GCSE Maths · Statistics

Cumulative Frequency GCSE Questions and Answers

Cumulative frequency is a running total of frequencies up to each class boundary. Plot upper class boundaries against cumulative totals, then use N ÷ 4, N ÷ 2 and 3N ÷ 4 to estimate the quartiles and median from the curve.

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Count everything up to a boundary

What you need to know about cumulative frequency

Suppose 4 journeys took up to 10 minutes and another 8 took between 10 and 20 minutes. Then 4 + 8 = 12 journeys took no more than 20 minutes. This running total is the cumulative frequency: it counts everything up to each boundary.

Running total → curve → estimates

Build the running total before drawing the graph

Start with the first frequency. For every later group, add its frequency to the total you already have.

Up to 104begin with the first frequency
Up to 204 + 8 = 12include both groups
Up to 3012 + 11 = 23keep accumulating
Journey times and cumulative-frequency plotting points
Time t (minutes)FrequencyCumulative frequencyPoint
0 < t ≤ 1044(10, 4)
10 < t ≤ 20812(20, 12)
20 < t ≤ 301123(30, 23)
30 < t ≤ 40932(40, 32)
40 < t ≤ 50840(50, 40)
At the upper boundary 20, the cumulative frequency 12 means 12 journeys lasted no more than 20 minutes. That is why the point is (20, 12), not a midpoint.
A cumulative frequency curve for 50 valuesThe increasing curve begins at zero and ends at cumulative frequency 50. A horizontal guide from 25 meets the curve and drops to an estimated median of about 27.0102030405001020304050median ≈ 27N ÷ 2 = 25valuecumulative frequency
For 50 values, read the median at cumulative frequency 25. Graph readings are estimates, so show guide lines and use suitable accuracy.
0102030405060020406080valuecumulative frequency
For N = 80, the guide levels are 20, 40 and 60. Read across to the curve, then down to estimate Q1, the median and Q3.
From problem to method

Typical problems you need to be able to solve

These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.

Complete cumulative totals

What the problem asks: A grouped frequency table has an empty cumulative-frequency column.

How to solve it: Add each new class frequency to the previous running total. Check the last total equals the sample size.

Recover class frequencies

What the problem asks: Cumulative totals are given but ordinary frequencies are missing.

How to solve it: Subtract consecutive cumulative totals. The first frequency equals the first cumulative total.

Plot a curve

What the problem asks: The question gives class intervals and cumulative totals.

How to solve it: Plot each cumulative total at the upper class boundary, include the lower starting boundary at zero, then join with a smooth increasing curve.

Estimate median and quartiles

What the problem asks: A curve and total frequency N are given.

How to solve it: Read Q1 at N/4, the median at N/2 and Q3 at 3N/4 using horizontal and vertical guide lines.

Estimate a frequency in an interval

What the problem asks: The question asks how many values lie between two boundaries.

How to solve it: Read both cumulative frequencies and subtract the lower cumulative total from the upper one.

A reliable routine

Method for quartiles from a cumulative-frequency curve

This routine applies when a graph asks for the lower quartile, median, upper quartile or interquartile range.

  1. Find the total frequency N from the final cumulative total.
  2. Calculate N ÷ 4, N ÷ 2 or 3N ÷ 4 for the requested position.
  3. Draw horizontally from that cumulative frequency to the curve, then vertically to the value axis.
  4. For IQR, subtract the lower-quartile estimate from the upper-quartile estimate.

Check: A cumulative-frequency curve must never fall. A steep section contains many values; a flatter section contains fewer. Graph readings are estimates because the exact positions of values inside each grouped class are unknown.

Fully worked

Cumulative frequency GCSE worked examples

Each solution explains the clue, why the method fits and how to check the result.

Example 1

Build a cumulative-frequency column

4 marks
Question

Complete the cumulative-frequency column in the table.

Journey times and cumulative-frequency plotting points
Time t (minutes)FrequencyCumulative frequencyPoint
0 < t ≤ 1044(10, 4)
10 < t ≤ 208??
20 < t ≤ 3011??
30 < t ≤ 409??
40 < t ≤ 508??
At the upper boundary 20, the cumulative frequency 12 means 12 journeys lasted no more than 20 minutes. That is why the point is (20, 12), not a midpoint.

