Statistical diagrams, histograms & cumulative frequency – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

The pictogram shows the number of cups of hot chocolate a café sold on Monday, Tuesday and Wednesday.

(a) How many cups were sold on Monday? [1]

(b) How many more cups were sold on Tuesday than on Wednesday? [1]

The café sold 30 cups on Thursday.

(d) Work out the total number of cups sold from Monday to Thursday. [1]

(a) 24 caoB1
(b) 6 cao (Tuesday 20, Wednesday 14)B1
(c) 3 full circles and a three-quarter circle (3¾ symbols); accept a correct drawing described in wordsB1
(d) 88 cao (24 + 20 + 14 + 30); ft their (a) and Tuesday/Wednesday readings if clearly shownB1
Answer: (a) 24 (b) 6 (c) 3¾ circles (d) 88
Examiner tip: Work out what a part-symbol is worth from the key first: a quarter of a circle is 2 cups, so three-quarters is 6, not 3.

2Grade 1–3FoundationCalculator4 marks

80 students on a school trip each chose one activity: kayaking, climbing or archery.

The two-way table gives some information about their choices.

	Kayaking	Climbing	Archery	Total
Year 10	17		12	44
Year 11		9
Total	30		26	80

(a) Complete the two-way table. [3]

(b) What fraction of the students who chose kayaking were in Year 11? [1]

(a) at least 1 correct missing entry (Year 10 climbing 15, Year 11 kayaking 13, Year 11 archery 14, Year 11 total 36, climbing total 24)B1
(a) at least 3 of the 5 missing entries correctB1
(a) all 5 missing entries correct: 15, 13, 14, 36, 24 (B3 fully correct, B2 for 3 or 4 correct, B1 for 1 or 2 correct)B1
(b) $\frac{13}{30}$ oe; ft their Year 11 kayaking entry over 30; do not accept 13 : 30 or 13 out of 30 written as a ratioB1
Answer: (a) Year 10: climbing 15. Year 11: kayaking 13, climbing 9, archery 14, total 36. Climbing total 24. (b) $\frac{13}{30}$
Examiner tip: Check every row and every column adds to its total before moving on; one wrong entry usually spoils two others.

3Grade 4–5FoundationCalculator5 marks

Ninety people were asked how they usually travel to work. The table shows the results.

Method	Frequency	Angle
Bus	27
Car	21
Train	15
Bicycle	9
Walk	18

(a) Complete the table to show the angle needed for each sector of a pie chart. [3]

Ella says, “More than a quarter of these people travel to work by bus.”

(b) Is Ella correct? You must show your working. [2]

(a) $360 \div 90 = 4$ (degrees per person) or e.g. $\frac{27}{90}\times 360$ seenM1
(a) at least two correct angles from 108°, 84°, 60°, 36°, 72°A1
(a) all five angles correct: Bus 108°, Car 84°, Train 60°, Bicycle 36°, Walk 72°A1
(b) a correct comparison method, e.g. $\frac{27}{90} = 0.3$ (30%) compared with 25%, or $\frac{1}{4}\times 90 = 22.5$ compared with 27, or 108° compared with 90°M1
(b) Yes, with correct supporting figures (e.g. 30% > 25%, or 27 > 22.5, or 108° > 90°)C1
Answer: (a) 108°, 84°, 60°, 36°, 72° (b) Yes: 27 out of 90 is 30%, which is more than 25%
Examiner tip: Divide 360 by the total number of people, not by the number of categories, to get the angle for one person.

Stuck on Statistical diagrams, histograms & cumulative frequency?

Get it sorted in a free 30-minute session

We work through your weakest questions on this topic on Zoom. I reply the same day.

Free 30-min session on this topic Send me 5 questions a day

4Grade 4–5Calculator4 marks

The pie chart gives information about the favourite sport of each student in a group at School A.

27 of the students chose football.

(a) Work out the total number of students in the group. [2]

(b) Work out the number of students who chose tennis. [1]

A pie chart for a group at School B has a bigger sector for swimming than the one shown here.

Ravi says, “This means more students chose swimming at School B than at School A.”

