $1.2$

Work out the height of the bar for $20 < m \leqslant 25$ . [4]
Step-by-step solution
1. Step 1
  M1 Find the frequencies of the four drawn bars.
  $2 \times 10 = 20, \quad 5.5 \times 10 = 55, \quad 3 \times 15 = 45, \quad 1.2 \times 20 = 24$
2. Step 2
  M1 The rest of the $200$ bags are in the missing class.
  $200 - (20 + 55 + 45 + 24) = 200 - 144 = 56$
3. Step 3
  M1 Frequency density = frequency ÷ class width.
  $56 \div 5$
4. Step 4
  A1
  $= 11.2$
Answer
Frequency density $11.2$
Examiner tip
Examiner tip: The question asks for a height, so the final step is a frequency density, not a frequency. Check: $11.2 \times 5 = 56$ and $144 + 56 = 200$ .

Common mistake: Stopping at $56$ , the frequency. On a histogram grid a bar drawn at height $56$ would be far off the scale.
10Bar heights in a ratio and a known total [5 marks]
Stretch• Paper 1H/2H style — 5 marks• histograms, ratio
▼
Question
A histogram shows the distances, $d$ metres, that $210$ students threw a ball. It has four bars:

$0 \leqslant d < 10$

$10 \leqslant d < 15$

$15 \leqslant d < 25$

$25 \leqslant d < 45$

The heights of the bars are in the ratio $2 : 5 : 4 : 1$ .

(a) Work out the number of students with $15 \leqslant d < 25$ . [3]

(b) Work out the number of students who threw less than $20$ metres. [2]
Step-by-step solution
1. Step 1
  M1 (a) Let the heights be $2k$ , $5k$ , $4k$ and $k$ . Frequency = height × width for each bar.
  $2k \times 10 + 5k \times 5 + 4k \times 10 + k \times 20 = 20k + 25k + 40k + 20k = 105k$
2. Step 2
  M1 The total frequency is $210$ .
  $105k = 210 \quad\Rightarrow\quad k = 2$
3. Step 3
  A1
  $40k = 40 \times 2 = 80$
4. Step 4
  M1 (b) The first two bars hold $20k = 40$ and $25k = 50$ . From $15$ to $20$ is half of the $15 \leqslant d < 25$ bar.
  $40 + 50 + \frac{1}{2} \times 80$
5. Step 5
  A1
  $= 130$
Answer
(a) $80$ (b) $130$
Examiner tip
Examiner tip: The ratio is for heights, but the total $210$ is shared by AREA. Multiply each part of the ratio by its class width before you share.

Common mistake: Sharing $210$ in the ratio $2 : 5 : 4 : 1$ directly, which gives $70$ for the third class. That ignores the unequal widths.

Question 1
Paper 1H/2H style — 1 mark1 mark
Give a reason why there are no gaps between the bars of a histogram for the times, $t$ minutes, taken by some runners, grouped as $20 \leqslant t < 25$ , $25 \leqslant t < 30$ and so on. [1]
Model answer
Time is continuous data, and each class ends exactly where the next one begins (for example at $t = 25$ ), so each bar ends where the next bar starts.
Why this scores
Mark scheme:
- B1 for a reason referring to continuous data or to the classes having no gaps between them oe
Why this scores full marks: It names the type of data (continuous) and links it to the class boundaries, which is exactly why the bars touch. "Because it is a histogram" does not explain anything.
Question 2
Paper 1H/2H style — 2 marks2 marks
A histogram shows the ages, $a$ years, of $60$ people at a concert. The bar for $20 \leqslant a < 25$ has frequency density $1.8$ .

Work out the fraction of the people who are aged $20 \leqslant a < 25$ . Give your answer in its simplest form. [2]
Model answer
Frequency $= 1.8 \times 5 = 9$

$\dfrac{9}{60} = \dfrac{3}{20}$

Answer: $\dfrac{3}{20}$
Why this scores
Mark scheme:
- M1 for $1.8 \times 5$ ( $= 9$ )
- A1 for $\frac{3}{20}$ cao
Why this scores full marks: The candidate turns the height into a frequency (height × class width) before dividing by the total, and simplifies as asked. Using the height as the frequency, $\frac{1.8}{60}$ , is the usual wrong answer.
Question 3
Paper 1H/2H style — 2 marks2 marks
A histogram shows the times, $t$ minutes, that some customers spent in a shop. The bars are:
- $0 \leqslant t < 5$ : frequency density $2$
- $5 \leqslant t < 10$ : frequency density $3.6$
- $10 \leqslant t < 20$ : frequency density $2.2$
- $20 \leqslant t < 40$ : frequency density $0.8$
Find the class that contains the median. [2]
Model answer
Frequencies: $2 \times 5 = 10$ , $3.6 \times 5 = 18$ , $2.2 \times 10 = 22$ , $0.8 \times 20 = 16$

Total $= 66$ , so the median is between the $33$ rd and $34$ th values.