Keep a running total.

44

4+8=124+8=12

12+11=2312+11=23

23+9=3223+9=32

32+8=4032+8=40

4, 12, 23, 32, 40\boxed{4,\ 12,\ 23,\ 32,\ 40}

Example 2

Recover ordinary frequencies

3 marks
Question

The cumulative frequencies are 4,11,19,274, 11, 19, 27. Find the four class frequencies.

The first frequency is 44.

114=711-4=7

1911=819-11=8

2719=827-19=8

4, 7, 8, 8\boxed{4,\ 7,\ 8,\ 8}

Example 3

Find quartile positions

3 marks
Question

A cumulative-frequency graph represents 6060 values. State the cumulative frequencies used to read Q1Q_1, the median and Q3Q_3.

Q1: 604=15Q_1:\ \frac{60}{4}=15

Median: 602=30\text{Median}:\ \frac{60}{2}=30

Q3: 3×604=45Q_3:\ \frac{3\times60}{4}=45

15, 30, 45\boxed{15,\ 30,\ 45}

Example 4

Estimate an interquartile range

4 marks
Question

From a cumulative-frequency curve, the lower quartile is estimated as 1818 and the upper quartile as 4343. Estimate the IQR.

IQR=Q3Q1\text{IQR}=Q_3-Q_1

=4318=43-18

25\boxed{25}

Example 5

Estimate a frequency between two values

4 marks
Question

A curve gives cumulative frequency 1818 at 2020 and cumulative frequency 7272 at 4545. Estimate how many values are greater than 2020 and no more than 4545.

The total up to 4545 includes the total up to 2020, so subtract it.

7218=5472-18=54

54 values\boxed{54\text{ values}}

Example 6

Compare two distributions

4 marks
Question

The graph shows cumulative-frequency curves for Groups A and B. At cumulative frequency 4040, A has value 3333 and B has value 2424. The estimated IQRs are 2020 for A and 3131 for B. Compare the distributions.

0102030405060020406080valuecumulative frequencyGroup AGroup B
Compare the horizontal readings at the same cumulative-frequency levels.

Distribution A has the greater median, so its typical value is higher.

33>2433>24

Distribution A also has the smaller IQR because 20<3120<31, so its middle half is less spread out.

A typically has larger values and is more consistent than B.\boxed{\text{A typically has larger values and is more consistent than B.}}

Example 7

Plot a complete cumulative-frequency curve

5 marks
Question

Use the completed table to plot a cumulative-frequency curve. Include the starting point.

Journey times and cumulative-frequency plotting points
Time t (minutes)FrequencyCumulative frequencyPoint
0 < t ≤ 1044(10, 4)
10 < t ≤ 20812(20, 12)
20 < t ≤ 301123(30, 23)
30 < t ≤ 40932(40, 32)
40 < t ≤ 50840(50, 40)
At the upper boundary 20, the cumulative frequency 12 means 12 journeys lasted no more than 20 minutes. That is why the point is (20, 12), not a midpoint.

The lower boundary is 00, so begin at (0,0)(0,0). Plot the cumulative totals at upper class boundaries:

(0,0), (10,4), (20,12), (30,23), (40,32), (50,40)\boxed{(0,0),\ (10,4),\ (20,12),\ (30,23),\ (40,32),\ (50,40)}

Join the points with a smooth increasing curve rather than separate straight bars.

01020304050010203040journey time (minutes)cumulative frequency
The curve begins at the lower boundary with cumulative frequency zero and passes through every upper-boundary total from the table.
Example 8

Convert quartile readings to a box plot

4 marks
Question

A cumulative-frequency curve gives minimum 44, lower quartile 1818, median 2929, upper quartile 4343 and maximum 5858. Draw the corresponding box plot.

Use one value scale. Draw the box from 1818 to 4343, the median line at 2929, and whiskers to 44 and 5858.

