Cumulative frequency & box plots (median, IQR) – GCSE Maths exam questions with mark scheme

1Grade 4–5HigherNon-calculator3 marks

Here are the numbers of emails received by 11 people one morning, written in order.

3 5 7 8 10 12 13 15 18 20 24

(a) Write down the median. [1]

(b) Work out the interquartile range. [2]

(a) 12B1
(b) lower quartile 7 and upper quartile 18 identified (3rd and 9th values)M1
(b) 11A1
Answer: (a) 12 (b) 11
Examiner tip: With 11 values the quartiles are the 3rd and 9th values – IQR is upper quartile minus lower quartile, not max minus min.

2Grade 4–5HigherNon-calculator4 marks

The box plot shows information about the journey times, in minutes, of some workers.

(a) Write down the median journey time. [1]

(b) Work out the interquartile range. [1]

(d) What fraction of the workers had a journey time of more than 30 minutes? [1]

(a) 21 (minutes)B1
(b) 15 (minutes)B1
(c) 36 (minutes)B1
(d) $\frac{1}{4}$ (oe, e.g. 25%)B1
Answer: (a) 21 (b) 15 (c) 36 (d) $\frac{1}{4}$
Examiner tip: Each section of a box plot (whisker, half-box, half-box, whisker) holds a quarter of the data.

3Grade 6–7HigherNon-calculator3 marks

The box plots show information about the test scores of two groups of students.

(a) Work out the interquartile range for Group A. [1]

(b) Compare the distributions of the scores of the two groups. [2]

(a) 22B1
(b) correct comparison of the medians, e.g. Group B had a higher median score (56 v 50), so on average Group B scored higher. Accept equivalent wording (e.g. 'Group A did worse on average'). Do not accept a comparison of maximum or minimum values.C1
(b) correct comparison of the IQRs or ranges, e.g. Group B had a smaller IQR (14 v 22) [or range 40 v 56], so Group B's scores were more consistent / less spread out. Accept 'Group A's scores were more varied'. Do not accept 'Group B's scores were more spread out'. For both marks at least one of the two comparisons must be interpreted in context (about the scores); two correct comparisons with neither in context score C1 C0.C1
Answer: (a) 22 (b) Group B scored higher on average (median 56 v 50) and more consistently (IQR 14 v 22)
Examiner tip: One mark is for an average, one for a spread – and each comparison must be about the scores, not just the numbers.

4Grade 6–7HigherCalculator6 marks

The table shows information about the heights, $h$ cm, of 80 plants.

Height ($h$ cm)	Frequency	Cumulative frequency
$0 \lt h \le 10$	6
$10 \lt h \le 20$	14
$20 \lt h \le 30$	28
$30 \lt h \le 40$	20
$40 \lt h \le 50$	12

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

The cumulative frequency graph for this information has been drawn.

(b) Use the graph to find an estimate for the median height. [1]

(d) Use the graph to estimate the number of plants taller than 35 cm. [2]

(a) 6, 20, 48, 68, 80B1
(b) answer in the range 26 to 28 (graph gives 27.1)B1
(c) readings from the graph at cumulative frequency 20 and 60 (lower quartile 20, allow 19.5 to 20.5; upper quartile in range 34.5 to 36.5)M1
(c) answer in the range 14.5 to 16.5 (graph gives 15.5)A1
(d) reading from the graph at 35 cm (in range 57 to 61) and subtracting from 80M1
(d) answer in the range 19 to 23 (graph gives 21)A1
Answer: (a) 6, 20, 48, 68, 80 (b) about 27 cm (c) about 15.5 cm (d) about 21
Examiner tip: Read the quartiles at $\frac{1}{4}$ and $\frac{3}{4}$ of the TOTAL frequency (20th and 60th), and for ‘taller than’ subtract your reading from 80.

5Grade 6–7HigherCalculator5 marks

The cumulative frequency graph shows information about the times taken by 120 runners to complete a race.

