Cumulative Frequency · Study Notes · Edexcel IGCSE 4MA1 Higher

Cumulative frequency curves, quartiles, IQR, box plots. The S-curve and box-plot comparisons are tested every series.

At a glance

Cumulative frequency = running total.
Plot at UPPER BOUND of each class.
Curve is smooth S-shape (ogive).
Median at $n/2$ .
$Q_1$ at $n/4$ , $Q_3$ at $3n/4$ .
IQR = $Q_3 - Q_1$ .
Box plot = 5-number summary.
IQR > range? Never. IQR is part of the range.

What you’ll learn

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

6.4 — Construct a cumulative frequency table from grouped data.
6.4 — Plot and interpret a cumulative frequency curve.
6.4 — Find the median and quartiles from a cumulative frequency curve.
6.4 — Calculate and interpret the interquartile range.
6.4 — Construct and compare box plots.

Cumulative frequency tables

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Running total. Plot at the UPPER BOUND.

Cumulative frequency is a running total — the count of values up to and including a given point.

Worked example. Times: $0$ – $10$ (5), $10$ – $20$ (12), $20$ – $30$ (8), $30$ – $40$ (5). Total $n = 30$ .

Cumulative table:

Time $\le t$	Cumulative freq
$\le 10$	5
$\le 20$	17
$\le 30$	25
$\le 40$	30

KEY RULE. Plot cumulative frequency at the UPPER BOUND of each class — NOT the midpoint, NOT the lower bound.

Why? Because "cumulative frequency at $t = 20$ " means "how many had time UP TO 20". The class $10$ – $20$ contributes its full 12 to that count.

Edexcel tip. Always start the curve at $(0, 0)$ if there are no values below the first class — otherwise the curve doesn't pass through the origin.

Running total.
Plot at upper bound.
Start curve at zero where appropriate.
Final cumulative = total $n$ .

The cumulative frequency curve

▼

Smooth S-shape (ogive). Read off median and quartiles.

The curve. Plot the cumulative frequencies at the upper bounds, then connect with a SMOOTH CURVE (not straight lines). The shape is an "S" or "ogive".

Median from the curve. Median position is $n/2$ .

Worked example. $n = 60$ . Median position = $30$ .

Find $30$ on the y-axis.
Read across to the curve.
Read down to the x-axis.
That x-value is the median.

Read across from n/2 to the curve, then down to the axis to find the median.

Quartiles from the curve.

Quartile	Position
$Q_1$ (lower quartile)	$n/4$
Median ( $Q_2$ )	$n/2$
$Q_3$ (upper quartile)	$3n/4$

Worked example. $n = 60$ . $Q_1$ at $y = 15$ ; median at $y = 30$ ; $Q_3$ at $y = 45$ .

Interquartile range (IQR). $\text{IQR} = Q_3 - Q_1$ . Measures the spread of the middle 50% — robust to outliers.

Worked example. $Q_1 = 18$ , $Q_3 = 32$ . IQR $= 14$ .

Edexcel tip. Mark schemes give ±1 unit tolerance for graph readings (because precision depends on graph scale). Always show your construction lines on the graph — examiners credit the method.

Connect points with a smooth curve.
Median at $y = n/2$ .
$Q_1$ at $y = n/4$ ; $Q_3$ at $y = 3n/4$ .
IQR = $Q_3 - Q_1$ .
Show construction lines.

See the full worked example for cumulative frequency →

Box plots

▼

Five-number summary as a picture.

A box plot displays five numbers:

Minimum
Lower quartile $Q_1$
Median
Upper quartile $Q_3$
Maximum

Construction.

Box from $Q_1$ to $Q_3$ with a line at the median.
Whiskers from min to $Q_1$ and from $Q_3$ to max.

Worked example. $\min = 5$ , $Q_1 = 12$ , median $= 18$ , $Q_3 = 24$ , $\max = 40$ .

Box: 12 to 24.
Median line at 18.
Left whisker: 5 to 12.
Right whisker: 24 to 40.

The box spans the quartiles with the median inside; whiskers reach the minimum and maximum.

Skewness. Compare the two whisker lengths and the position of the median in the box.

Right whisker much longer → POSITIVE skew (long tail to the right).
Left whisker much longer → NEGATIVE skew.
Symmetric: median in middle of box, whiskers similar length.