Running totals: $10$ , $28$ , $50$ , $66$

Answer: $10 \leqslant t < 20$
Why this scores
Mark scheme:
- M1 for finding all four frequencies ( $10$ , $18$ , $22$ , $16$ )
- A1 for $10 \leqslant t < 20$ cao
Why this scores full marks: It turns the heights into frequencies first, then uses running totals. Picking the tallest bar ( $5 \leqslant t < 10$ ) is the usual wrong answer.
Question 4
Paper 1H/2H style — 2 marks2 marks
On a histogram, the bar for $10 \leqslant x < 15$ has frequency density $4$ and the bar for $15 \leqslant x < 25$ has frequency density $1.6$ .

The two classes are combined into one class, $10 \leqslant x < 25$ . Work out the frequency density of the new bar. [2]
Model answer
Frequencies: $4 \times 5 = 20$ and $1.6 \times 10 = 16$

$\dfrac{20 + 16}{15} = \dfrac{36}{15}$

Answer: $2.4$
Why this scores
Mark scheme:
- M1 for $4 \times 5 + 1.6 \times 10$ ( $= 36$ ) oe
- A1 for $2.4$ cao
Why this scores full marks: The candidate finds the frequencies first, adds them, and divides by the new class width, $15$ . Averaging the two heights, $\frac{4 + 1.6}{2} = 2.8$ , is the usual wrong answer; it ignores the different widths.
Question 5
Paper 1H/2H style — 3 marks3 marks
A histogram shows the heights, $h$ cm, of some plants. Key: $1$ cm² of area represents $4$ plants.

The bar for $10 \leqslant h < 25$ is $3$ cm wide and $4.5$ cm tall. The bar for $25 \leqslant h < 30$ is $1$ cm wide and $6$ cm tall.

Work out the number of plants with $10 \leqslant h < 30$ . [3]
Model answer
Area of first bar $= 3 \times 4.5 = 13.5$ cm²

Area of second bar $= 1 \times 6 = 6$ cm²

$(13.5 + 6) \times 4 = 19.5 \times 4$

Answer: $78$
Why this scores
Mark scheme:
- M1 for the area of one bar, $13.5$ or $6$
- M1 for $(13.5 + 6) \times 4$ oe
- A1 for $78$ cao
Why this scores full marks: The key is about area, so the candidate multiplies width by height for each bar before using it. Comparing heights ( $4.5$ and $6$ ) alone would give the wrong idea that the second class is larger.
Question 6
Paper 1H/2H style — 3 marks3 marks
A histogram shows the times, $t$ minutes, of the runners in a fun run. The bars are:
- $15 \leqslant t < 20$ : frequency density $1.6$
- $20 \leqslant t < 30$ : frequency density $3.8$
- $30 \leqslant t < 35$ : frequency density $4.4$
- $35 \leqslant t < 50$ : frequency density $0.8$
Estimate the fraction of the runners who took less than $25$ minutes. [3]
Model answer
Frequencies: $1.6 \times 5 = 8$ , $3.8 \times 10 = 38$ , $4.4 \times 5 = 22$ , $0.8 \times 15 = 12$

Total $= 80$

Less than $25$ : $8 + 3.8 \times 5 = 8 + 19 = 27$

Answer: $\dfrac{27}{80}$
Why this scores
Mark scheme:
- M1 for $3.8 \times 5$ ( $= 19$ ) oe
- M1 for the total $80$ from all four bars
- A1 for $\frac{27}{80}$ oe
Why this scores full marks: Only half of the $20$ – $30$ bar lies below $25$ , and the total is not given, so the candidate finds it from every bar.
Question 7
Paper 1H/2H style — 4 marks4 marks
A histogram shows the times, $x$ minutes, that some people took to walk to work. The bars are:
- $0 \leqslant x < 10$ : frequency density $0.8$
- $10 \leqslant x < 20$ : frequency density $p$
- $20 \leqslant x < 40$ : frequency density $0.6$
An estimate for the mean time is $17$ minutes.

Work out the value of $p$ . [4]
Model answer
Frequencies: $0.8 \times 10 = 8$ , $10p$ , $0.6 \times 20 = 12$

$\frac{8 \times 5 + 10p \times 15 + 12 \times 30}{8 + 10p + 12} = 17$

$400 + 150p = 17(20 + 10p) = 340 + 170p$

$60 = 20p$

Answer: $p = 3$
Why this scores
Mark scheme:
- M1 for the frequencies $8$ , $10p$ and $12$
- M1 for $\sum fx = 40 + 150p + 360$ oe, using the midpoints $5$ , $15$ and $30$
- M1 for a correct equation, e.g. $\frac{400 + 150p}{20 + 10p} = 17$ oe
- A1 for $p = 3$ cao
Why this scores full marks: Frequencies come first (height × width), including $10p$ for the unknown bar, then the grouped-data mean is set equal to $17$ and solved. The candidate can check: frequencies $8$ , $30$ , $12$ give $\sum fx = 850$ and $850 \div 50 = 17$ . Using $p$ itself as the frequency of the middle class is the usual error.
Question 8
Paper 1H/2H style — 4 marks4 marks
A histogram shows the masses, $m$ kg, of some adults. The bars are:
- $40 \leqslant m < 50$ : frequency density $1.5$
- $50 \leqslant m < 55$ : frequency density $5$
- $55 \leqslant m < 60$ : frequency density $6.4$
- $60 \leqslant m < 70$ : frequency density $2.2$
- $70 \leqslant m < 90$ : frequency density $0.3$
Work out an estimate for the median mass. [4]
Model answer
Frequencies: $15$ , $25$ , $32$ , $22$ , $6$ ; total $100$

Median is the $50$ th value. Running totals: $15$ , $40$ , $72$ , so it is in $55 \leqslant m < 60$ , $10$ values into a class of $32$ .