(a) Runners who finished in less than 35 minutes received a medal. What percentage of the runners received a medal? [3]

(b) The slowest 10% of runners were given a ‘finisher’ badge. Use the graph to estimate the shortest time taken by a runner who was given a ‘finisher’ badge. [2]

(a) reads 72 from the graph at 35 minutes (allow 71 to 73)M1
(a) $\frac{72}{120} \times 100$ (ft their reading)M1
(a) 60(%)A1
(b) $0.9 \times 120$ (= 108) or 12 runners identified, and reading taken at cumulative frequency 108M1
(b) answer in the range 42 to 43.5 minutes (graph gives 42.4)A1
Answer: (a) 60% (b) about 42.4 minutes (accept 42 to 43.5)
Examiner tip: The slowest 10% are at the TOP of a cumulative frequency graph – read across from 108, not from 12.

6Grade 8–9HigherNon-calculator4 marks

Seven whole numbers are written in order of size.

The smallest number is 5 and the range is 15.
The lower quartile is 7, the median is 11 and the upper quartile is 16.
The mean is 11.

Find all seven numbers. [4]

5 and 20 as the smallest and largest numbersB1
7, 11 and 16 placed as the 2nd, 4th and 6th numbersB1
$7 \times 11 = 77$ and the two missing numbers sum to $77 - 59 = 18$M1
5, 7, 7, 11, 11, 16, 20A1
Answer: 5, 7, 7, 11, 11, 16, 20
Examiner tip: With 7 values the quartiles are the 2nd and 6th values; then use the order (3rd between 7 and 11, 5th between 11 and 16) to pin down the gaps.

7Grade 6–7HigherNon-calculator4 marks

Here are the numbers of minutes that 11 buses were late, written in order.

4 5 5 6 7 7 8 8 9 10 41

(a) Work out the range. [1]

(b) Work out the interquartile range. [2]

(c) Which of the range and the interquartile range is the better measure of spread for this data? Give a reason for your answer. [1]

(a) 37B1
(b) either quartile correctly identified: lower quartile 5 (3rd value) or upper quartile 9 (9th value); or $9-5$ seenM1
(b) 4A1
(c) interquartile range, because it is not affected by the extreme value (outlier) 41. Accept 'the range is distorted by 41' or 'IQR uses only the middle 50%'. Must have a reason; 'IQR' alone scores 0.C1
Answer: (a) 37 (b) 4 (c) IQR – it is not affected by the outlier 41
Examiner tip: The IQR uses only the middle 50% of the data, so one extreme value cannot distort it.

8Grade 1–3Non-calculator3 marks

Ten students each did as many press-ups as they could in one minute. Here are their results.

23 17 31 25 19 28 35 22 26 30

(a) Find the median number of press-ups. [2]

(b) Work out the range. [1]

(a) writes the numbers in order (allow one error), or identifies 25 and 26 as the middle pairM1
(a) 25.5 (oe, e.g. $25\frac{1}{2}$)A1
(b) 18 (35 − 17)B1
Answer: (a) 25.5 (b) 18
Examiner tip: With an even number of values there are two middle numbers – the median is halfway between them, and the list must be put in order first.

9Grade 4–5HigherNon-calculator4 marks

The stem and leaf diagram shows the ages, in years, of 19 people at a climbing centre.

1	6 8 9
2	0 3 4 5 7 7
3	1 2 6 8 9
4	2 4 5
5	0 3

Key: 1 | 6 represents 16 years

(a) On the grid, draw a box plot for this information. [3]

(b) Work out the interquartile range of the ages. [1]

(a) median = 31 (the 10th value)B1
(a) lower quartile = 23 and upper quartile = 42 (the 5th and 15th values)B1
(a) fully correct box plot: box from 23 to 42 with median line at 31, whiskers to 16 and 53 (ft their quartiles and median if both B marks not scored, provided whiskers 16 and 53 are correct)B1
(b) 19 (ft their UQ − LQ)B1
Answer: (a) box plot with minimum 16, LQ 23, median 31, UQ 42, maximum 53 (b) 19 years
Examiner tip: With 19 values the quartiles are the 5th and 15th values – count along the leaves carefully and remember the key turns 1 | 6 into 16.