Comparing box plots. Edexcel asks "compare the two distributions". TWO statements needed:

One about CENTRE (compare medians).
One about SPREAD (compare IQRs).

Worked example. Group A: median 12, IQR 6. Group B: median 18, IQR 4.

"Group B has a higher median (18 vs 12), so on average Group B values are larger."
"Group A has a larger IQR (6 vs 4), so Group A values are more spread out."

Edexcel tip. Always state comparisons IN CONTEXT — say "students" or "marks", not just numbers.

Five-number summary.
Box: $Q_1$ to $Q_3$ with median.
Whiskers: min to $Q_1$ , $Q_3$ to max.
Compare medians AND IQRs in context.

See the full worked example for cumulative frequency →

Quick recap

Cumulative frequency = running total. Plot at UPPER BOUND.
Curve is a smooth S (ogive).
Median at $n/2$ . $Q_1$ at $n/4$ . $Q_3$ at $3n/4$ .
IQR = $Q_3 - Q_1$ (middle 50% spread).
Box plot: 5-number summary.
Compare TWO things: medians + IQRs.
Always in context.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Plot at UPPER BOUND.
Median at $y = n/2$ .
$Q_1$ at $y = n/4$ ; $Q_3$ at $y = 3n/4$ .
IQR = $Q_3 - Q_1$ .
Compare: median (centre) + IQR (spread), in context.

How it’s examined

Cumulative frequency + box plots is a 6-8 mark question on Paper 2H every series. Compare-two-distributions is the most common follow-up. Examiner reports flag: (1) plotting at midpoints, (2) drawing straight lines instead of a curve, (3) comparison statements without context.

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 cumulative frequency, written exactly the way a tutor would explain them at the board.

1Build a cumulative frequency table
Foundation• cumulative frequency
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Question
Times: $0 \le t < 10$ (5), $10 \le t < 20$ (12), $20 \le t < 30$ (8), $30 \le t < 40$ (5). Construct cumulative frequency.
Step-by-step solution
1. Step 1
  Cumulative = running total of frequencies.
2. Step 2
  $5, 5+12=17, 17+8=25, 25+5=30$ .
Answer
$5, 17, 25, 30$
Examiner tip
Plot cumulative frequency at the UPPER BOUND of each class.
2Median from cumulative frequency curve
Higher• Adapted from 4MA1/2H May/Jun 2024 Q17• median
▼
Question
60 students. Read the median from the curve.
Step-by-step solution
1. Step 1
  Median position: $n/2 = 30$ on the y-axis.
2. Step 2
  Read across to the curve, then down to the x-axis. The x-value is the median.
Answer
Median = (whatever the x-value is at $y = 30$ ).
Examiner tip
Use $n/2$ for cumulative frequency curves (not $(n+1)/2$ ). Mark schemes accept ±1 unit tolerance for graph readings.
3Interquartile range
Higher• IQR
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Question
$n = 60$ . Lower quartile from $y = 15$ , upper quartile from $y = 45$ . If $Q_1 = 18$ and $Q_3 = 32$ , find IQR.
Step-by-step solution
1. Step 1
  IQR = $Q_3 - Q_1$ .
2. Step 2
  $32 - 18 = 14$ .
Answer
IQR = $14$
4Box plot from a five-number summary
Higher• Adapted from 4MA1/1H Jan 2024 Q21• box plot
▼
Question
Min = $5$ , $Q_1 = 12$ , median = $18$ , $Q_3 = 24$ , max = $40$ . Describe the box plot.
Step-by-step solution
1. Step 1
  Whiskers from 5 to 12 and 24 to 40. Box from 12 to 24, median at 18.
2. Step 2
  Note: the right whisker is much longer — distribution skewed right (positive skew).
Answer
Box: 12–24 with median at 18. Left whisker: 5–12. Right whisker: 24–40.
Examiner tip
Box plot summarises 5 numbers: min, $Q_1$ , median, $Q_3$ , max. Compare two box plots: comment on median (centre) AND IQR (spread).
5Where to read the upper quartile [1 mark]
Getting started• Paper 1H/2H style — 1 mark• cumulative frequency, quartiles
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Question
A cumulative frequency graph shows information about the masses of $240$ parcels.