Box plot from curve07.51522.53037.54552.560
Read each vertical line against the shared linear scale.
15 original questions · total 43 marks

Cumulative frequency GCSE exam-style questions

Try each question before opening its fully worked answer. The difficulty rises through the set.

Before you startAllow about 50 minutes · show graph guide lines and label estimated answers · answers start collapsed
1

Continue a running total

3 marks

The first four class frequencies are 6,9,4,116, 9, 4, 11. Find the cumulative frequencies.

Show worked answer

6,6+9=15,15+4=19,19+11=306,\quad6+9=15,\quad15+4=19,\quad19+11=30

6, 15, 19, 30\boxed{6,\ 15,\ 19,\ 30}

2

Check the sample size

1 mark

The final cumulative frequency in a table is 8484. How many values are in the data set?

Show worked answer

The final running total counts every value.

84\boxed{84}

3

Find frequencies by subtraction

3 marks

The cumulative totals are 3,10,24,303, 10, 24, 30. Find the ordinary frequencies.

Show worked answer

33

103=710-3=7

2410=1424-10=14

3024=630-24=6

3, 7, 14, 6\boxed{3,\ 7,\ 14,\ 6}

4

Choose plotting coordinates

2 marks

For the class 20<x3020<x\le30, the cumulative frequency is 1717. Which point should be plotted?

Show worked answer

Cumulative frequency is plotted against the upper class boundary.

(30,17)\boxed{(30,17)}

5

Calculate quartile levels

3 marks

A graph represents 9696 values. Find the cumulative-frequency levels for Q1Q_1, the median and Q3Q_3.

Show worked answer

Q1=964=24Q_1=\frac{96}{4}=24

Median=962=48\text{Median}=\frac{96}{2}=48

Q3=3×964=72Q_3=\frac{3\times96}{4}=72

24, 48, 72\boxed{24,\ 48,\ 72}

6

Use two graph readings

2 marks

A graph shows cumulative frequency 2121 at 3030 and 6767 at 5555. Estimate the number of values in 30<x5530<x\le55.

Show worked answer

Subtract the number at or below 3030 from the number at or below 5555.

6721=4667-21=\boxed{46}

7

Estimate a number above a value

2 marks

There are 9090 values. A curve shows cumulative frequency 5858 at 4040. Estimate how many values are greater than 4040.

Show worked answer

9058=3290-58=\boxed{32}

8

Find an IQR from graph estimates

2 marks

The graph estimates Q1=24.5Q_1=24.5 and Q3=61.0Q_3=61.0. Estimate the IQR.

Show worked answer

IQR=61.024.5\text{IQR}=61.0-24.5

36.5\boxed{36.5}

9

Compare two cumulative-frequency curves

4 marks

The graph shows Groups A and B. At cumulative frequency 4040, A has value 3333 and B has value 2424. Their estimated IQRs are 2020 and 3131 respectively. Compare them.

0102030405060020406080valuecumulative frequencyGroup AGroup B
Compare the horizontal readings at the same cumulative-frequency levels.
Show worked answer

Group A has the higher median, so its typical value is greater.

Group A also has the smaller IQR, so its middle half is less spread out.

A typically has higher values and is more consistent.\boxed{\text{A typically has higher values and is more consistent.}}

10

Interpret a complete graph

4 marks

A cumulative-frequency graph for 120120 journey times gives Q1=18Q_1=18 minutes, median 2626 minutes and Q3=41Q_3=41 minutes. Estimate the IQR and the number of journeys lasting no more than the median.

Show worked answer

IQR=4118=23 minutes\text{IQR}=41-18=23\text{ minutes}

The median is at half the total frequency.

1202=60\frac{120}{2}=60

IQR=23 minutes; about 60 journeys\boxed{\text{IQR}=23\text{ minutes; about }60\text{ journeys}}

11

Complete a table and plot the curve

5 marks

Complete the cumulative-frequency column, list the plotting coordinates and draw a smooth curve.