10Grade 6–7HigherCalculator6 marks

The table shows information about the masses, $m$ grams, of 80 eggs.

Mass ($m$ grams)	Frequency	Cumulative frequency
$ 40 \lt m \le 45$	4
$ 45 \lt m \le 50$	11
$ 50 \lt m \le 55$	23
$ 55 \lt m \le 60$	26
$ 60 \lt m \le 65$	12
$ 65 \lt m \le 70$	4

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

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

(c) Eggs with a mass of more than 62 grams are sold as ‘large’. Use your graph to estimate the number of eggs that are large. [2]

(d) Sam draws his cumulative frequency graph by plotting the points (42.5, 4), (47.5, 15), (52.5, 38) and so on. Explain what Sam has done wrong. [1]

(a) 4, 15, 38, 64, 76, 80B1
(b) at least 5 of the 6 points plotted correctly at the upper class boundaries (45, 4), (50, 15), (55, 38), (60, 64), (65, 76), (70, 80), ft their tableB1
(b) fully correct graph: all points plotted at upper class boundaries and joined with a smooth curve or straight line segments (ignore a point at (40, 0) or the curve starting at (40, 0))B1
(c) reads their graph at $m = 62$ (about 68 or 69) or shows 80 − their readingM1
(c) answer in the range 9 to 14 (ft 80 − their reading if M1 awarded); accept a non-integer value in range, e.g. 11.2A1
(d) he has plotted at the midpoints of the classes; cumulative frequencies must be plotted at the upper class boundaries (e.g. 4 eggs have a mass of up to 45 g, not 42.5 g) – accept ‘should be plotted at the end of each class’C1
Answer: (a) 4, 15, 38, 64, 76, 80 (b) points at (45,4), (50,15), (55,38), (60,64), (65,76), (70,80) joined (c) about 11 (accept 9 to 14) (d) he plotted at midpoints instead of upper class boundaries
Examiner tip: For ‘more than’ questions read the graph at 62 g, then subtract that reading from the total – the reading itself counts the eggs that are NOT large.

11Grade 6–7HigherNon-calculator4 marks

The cumulative frequency graph shows information about the lengths, in minutes, of 60 phone calls made from an office.

The shortest call lasted 1 minute. The longest call lasted 29 minutes.

(a) Use this information and the cumulative frequency graph to draw a box plot on the grid below the graph. [3]

Calls that last more than 20 minutes are charged at a higher rate.

(b) Mia says, ‘Fewer than a quarter of the calls were charged at the higher rate.’ Is Mia correct? Give a reason for your answer. [1]

(a) median read from the graph at cumulative frequency 30: answer in the range 14 to 15B1
(a) lower quartile 10 and upper quartile 20 read at cumulative frequencies 15 and 45 (allow ±0.5)B1
(a) fully correct box plot with whiskers at 1 and 29, box from LQ to UQ and median line drawn (ft their median and quartiles)B1
(b) No, with a reason, e.g. 60 − 45 = 15 calls lasted more than 20 minutes and 15 is exactly $\frac{1}{4}$ of 60 / 20 minutes is the upper quartile so a quarter of calls were longerC1
Answer: (a) box plot: minimum 1, LQ 10, median about 14.4 (accept 14 to 15), UQ 20, maximum 29 (b) No – 15 of the 60 calls, exactly a quarter, were longer than 20 minutes
Examiner tip: Read the quartiles across from $\frac{1}{4}$, $\frac{1}{2}$ and $\frac{3}{4}$ of the total (15, 30, 45) – the whisker ends come from the extra information, not from the graph.

12Grade 6–7HigherCalculator4 marks

A shop tested 100 batteries of Brand P and 100 batteries of Brand Q. The cumulative frequency graphs show information about how long each battery lasted, in hours.