Write down the cumulative frequency from which you would read across to estimate the upper quartile. [1]
Step-by-step solution
1. Step 1
  The upper quartile has three quarters of the data at or below it. For a cumulative frequency graph, use three quarters of the total frequency.
2. Step 2
  B1 Three quarters of $240$ :
  $\tfrac{3}{4} \times 240 = 180$
Answer
$180$
Examiner tip
Examiner tip: On a cumulative frequency graph use $\tfrac{n}{4}$ , $\tfrac{n}{2}$ and $\tfrac{3n}{4}$ for the lower quartile, median and upper quartile. Here those are $60$ , $120$ and $180$ .

Common mistake: Giving $60$ , which is the position for the lower quartile, or reading across from $\tfrac{3}{4}$ of the way along the horizontal axis.
6Number between two class boundaries [2 marks]
Getting started• Paper 1H/2H style — 2 marks• cumulative frequency, reading tables
▼
Question
The cumulative frequency table gives information about the heights, $h$ cm, of $80$ students.

$140 < h \leqslant 150$ : $11$

$140 < h \leqslant 160$ : $29$

$140 < h \leqslant 170$ : $58$

$140 < h \leqslant 180$ : $74$

$140 < h \leqslant 190$ : $80$

Work out the number of students with a height greater than $150$ cm but no more than $170$ cm. [2]
Step-by-step solution
1. Step 1
  M1 $58$ students are $170$ cm or less, and $11$ of them are $150$ cm or less. Subtract to leave those in between.
  $58 - 11$
2. Step 2
  A1
  $58 - 11 = 47$
Answer
$47$
Examiner tip
Examiner tip: A number between two values is always a difference of two cumulative frequencies. Here both values are class boundaries, so the table gives the answer exactly; on a curve you would read at both values.

Common mistake: Adding the entries $11 + 29 + 58$ , treating cumulative frequencies as if they were frequencies.
7Testing a claim about half of the group [2 marks]
Getting started• Paper 1H/2H style — 2 marks• cumulative frequency, median
▼
Question
A cumulative frequency graph shows the times taken by $120$ runners to finish a race. The graph shows that $58$ runners took $30$ minutes or less.

Ben says, "More than half of the runners finished in $30$ minutes or less."

Is Ben correct? You must give a reason for your answer. [2]
Step-by-step solution
1. Step 1
  M1 Half of the runners is
  $\tfrac{1}{2} \times 120 = 60$
2. Step 2
  A1 Only $58$ runners took $30$ minutes or less, and $58 < 60$ . So Ben is not correct. (This also means the median time is more than $30$ minutes.)
Answer
No, because $58$ is less than $60$ , which is half of $120$ .
Examiner tip
Examiner tip: "You must give a reason" means the numbers must appear in your answer. Compare the reading with half the total, then state the decision.

Common mistake: Saying "yes" because $58$ is close to $60$ , or writing "No" with no working, which scores nothing.
8Percentage above a reading on the curve [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• cumulative frequency, percentages
▼
Question
The speeds of $250$ cars on a road were recorded and a cumulative frequency graph was drawn. Reading from the curve, the cumulative frequency at a speed of $70$ km/h is $190$ .

Estimate the percentage of the cars that were travelling faster than $70$ km/h. [3]
Step-by-step solution
1. Step 1
  M1 The curve counts cars at $70$ km/h or less, so the cars travelling faster are
  $250 - 190 = 60$
2. Step 2
  M1 Write this as a percentage of all $250$ cars.
  $\frac{60}{250} \times 100$
3. Step 3
  A1
  $\frac{60}{250} \times 100 = 24\%$
Answer
$24\%$
Examiner tip
Examiner tip: "Faster than" or "more than" means subtract the reading from the total before working out the percentage.

Common mistake: Giving $\frac{190}{250} \times 100 = 76\%$ , which is the percentage at or below $70$ km/h.
9Comparing two groups of different sizes from readings [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• cumulative frequency, comparing distributions
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Question
Cumulative frequency graphs are drawn for the times, in seconds, that two groups took to type a paragraph. Group A has $80$ people and Group B has $120$ people. These readings are taken from the two curves.