Journey times and cumulative-frequency plotting points
Time t (minutes)FrequencyCumulative frequencyPoint
0 < t ≤ 1044(10, 4)
10 < t ≤ 208??
20 < t ≤ 3011??
30 < t ≤ 409??
40 < t ≤ 508??
At the upper boundary 20, the cumulative frequency 12 means 12 journeys lasted no more than 20 minutes. That is why the point is (20, 12), not a midpoint.
Show worked answer

The cumulative totals are

4, 12, 23, 32, 404,\ 12,\ 23,\ 32,\ 40

Include the starting point at the lower boundary.

(0,0), (10,4), (20,12), (30,23), (40,32), (50,40)\boxed{(0,0),\ (10,4),\ (20,12),\ (30,23),\ (40,32),\ (50,40)}

01020304050010203040journey time (minutes)cumulative frequency
The curve begins at the lower boundary with cumulative frequency zero and passes through every upper-boundary total from the table.
12

Explain the upper-boundary rule

2 marks

Why is cumulative frequency 1212 for 10<t2010<t\le20 plotted at (20,12)(20,12) rather than (15,12)(15,12)?

Show worked answer

The total 1212 counts all values no more than 2020. That statement is known at the upper boundary, while the midpoint 1515 does not represent the end of the class.

(20,12)\boxed{(20,12)}

13

Draw a box plot from graph readings

4 marks

A curve gives minimum 22, Q1=17Q_1=17, median 2828, Q3=44Q_3=44 and maximum 5959. Draw the corresponding box plot.

Show worked answer

Draw the box from 1717 to 4444, place the median at 2828, and extend the whiskers to 22 and 5959.

Box plot from curve07.51522.53037.54552.560
Read each vertical line against the shared linear scale.
14

Compare median and spread

4 marks

Curve R gives median 3535 and IQR 1616. Curve S gives median 3131 and IQR 2424. Compare the distributions.

Show worked answer

R has the higher median, so its typical value is higher.

R has the smaller IQR, so its middle half is less spread out.

R typically has higher values and is more consistent than S.\boxed{\text{R typically has higher values and is more consistent than S.}}

15

Interpret steep and flat sections

2 marks

A cumulative-frequency curve is steep between 2020 and 3030 but almost flat between 5050 and 6060. What does this show?

Show worked answer

The running total rises quickly from 2020 to 3030, so many values lie there. It rises only slightly from 5050 to 6060, so few values lie there.

many values from 20–30; few from 50–60\boxed{\text{many values from 20–30; few from 50–60}}

Examiner-style feedback

Common cumulative frequency mistakes

Plotting class midpoints

Cumulative totals are plotted at upper class boundaries, not at class midpoints.

Missing the zero starting point

Begin at the lower boundary of the first class with cumulative frequency 0 before plotting the upper-boundary totals.

Resetting each total

Every cumulative entry includes all earlier classes. It is a running total, not an ordinary frequency.

Using value-axis quarters

Quartile positions come from the total frequency on the vertical axis, not from quartering the horizontal scale.

Adding for an interval count

To count between two boundaries, subtract the lower cumulative total from the upper cumulative total.

30-second recap

Choose, carry out, check

Translate the wording before you reach for a rule.

  1. Add frequencies to create a running total.
  2. Plot totals at upper class boundaries.
  3. Use N/4, N/2 and 3N/4 for quartiles.
  4. Subtract graph readings for interval frequencies or IQR.
Quick answers

Cumulative frequency FAQ

Why does a cumulative-frequency graph always rise?

The running total can stay the same or increase as more classes are included; it cannot decrease.

Where do I plot each cumulative frequency?

At the upper boundary of its class interval.

How do I find the median?

Use half the total frequency, move across to the curve and then down to the value axis.

Are answers from a cumulative-frequency graph exact?

Usually not. They are estimates based on the drawn curve and should be given to sensible accuracy.

Build connected skills

What to revise next

Display the summary

Box plots

Turn quartile estimates into a compact comparison diagram.

Revise box plots
Content standards

Curriculum and rights review

Curriculum references checked 4 September 2026. Constructing and interpreting cumulative-frequency graphs, estimating quartiles and comparing distributions are shared Higher-tier GCSE Mathematics skills. All questions, diagrams, data and wording are original Pass an Exam material.