(a) $135 \div 27 = 5$ (degrees per student) or $\frac{360}{135}\times 27$ oeM1
(a) 72 caoA1
(b) 11 cao ($55 \div 5$); ft their (a): $\frac{55}{360}\times$ their totalB1
(c) a pie chart shows proportions (fractions) only, not actual numbers, and the total number of students at School B is not known (or could be smaller); accept equivalent wordingC1
Answer: (a) 72 (b) 11 (c) Pie charts show proportions, not frequencies; School B may have fewer students in total.
Examiner tip: In part (c) you must mention that the totals of the two groups may be different; just saying 'it is not accurate' scores nothing.

5Grade 4–5FoundationCalculator4 marks

Here are the times, in minutes, that 24 students took to travel to school.

12 25 8 31 17 20 5 28 14 22 36 19
10 26 30 15 9 23 18 33 27 11 20 16

Time ($t$ minutes)	Frequency
$0 < t \leqslant 10$
$10 < t \leqslant 20$
$20 < t \leqslant 30$
$30 < t \leqslant 40$

(a) Complete the frequency table. [2]

(b) Write down the modal class. [1]

Kofi says, “The two times of 20 minutes go in the $20 < t \leqslant 30$ class.”

(a) at least two frequencies correct from 4, 10, 7, 3B1
(a) all four correct: 4, 10, 7, 3 (total 24)B1
(b) $10 < t \leqslant 20$; ft their table; do not accept 10 alone (the frequency)B1
(c) $20 < t$ means greater than 20 (20 is not included), so 20 belongs in $10 < t \leqslant 20$ where $t \leqslant 20$ includes 20; accept equivalent explanationC1
Answer: (a) 4, 10, 7, 3 (b) $10 < t \leqslant 20$ (c) 20 is not greater than 20, so it goes in $10 < t \leqslant 20$.
Examiner tip: The modal class is the class interval, not its frequency; write the inequality, not the number 10.

6Grade 6–7HigherCalculator7 marks

The table and the histogram give information about the masses, $m$ kg, of 120 parcels.

Mass ($m$ kg)	Frequency
$0 < m \leqslant 2$
$2 < m \leqslant 5$	36
$5 < m \leqslant 10$
$10 < m \leqslant 20$	22
$20 < m \leqslant 30$

(a) Use the histogram to complete the table. [2]

(b) Use the table to complete the histogram. Give the frequency density of each bar you draw. [2]

(a) at least one of 14, 40, 8 correct (frequency = frequency density × class width)B1
(a) all three correct: 14, 40, 8 (check total 120)B1
(b) bar for $2 < m \leqslant 5$ drawn with frequency density 12B1
(b) bar for $10 < m \leqslant 20$ drawn with frequency density 2.2 (B1 for one bar correct)B1
(c) $\frac{2}{5}\times 40$ (= 16) or $2 \times 8$ for the part of the $5 < m \leqslant 10$ class above 8 kgM1
(c) their 16 + 22 + their 8 (adding the complete classes above 10 kg)M1
(c) 46 caoA1
Answer: (a) 14, 40, 8 (b) bars of height 12 (2 to 5 kg) and 2.2 (10 to 20 kg) (c) 46
Examiner tip: Frequency is the area of the bar, so multiply the frequency density by the class width; reading the bar height as the frequency loses every mark.

7Grade 8–9HigherCalculator5 marks

The histogram shows information about the times, $t$ minutes, that some runners took to complete a race.

The frequency density scale is not labelled.

15 runners took between 20 and 25 minutes ($20 < t \leqslant 25$).

(a) Work out the total number of runners. [3]

(b) Work out an estimate for the number of runners who took between 27 and 35 minutes. [2]

(a) frequency density for $20 < t \leqslant 25$ is $15 \div 5 = 3$, so one large square (vertically) = 1 runner per minute; or 1 small square = 0.4 runners oe (finding the scale)M1
(a) correct frequencies for at least two other classes from 12, 22, 18, 8M1
(a) 75 caoA1
(b) $\frac{3}{5}\times 22$ (= 13.2) or $\frac{5}{10}\times 18$ (= 9); ft their frequenciesM1
(b) 22 (accept 22.2)A1
Answer: (a) 75 (b) 22
Examiner tip: When the vertical scale is missing, use the one class you know to find the frequency density scale (or the value of one square), then work out every other bar from its area.

8Grade 6–7HigherCalculator6 marks

The table gives information about the heights, $h$ cm, of 80 seedlings.