Compare the distributions of the battery lives for Brand P and Brand Q. [4]

reads both medians at cumulative frequency 50: Brand P in the range 21.5 to 23 and Brand Q in the range 25.5 to 26.5M1
reads quartiles at cumulative frequencies 25 and 75 for at least one brand, or works out an interquartile range (P about 6, accept 5 to 7; Q about 8, accept 7.5 to 9)M1
correct comparison of medians in context, e.g. on average Brand Q batteries last longer because its median is higherC1
correct comparison of interquartile ranges in context, e.g. Brand P battery lives are more consistent because its IQR is smaller (dependent on IQRs for both brands found)C1
Answer: Median P ≈ 22 h, Q ≈ 26 h, so Brand Q batteries last longer on average; IQR P ≈ 6 h, Q ≈ 8 h, so Brand P is more consistent.
Examiner tip: A comparison needs one average AND one measure of spread, each compared in context – just quoting the numbers scores no C marks.

13Grade 6–7HigherCalculator5 marks

Here are the prices, in pounds, of 15 second-hand bikes for sale online.

85 120 95 140 110 300 130 90 105 150 125 100 115 135 145

An outlier is a value that is

more than 1.5 × interquartile range above the upper quartile, or
more than 1.5 × interquartile range below the lower quartile.

(a) Show that exactly one of these prices is an outlier. [4]

The bike that is an outlier is removed from the list.

(b) Work out the median price of the remaining bikes. [1]

(a) prices put in order and lower quartile £100 or upper quartile £140 found (4th or 12th value)M1
(a) interquartile range 140 − 100 = 40 and 1.5 × 40 = 60M1
(a) upper limit 140 + 60 = 200 and lower limit 100 − 60 = 40M1
(a) conclusion: £300 is above 200 so is an outlier, and no price is below 40 (lowest is £85) and no other price is above 200, so exactly one outlierC1
(b) £117.50 (accept 117.5)B1
Answer: (a) LQ £100, UQ £140, IQR £40; limits £40 and £200; only £300 lies outside (b) £117.50
Examiner tip: ‘Show that exactly one’ needs both limits checked – say why £300 is an outlier AND why the lowest price, £85, is not.

14Grade 6–7HigherNon-calculator4 marks

The incomplete box plot shows information about the times, in minutes, that 120 visitors spent in a museum. It shows the shortest time, the lower quartile and the median.

For these times

the interquartile range is 15 minutes
the range is 3 times the interquartile range.

(a) Complete the box plot. [3]

(b) Estimate the number of visitors who spent between 20 minutes and the upper quartile time in the museum. [1]

(a) upper quartile = 20 + 15 = 35 (lower quartile 20 and minimum 12 read from the diagram)M1
(a) range = 3 × 15 = 45 and maximum = 12 + 45 = 57M1
(a) box plot completed correctly: box closed at 35 (median line at 26 kept) and whisker from 35 to 57A1
(b) 60 (half of 120, since 20 is the lower quartile)B1
Answer: (a) upper quartile 35 minutes, maximum 57 minutes (b) 60 visitors
Examiner tip: The range is highest − lowest, so add the range to the minimum (12) – adding it to the upper quartile is the most common slip.

15Grade 8–9HigherCalculator5 marks

The cumulative frequency graph shows information about the masses, in grams, of 200 apples delivered to a shop.

The shop owner

throws away the lightest 10% of the apples
sells each of the other apples with a mass of 150 grams or less for 25p
sells each apple with a mass of more than 150 grams for 40p.

All the apples that are not thrown away are sold.

Work out an estimate for the total amount of money the shop owner gets from selling these apples. [5]

reads the graph at 150 g: cumulative frequency in the range 104 to 110M1
number of apples more than 150 g = 200 − their reading (e.g. 93)M1
10% of 200 = 20 apples thrown away and number sold at 25p = their reading − 20 (e.g. 87)M1
complete method for income, e.g. 87 × 0.25 + 93 × 0.40 (ft their counts)M1
answer in the range £58.50 to £59.40 (e.g. £58.95), given in correct money notationA1
Answer: about £58.95 (accept £58.50 to £59.40)
Examiner tip: The lightest 10% is a number of apples (20), not a mass – you do not need to read a mass from the graph for it; just subtract 20 from the 25p group.

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