$55 + \frac{10}{32} \times 5 = 56.5625$

Answer: $56.6$ kg
Why this scores
Mark scheme:
- M1 for the frequencies $15$ , $25$ , $32$ , $22$ , $6$
- M1 for identifying the $50$ th (or $50.5$ th) value in $55 \leqslant m < 60$
- M1 for $55 + \frac{10}{32} \times 5$ oe
- A1 for an answer in the range $56.5$ to $56.7$
Why this scores full marks: The median is where half of the total area lies on each side. The candidate finds how far into the bar that is and moves across the $5$ kg class by the same fraction, instead of just stating the class.
Question 9
Paper 1H/2H style — 5 marks5 marks
The table gives information about the heights, $h$ cm, of $87$ tomato plants.
- $20 < h \leqslant 30$ : frequency $15$
- $30 < h \leqslant 40$ : frequency $26$
- $40 < h \leqslant 60$ : frequency $34$
- $60 < h \leqslant 100$ : frequency $12$
(a) Work out the frequency density for each class. [2]

(b) Work out an estimate for the number of plants taller than $47$ cm. [3]
Model answer
(a) $15 \div 10 = 1.5$ , $26 \div 10 = 2.6$ , $34 \div 20 = 1.7$ , $12 \div 40 = 0.3$

Answer: $1.5$ , $2.6$ , $1.7$ , $0.3$

(b) From $47$ to $60$ is $13$ cm of the $20$ cm class: $1.7 \times 13 = 22.1$

$22.1 + 12 = 34.1$

Answer: $34$
Why this scores
Mark scheme:
- B2 for all four frequency densities correct (B1 for two or three correct)
- M1 for $1.7 \times 13$ or $\frac{13}{20} \times 34$ oe
- M1 for adding $12$
- A1 for $34$ (accept $34.1$ )
Why this scores full marks: The frequency density column is set out first, so part (b) is just an area: $13$ cm of the $40$ – $60$ bar plus all of the last class. The candidate uses $60 - 47 = 13$ , not $47 - 40 = 7$ , because the question asks for plants taller than $47$ cm.
Question 10
Paper 1H/2H style — 5 marks5 marks
A histogram shows the times, $t$ minutes, that some swimmers took to finish an open-water race. The bars are:
- $20 \leqslant t < 30$ : frequency density $4.5$
- $30 \leqslant t < 35$ : frequency density $6$
- $35 \leqslant t < 45$ : frequency density $3.5$
- $45 \leqslant t < 60$ : frequency density $1$
Ali says, "More than $75\%$ of the swimmers finished in less than $40$ minutes."

Use the histogram to work out an estimate for the percentage of the swimmers who finished in less than $40$ minutes. Does your estimate support what Ali says? You must show your working. [5]
Model answer
Frequencies: $4.5 \times 10 = 45$ , $6 \times 5 = 30$ , $3.5 \times 10 = 35$ , $1 \times 15 = 15$

Total $= 125$

Less than $40$ : $45 + 30 + 3.5 \times 5 = 45 + 30 + 17.5 = 92.5$

$\dfrac{92.5}{125} \times 100 = 74\%$

Answer: No. The estimate, $74\%$ , is less than $75\%$ , so the histogram does not support what Ali says.
Why this scores
Mark scheme:
- M1 for a frequency from any bar, e.g. $4.5 \times 10$
- M1 for the total $125$
- M1 for $3.5 \times 5$ ( $= 17.5$ ) oe
- M1 for $\frac{92.5}{125}$ oe, or $0.75 \times 125$ ( $= 93.75$ ) to compare with $92.5$
- A1 for $74\%$ (or $92.5 < 93.75$ ) and the conclusion that the estimate does not support Ali oe
Why this scores full marks: Every step is shown, and the final line compares two numbers and gives a decision. "No" with no comparison would lose the A mark, as would a correct $74\%$ with no conclusion. The conclusion is about the estimate, not a certainty: the $17.5$ assumes the times in $35 \leqslant t < 45$ are spread evenly, and $74\%$ is close to $75\%$ , so the candidate says the histogram does not support Ali rather than that Ali is definitely wrong.

4MA1 Histograms

At a glance

What you’ll learn

Why frequency density?

Area = Frequency

Drawing a histogram from a table

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Calculate frequency density

2Build histogram from a frequency table

3Find frequency from histogram

4Estimate a partial-bar frequency

5Two bars of the same height [1 mark]

6Total frequency from every bar [2 marks]

7Estimating over three bars [3 marks]

8Scale from a known bar measured in centimetres [3 marks]

9Height of a missing bar from the total [4 marks]

10Bar heights in a ratio and a known total [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10