Height ($h$ cm)	Frequency	Cumulative frequency
$0 < h \leqslant 4$	6
$4 < h \leqslant 8$	14
$8 < h \leqslant 12$	25
$12 < h \leqslant 16$	21
$16 < h \leqslant 20$	10
$20 < h \leqslant 24$	4

(a) Complete the cumulative frequency column. [1]

(b) On the grid, draw a cumulative frequency graph for your table. [2]

(d) Use your graph to find an estimate for the number of seedlings taller than 15 cm. [2]

(a) 6, 20, 45, 66, 76, 80 caoB1
(b) at least 4 points plotted correctly at the upper class boundaries (4, 6), (8, 20), (12, 45), (16, 66), (20, 76), (24, 80), ft their (a); points plotted at midpoints score B0B1
(b) all 6 points plotted correctly and joined with a smooth curve or straight line segments; joining to (0, 0) is not required (ignore any part of the graph below $h = 4$)B1
(c) answer in the range 10.8 to 11.6 (reading at cumulative frequency 40); ft their graphB1
(d) reading their graph at $h = 15$ (e.g. a value in the range 59 to 63 seen) or a vertical line drawn at $h = 15$M1
(d) answer in the range 17 to 21 (from 80 − their reading)A1
Answer: (a) 6, 20, 45, 66, 76, 80 (c) about 11.2 cm (d) about 19
Examiner tip: Plot each cumulative frequency at the END of its class (the upper class boundary), not at the midpoint, and remember the question asks for 'taller than', so subtract your reading from 80.

9Grade 8–9HigherCalculator7 marks

The cumulative frequency graph shows information about the times, in minutes, that 160 workers in Town A take to travel to work.

(a) Use the graph to find an estimate for the median time. [1]

(b) Use the graph to find an estimate for the interquartile range. [2]

For workers in Town B, the median time is 28 minutes and the interquartile range is 25 minutes.

(d) Compare the distributions of the times for Town A and Town B. [2]

(a) answer in the range 30 to 32B1
(b) readings at cumulative frequency 40 (about 20.5) and 120 (about 43) seen, or lines drawn on the graph at 40 and 120M1
(b) answer in the range 20 to 25A1
(c) reading at 45 minutes (about 124 to 127) seen, or a line drawn at 45 minutesM1
(c) answer in the range 33 to 36A1
(d) a correct comparison of the averages: e.g. the median for Town A is higher, so on average Town A workers take longer; ft their (a)C1
(d) a correct comparison of the spread: e.g. the IQR for Town B is larger, so Town B times are more varied (less consistent); ft their (b). Statements must compare and be in context; just quoting values scores C0C1
Answer: (a) about 31 minutes (b) about 22 minutes (c) about 35 (d) Town A has the higher median so its workers take longer on average; Town B has the larger IQR so its times are more varied.
Examiner tip: When comparing, give one statement about an average (median) and one about spread (IQR), and say what each means in context.

10Grade 8–9HigherCalculator6 marks

The table gives information about the lengths, $l$ cm, of 200 fish.

Length ($l$ cm)	Frequency
$10 < l \leqslant 20$	28
$20 < l \leqslant 25$	$x$
$25 < l \leqslant 30$	$2x$
$30 < l \leqslant 40$	46
$40 < l \leqslant 60$	24

A histogram is drawn for this information.

(a) Work out the frequency density of the tallest bar. [4]

One of the fish is chosen at random.

(b) Work out an estimate for the probability that its length is more than 35 cm. [2]

(a) $28 + x + 2x + 46 + 24 = 200$ oeM1
(a) $x = 34$A1
(a) a correct method for a frequency density, e.g. $68 \div 5$ or $34 \div 5$ (= 6.8) or $28 \div 10$ (= 2.8); ft their $x$M1
(a) 13.6 cao (from $25 < l \leqslant 30$; all others are 2.8, 6.8, 4.6, 1.2)A1
(b) $\frac{1}{2}\times 46 + 24$ (= 47) oeM1
(b) $\frac{47}{200}$ oe (0.235)A1
Answer: (a) 13.6 (b) $\frac{47}{200}$ (0.235)
Examiner tip: Use the total of 200 to find $x$ first, then divide each frequency by its class width: the tallest bar has the greatest frequency density, so check every class rather than assuming.

Stuck on this topic?

Book a free 30-minute session on WhatsApp and we'll find exactly which step is costing you the marks. Or build a timed paper on it.

Free 30-minute session