Group A:

cumulative frequency $20$ : $34$ seconds

cumulative frequency $30$ : $38$ seconds

cumulative frequency $40$ : $41$ seconds

cumulative frequency $60$ : $50$ seconds

Group B:

cumulative frequency $30$ : $38$ seconds

cumulative frequency $40$ : $40$ seconds

cumulative frequency $60$ : $43$ seconds

cumulative frequency $90$ : $47$ seconds

Compare the distributions of the times for the two groups. [4]
Step-by-step solution
1. Step 1
  B1 Group A ( $n = 80$ ): median at $40$ is $41$ s; quartiles at $20$ and $60$ are $34$ s and $50$ s, so the interquartile range is $50 - 34 = 16$ s.
2. Step 2
  B1 Group B ( $n = 120$ ): median at $60$ is $43$ s; quartiles at $30$ and $90$ are $38$ s and $47$ s, so the interquartile range is $47 - 38 = 9$ s.
3. Step 3
  B1 Group A has the lower median ( $41$ s against $43$ s), so on average Group A typed the paragraph faster.
4. Step 4
  B1 Group B has the smaller interquartile range ( $9$ s against $16$ s), so Group B's times were more consistent.
Answer
A: median $41$ s, IQR $16$ s. B: median $43$ s, IQR $9$ s. Group A was faster on average; Group B's times were more consistent.
Examiner tip
Examiner tip: The two groups have different totals, so the reading positions are different: $20$ , $40$ , $60$ for Group A but $30$ , $60$ , $90$ for Group B. Work these out before choosing readings.

Common mistake: Using a cumulative frequency of $40$ for both medians, which gives $40$ s for Group B, the wrong value. A lower median for times means faster, not slower; say what the numbers mean in context.
10Combining two cumulative frequency tables [5 marks]
Stretch• Paper 1H/2H style — 5 marks• cumulative frequency, median
▼
Question
The cumulative frequency tables give information about the test scores, $s$ , of the $50$ students in Class X and the $70$ students in Class Y.

Class X:

$0 < s \leqslant 20$ : $4$

$0 < s \leqslant 40$ : $15$

$0 < s \leqslant 60$ : $33$

$0 < s \leqslant 80$ : $46$

$0 < s \leqslant 100$ : $50$

Class Y:

$0 < s \leqslant 20$ : $3$

$0 < s \leqslant 40$ : $17$

$0 < s \leqslant 60$ : $41$

$0 < s \leqslant 80$ : $62$

$0 < s \leqslant 100$ : $70$

(a) Work out the cumulative frequencies for all $120$ students together. [2]

(b) Write down the class interval that contains the median score of all $120$ students. [1]

(c) Work out the number of the $120$ students who scored more than $60$ . [2]
Step-by-step solution
1. Step 1
  M1 (a) At each upper boundary, the number of students up to that score is the Class X count plus the Class Y count.
  $4 + 3,\quad 15 + 17,\quad 33 + 41,\quad 46 + 62,\quad 50 + 70$
2. Step 2
  A1
  $7,\ 32,\ 74,\ 108,\ 120$
3. Step 3
  B1 (b) The median is read at $\tfrac{1}{2} \times 120 = 60$ . The combined cumulative frequency passes $60$ between $32$ (at $40$ ) and $74$ (at $60$ ), so the median is in $40 < s \leqslant 60$ .
4. Step 4
  M1 (c) $74$ students scored $60$ or less, so those scoring more than $60$ are
  $120 - 74$
5. Step 5
  A1
  $120 - 74 = 46$
Answer
(a) $7$ , $32$ , $74$ , $108$ , $120$ (b) $40 < s \leqslant 60$ (c) $46$
Examiner tip
Examiner tip: Cumulative frequencies for the same class boundaries can be added across groups, because each one counts students up to the same score. You cannot combine two medians or two interquartile ranges this way.

Common mistake: Averaging the two classes' medians, or finding the median class of each class separately. The median of all $120$ students must come from the combined running totals.

Question 1
Paper 1H/2H style — 1 mark1 mark
Nadia is drawing a cumulative frequency graph. In her table, the class $20 < m \leqslant 30$ has a cumulative frequency of $47$ .

