What is S3: Histograms and cumulative frequency?

Higher tier only. In a histogram, bar area represents frequency and bar height is frequency density: frequency density=frequency/class width.

How many marks are S3 questions worth?

In verified past-paper appearances, S3 questions on Edexcel GCSE Maths 1MA1 have been worth 1–7 marks.

Which past exam papers has S3 appeared in?

S3 has appeared in 21 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Construct and interpret diagrams for grouped discrete and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs [Higher only]

Higher only

Histograms and cumulative frequency

Worked answers, methods and verified real exam appearances for S3 on Edexcel GCSE Maths 1MA1.

Explanation

Higher tier only. In a histogram, bar area represents frequency and bar height is frequency density: $\text{frequency density}=\dfrac{\text{frequency}}{\text{class width}}$ .
Therefore $\text{frequency}=\text{frequency density}\times\text{class width}$.
Use the class boundaries for each width, draw touching bars, and label both axes.
For a cumulative frequency graph, plot the lower boundary with cumulative frequency $0$ , then each upper class boundary against its running total.
Estimate the median at $n/2$ , the lower quartile at $n/4$ and the upper quartile at $3n/4$ .

Worked example

The classes $10<x\leq25$ and $25<x\leq35$ are adjacent. The first has frequency $30$ ; the second has histogram height $1.4$ . Find the first bar height, the second frequency and the cumulative-frequency points.

1.First width $=15$ , so its height is $30\div15=2$ .
2.Second width $=10$ , so its frequency is $1.4\times10=14$ .
3.The cumulative-frequency points are $(10,0)$ , $(25,30)$ and $(35,44)$ .

Answer: First height $2$ ; second frequency $14$ ; cumulative points $(10,0)$ , $(25,30)$ and $(35,44)$ .

Common mistakes

Don't fall into the trap of using frequency as the bar height instead of dividing by class width.
Don't fall into the trap of plotting cumulative frequencies at class midpoints instead of upper boundaries.

Exam tip

Write the frequency-density formula beside the histogram before calculating any height or area.

Worked practice

Tier 1 · Easy

A histogram class is

15<x\leq20

and has frequency

20

. Work out the frequency density.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$4$	2	The class width is $20-15=5$ . Frequency density is $20/5=4$ .

Tier 2 · Standard

Grouped data have classes

0<x\leq10

10<x\leq25

and

25<x\leq40

with frequencies

12

30

and

18

. Calculate the three frequency densities and identify the tallest histogram bar.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Frequency densities: $1.2$ , $2$ , $1.2$ . The class $10<x\leq25$ has the tallest bar.	3	Divide each frequency by its class width: $12/10=1.2$ , $30/15=2$ and $18/15=1.2$ . The greatest density, $2$ , gives the tallest bar.

Tier 3 · Hard

A cumulative frequency graph for

80

values passes through

(10,8)

(20,26)

(30,50)

(40,70)

and

(50,80)

. Use linear interpolation between the given points to estimate the median and the interquartile range.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	Median $\approx25.8$ . Lower quartile $\approx16.7$ and upper quartile $=35$ . Interquartile range $\approx18.3$ .	5	The median is the $40$ th value, between cumulative frequencies $26$ and $50$ : $20+\dfrac{40-26}{50-26}\times10\approx25.8$ . The lower quartile is the $20$ th value: $10+\dfrac{20-8}{26-8}\times10\approx16.7$ . The upper quartile is the $60$ th value: $30+\dfrac{60-50}{70-50}\times10=35$ . Hence $IQR\approx35-16.7=18.3$ .

Tier 1 · Easy

A histogram class has width

8

and frequency density

2.5

. Work out the frequency in the class.

(1)

(Total for Question 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$20$	1	Frequency is the area of the bar: $\text{frequency}=\text{class width}\times\text{frequency density}=8\times2.5=20$ .

Tier 2 · Standard

In a histogram, the class

5<x\leq15

has frequency

24

and its bar is

6\,\text{cm}

high. The bar for

15<x\leq21

7.5\,\text{cm}

high. Work out the frequency in the second class.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$18$	3	For $5<x\leq15$ , the class width is $10$ , so the frequency density is $24/10=2.4$ . A height of $6\,\text{cm}$ represents density $2.4$ , so $1\,\text{cm}$ represents $0.4$ . The second density is $7.5\times0.4=3$ , and its width is $6$ , giving frequency $3\times6=18$ .

Tier 3 · Hard

The class

12<x\leq b

has frequency

32

and frequency density

2

. The next class is

b<x\leq36

and has frequency

18

. The cumulative frequency at

x=12

15

. Work out

b

, the cumulative frequencies at

x=b

and

x=36

, and the frequency density of the second class.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$b=28$ . Cumulative frequency $=47$ at $x=28$ and $65$ at $x=36$ . The second frequency density is $2.25$ .	5	The first class width is $32/2=16$ , so $b=12+16=28$ . Add the first frequency to get cumulative frequency $15+32=47$ at $x=28$ , then add $18$ to get $65$ at $x=36$ . The second width is $36-28=8$ , so its frequency density is $18/8=2.25$ .