Explain why she should plot the point $(30, 47)$ and not $(25, 47)$ . [1]
Model answer
$47$ is the number of values that are $30$ or less. We only know that all $47$ values are at most $30$ ; some of them are between $25$ and $30$ , so $47$ values are not all $25$ or less.
Why this scores
Mark scheme:
- B1 for an explanation that the cumulative frequency counts all values up to the upper class boundary ( $30$ ), not the midpoint
Why this scores full marks: The candidate says what the $47$ counts and why that fixes the point at the end of the class. "Because you always use the upper bound" states the rule without explaining it and would not score.
Question 2
Paper 1H/2H style — 2 marks2 marks
The cumulative frequency table gives information about the masses, $m$ grams, of $64$ tomatoes.
- $50 < m \leqslant 60$ : $5$
- $50 < m \leqslant 70$ : $18$
- $50 < m \leqslant 80$ : $40$
- $50 < m \leqslant 90$ : $56$
- $50 < m \leqslant 100$ : $64$
What fraction of the tomatoes had a mass of more than $80$ grams? Give your answer in its simplest form. [2]
Model answer
$64 - 40 = 24$ tomatoes had a mass of more than $80$ g.

$\frac{24}{64} = \frac{3}{8}$

Answer: $\dfrac{3}{8}$
Why this scores
Mark scheme:
- M1 for $64 - 40$ ( $= 24$ ) or $\frac{24}{64}$ oe
- A1 for $\frac{3}{8}$ cao
Why this scores full marks: The candidate subtracts the cumulative frequency at $80$ g from the total, then simplifies fully. Leaving $\frac{24}{64}$ loses the A mark when the simplest form is asked for, and $\frac{40}{64}$ answers the opposite question.
Question 3
Paper 1H/2H style — 2 marks2 marks
A cumulative frequency graph shows the ages of the $300$ members of a sports club. These readings are taken from the curve.
- age $20$ : cumulative frequency $36$
- age $25$ : cumulative frequency $75$
- age $30$ : cumulative frequency $132$
- age $35$ : cumulative frequency $190$
- age $40$ : cumulative frequency $225$
- age $45$ : cumulative frequency $262$
Find an estimate for the interquartile range of the ages. [2]
Model answer
$\tfrac{1}{4} \times 300 = 75$ , which is at age $25$ .

$\tfrac{3}{4} \times 300 = 225$ , which is at age $40$ .

$40 - 25 = 15$

Answer: $15$ years
Why this scores
Mark scheme:
- M1 for using cumulative frequencies $75$ and $225$ to give ages $25$ and $40$
- A1 for $15$ cao
Why this scores full marks: The candidate works out the two reading positions first and then finds them in the list. The median reading ( $150$ ) is not needed, and neither are the other rows.
Question 4
Paper 1H/2H style — 3 marks3 marks
A cumulative frequency graph shows the masses of $160$ apples. Reading from the curve, $28$ apples have a mass of $120$ g or less, and $104$ apples have a mass of $150$ g or less.

(a) Estimate the number of apples with a mass of more than $120$ g but no more than $150$ g. [2]

(b) One of the $160$ apples is picked at random. Estimate the probability that its mass is more than $120$ g but no more than $150$ g. [1]
Model answer
(a) $104 - 28 = 76$

Answer: $76$

(b) $\frac{76}{160} = \frac{19}{40}$

Answer: $\dfrac{19}{40}$
Why this scores
Mark scheme:
- M1 for $104 - 28$
- A1 for $76$ cao
- B1 for $\frac{76}{160}$ oe, ft their answer to (a)
Why this scores full marks: The difference of two readings gives the number between the two masses, and the probability is that number out of all $160$ apples. The probability is written as a fraction, not as a ratio such as $76 : 160$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
Leon draws a cumulative frequency graph. To estimate the upper quartile, he reads across from a cumulative frequency of $105$ .

(a) Work out the number of values in Leon's data. [1]

(b) Write down the cumulative frequency he should use to estimate the median. [1]

(c) Leon's estimate of the lower quartile is $14$ minutes. Estimate the number of values greater than $14$ minutes. [1]
Model answer
(a) $\tfrac{3}{4}n = 105$ , so $n = 105 \div 3 \times 4 = 140$

Answer: $140$

(b) $\tfrac{1}{2} \times 140 = 70$

Answer: $70$

(c) The lower quartile is read at $\tfrac{1}{4} \times 140 = 35$ , so about $35$ values are $14$ minutes or less.