Tier 2 · Standard

Three adjacent classes in a histogram have class widths

4

6

and

10

. Their bars measure

6\,\text{cm}

5\,\text{cm}

and

3\,\text{cm}

high respectively. There are

84

values altogether and there are no other classes. Work out the frequency in each class.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	The frequencies are $24$ , $30$ and $30$ , respectively.	4	Histogram frequency is proportional to bar area. The relative areas are $4\times6=24$ , $6\times5=30$ and $10\times3=30$ , in the ratio $4 : 5 : 5$ . The $14$ ratio parts represent $84$ values, so each part represents $6$ . The frequencies are therefore $24$ , $30$ and $30$ .

Tier 3 · Hard

For

50

values, a student plots the cumulative-frequency points

(0,0)

(10,8)

(20,23)

(30,19)

(40,42)

and

(50,50)

. Explain why the point

(30,19)

cannot be correct. The frequency in the class

20<x\leq30

12

. Work out the correct point at

x=30

, the frequency in the class

30<x\leq40

, and the class containing the median.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	A cumulative frequency cannot decrease as $x$ increases, but the plotted value falls from $23$ to $19$ . The correct point is $(30,35)$ . The frequency in $30<x\leq40$ is $7$ . The median is in the class $20<x\leq30$ .	4	The cumulative frequency is already $23$ at $x=20$ , so it cannot be $19$ at the larger boundary $x=30$ . Add the class frequency $12$ to get $23+12=35$ , giving $(30,35)$ . The next class frequency is $42-35=7$ . The $25$ th and $26$ th values lie after cumulative frequency $23$ and no later than cumulative frequency $35$ , so the median lies in $20<x\leq30$ .

Tier 3 · Hard

The histogram class

4<x\leq16

has frequency

36

. The adjacent class

16<x\leq b

has the same bar area and a bar twice as tall. The class

b<x\leq34

has a bar one third as tall as the second bar. The bars use the same vertical scale. Work out

b

and the total frequency in the three classes.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$b=22$ and the total frequency is $96$ .	5	The first class has width $12$ , so its frequency density is $36/12=3$ . The second bar has density $6$ and, because it has the same area, frequency $36$ . Its width is therefore $36/6=6$ , giving $b=16+6=22$ . The third density is $6/3=2$ and its width is $34-22=12$ , so its frequency is $24$ . The total is $36+36+24=96$ .

Q10

Tier 3 · Hard

Higher only: A histogram has classes

0<x\leq8

8<x\leq20

20<x\leq32

and

32<x\leq42

. Their frequency densities are

1.5

2.5

3

and

1.2

respectively. Assuming values are evenly spread within each class, estimate how many values are greater than

26

. Also estimate the value below which

70\%

of the data lie.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	Estimated number of values greater than $26$ is $30$ . $70\%$ of the data lie below an estimated value of $27$ .	5	The class frequencies are $8(1.5)=12$ , $12(2.5)=30$ , $12(3)=36$ and $10(1.2)=12$ , so there are $90$ values. Above $26$ , the part of the third class has estimated frequency $(32-26)\times3=18$ , and the final class contributes $12$ , giving $30$ . The $70\%$ point is the $0.70\times90=63$ rd value. The cumulative frequency is $42$ at $x=20$ , so interpolate $21$ values into the third class: $20+\dfrac{63-42}{36}\times12=27$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2021-11	2H	Q8	1	Allowed	Higher	QP MS
2024-06	2H	Q11	7	Allowed	Higher	QP MS
2022-06	3H	Q5	2	Allowed	Higher	QP MS
2022-06	3H	Q17	5	Allowed	Higher	QP MS
2019-06	2H	Q11	6	Allowed	Higher	QP MS
2019-06	3H	Q21	4	Allowed	Higher	QP MS
2024-11	1H	Q9	6	Non-calculator	Higher	QP MS
2019-06	3H	Q3	3	Allowed	Higher	QP MS
2021-11	2H	Q17	4	Allowed	Higher	QP MS
2022-11	3H	Q13	2	Allowed	Higher	QP MS
2022-11	1H	Q14	3	Non-calculator	Higher	QP MS
2022-06	1H	Q10	6	Non-calculator	Higher	QP MS
2024-06	1H	Q13	5	Non-calculator	Higher	QP MS
2023-11	2H	Q15	3	Allowed	Higher	QP MS
2019-11	2H	Q18	3	Allowed	Higher	QP MS
2019-11	1H	Q10	6	Non-calculator	Higher	QP MS
2024-11	3H	Q1	2	Allowed	Higher	QP MS
2021-11	1H	Q12	3	Non-calculator	Higher	QP MS
2022-11	2H	Q11	3	Allowed	Higher	QP MS
2019-11	2H	Q1	2	Allowed	Higher	QP MS
2024-11	2H	Q16	4	Allowed	Higher	QP MS

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