$140 - 35 = 105$

Answer: $105$
Why this scores
Mark scheme:
- B1 for $140$
- B1 for $70$ ft their $n$
- B1 for $105$ ft their $n$
Why this scores full marks: The candidate reverses the quartile rule: $105$ is three quarters of the total, so the total is $\tfrac{4}{3}$ of $105$ . Dividing $105$ by $3$ and stopping, or multiplying by $0.75$ , are the usual slips. In (c) no graph is needed: a quarter of the values lie at or below the lower quartile, so three quarters lie above it.
Question 6
Paper 1H/2H style — 4 marks4 marks
Two machines, A and B, fill bottles that should contain $500$ ml of juice. A sample of bottles from each machine was measured and cumulative frequency graphs were drawn. The estimates from the graphs are:
- Machine A: lower quartile $497$ ml, median $501$ ml, upper quartile $506$ ml
- Machine B: lower quartile $492$ ml, median $499$ ml, upper quartile $511$ ml
The factory wants the volumes to be close to $500$ ml with as little variation as possible. Which machine should the factory use? Give reasons for your answer. [4]
Model answer
Machine A: $\text{IQR} = 506 - 497 = 9$ ml

Machine B: $\text{IQR} = 511 - 492 = 19$ ml

Both medians are within $1$ ml of $500$ ml ( $501$ and $499$ ), so on average both machines fill close to the target.

Machine A has the smaller interquartile range, so its volumes vary less.

Answer: Machine A, because its median is as close to $500$ ml as B's and its volumes are more consistent.
Why this scores
Mark scheme:
- B1 for the interquartile range of A, $9$
- B1 for the interquartile range of B, $19$
- B1 for a correct comment about the medians, e.g. both are within $1$ ml of $500$ ml
- B1 for Machine A with a reason based on its smaller interquartile range (more consistent volumes)
Why this scores full marks: The candidate uses both conditions in the question: the medians show both machines are on target, and the interquartile ranges decide between them. Choosing B "because its median is lower" ignores the variation, which is what separates the machines.
Question 7
Paper 1H/2H style — 4 marks4 marks
A cumulative frequency graph shows the times taken for $300$ downloads of an app. Reading from the curve, $249$ downloads took $8$ seconds or less.

The company claims, "Fewer than $15\%$ of downloads take longer than $8$ seconds."

(a) Is the company's claim correct? Show your working. [3]

(b) Explain why your answer to (a) is based on an estimate. [1]
Model answer
(a) $300 - 249 = 51$ downloads took longer than $8$ seconds.

$\frac{51}{300} \times 100 = 17\%$

$17\% > 15\%$ , so the claim is not correct.

(b) The $249$ was read from a curve drawn from grouped data, so the exact number of downloads taking $8$ seconds or less is not known.
Why this scores
Mark scheme:
- M1 for $300 - 249$ ( $= 51$ )
- M1 for $\frac{51}{300} \times 100$ oe, or $0.15 \times 300$ ( $= 45$ ) to compare with $51$
- A1 for $17\%$ (or $51 > 45$ ) and a correct conclusion that the claim is not correct
- B1 for a correct explanation, e.g. the reading comes from a curve drawn from grouped data
Why this scores full marks: The candidate compares like with like, percentage with percentage, and states a conclusion. Working out $\frac{249}{300} = 83\%$ and stopping answers a different question.
Question 8
Paper 1H/2H style — 5 marks5 marks
The table gives information about the ages, $a$ years, of $100$ people on a train.
- $0 < a \leqslant 20$ : frequency $16$
- $20 < a \leqslant 40$ : frequency $34$
- $40 < a \leqslant 60$ : frequency $28$
- $60 < a \leqslant 80$ : frequency $22$
(a) Complete the cumulative frequency table. [1]

(b) Write down the coordinates of the points you would plot to draw the cumulative frequency graph. [1]

(c) A cumulative frequency curve is drawn through the points. Reading from the curve, a cumulative frequency of $25$ is at age $26$ , $50$ is at age $40$ and $75$ is at age $58$ . Use these readings to estimate the median and the interquartile range of the ages. [3]
Model answer
(a) $16$ , $50$ , $78$ , $100$

(b) $(0, 0)$ , $(20, 16)$ , $(40, 50)$ , $(60, 78)$ , $(80, 100)$

(c) Median: $\tfrac{1}{2} \times 100 = 50$ , at age $40$

Quartiles: $\tfrac{1}{4} \times 100 = 25$ gives $26$ ; $\tfrac{3}{4} \times 100 = 75$ gives $58$

$58 - 26 = 32$

Answer: median $40$ years, interquartile range $32$ years
Why this scores
Mark scheme:
- B1 for $16$ , $50$ , $78$ , $100$ cao
- B1 for the points at the upper class boundaries, ft their table
- B1 for a median of $40$
- M1 for using the readings at $25$ and $75$
- A1 for $32$ cao
Why this scores full marks: Every point is at an upper class boundary, including the starting point $(0, 0)$ , and each reading is taken from the correct fraction of $100$ . Plotting at $10$ , $30$ , $50$ , $70$ is the classic slip in (b).
Question 9
Paper 1H/2H style — 5 marks5 marks
A factory tests a sample of $120$ light bulbs and draws a cumulative frequency graph of their lifetimes. Reading from the curve, $87$ bulbs in the sample lasted $900$ hours or less.

(a) The factory makes $5000$ bulbs. Estimate how many of the $5000$ bulbs will last more than $900$ hours. [4]

(b) State one assumption you have made. [1]
Model answer
(a) $120 - 87 = 33$ bulbs in the sample lasted more than $900$ hours.

$\frac{33}{120} = 0.275$

$0.275 \times 5000 = 1375$

Answer: $1375$

(b) The sample of $120$ bulbs is representative of all $5000$ bulbs.
Why this scores
Mark scheme:
- M1 for $120 - 87$ ( $= 33$ )
- M1 for $\frac{33}{120}$ oe
- M1 for their proportion $\times 5000$
- A1 for $1375$ cao
- B1 for a correct assumption, e.g. the sample is representative of all the bulbs
Why this scores full marks: The candidate turns the sample reading into a proportion and scales it up, without rounding $0.275$ first. Using $\frac{87}{120}$ gives $3625$ , the number lasting $900$ hours or less, which is the opposite of what was asked.
Question 10
Paper 1H/2H style — 5 marks5 marks
A cumulative frequency graph shows the journey times, $t$ minutes, of $200$ commuters. The points on the graph are $(0, 0)$ , $(10, 14)$ , $(20, 52)$ , $(30, 118)$ , $(40, 170)$ and $(50, 200)$ .

(a) Work out the frequency for each class $0 < t \leqslant 10$ , $10 < t \leqslant 20$ , $20 < t \leqslant 30$ , $30 < t \leqslant 40$ and $40 < t \leqslant 50$ . [2]

(b) Write down the class interval that contains the upper quartile. [1]

(c) One of the commuters is chosen at random. Find the probability that their journey took more than $40$ minutes. [2]
Model answer
(a) $14$ , $52 - 14 = 38$ , $118 - 52 = 66$ , $170 - 118 = 52$ , $200 - 170 = 30$

Answer: $14$ , $38$ , $66$ , $52$ , $30$

(b) $\tfrac{3}{4} \times 200 = 150$ , which is between $118$ and $170$ .

Answer: $30 < t \leqslant 40$

(c) $\frac{30}{200} = \frac{3}{20}$

Answer: $\dfrac{3}{20}$
Why this scores
Mark scheme:
- M1 for subtracting consecutive cumulative frequencies, at least two correct
- A1 for $14$ , $38$ , $66$ , $52$ , $30$ cao
- B1 for $30 < t \leqslant 40$
- M1 for $200 - 170$ or $\frac{30}{200}$ oe
- A1 for $\frac{3}{20}$ oe (e.g. $0.15$ )
Why this scores full marks: The candidate works back from the running totals, uses $\tfrac{3}{4}$ of $200$ to place the upper quartile, and gives the probability as a fraction of all $200$ commuters. Giving the class with the greatest frequency ( $20 < t \leqslant 30$ ) in (b) confuses the upper quartile with the mode.

4MA1 Cumulative Frequency

At a glance

What you’ll learn

Cumulative frequency tables

The cumulative frequency curve

Box plots

Quick recap

Memorise this

How it’s examined

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

1Build a cumulative frequency table

2Median from cumulative frequency curve

3Interquartile range

4Box plot from a five-number summary

5Where to read the upper quartile [1 mark]

6Number between two class boundaries [2 marks]

7Testing a claim about half of the group [2 marks]

8Percentage above a reading on the curve [3 marks]

9Comparing two groups of different sizes from readings [4 marks]

10Combining two cumulative frequency tables [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