Averages & range (incl. from frequency tables and grouped data estimates) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

Here are eight numbers.

4 9 2 7 9 12 5 8

(a) Write down the mode. [1]

(b) Work out the median. [2]

(a) 9B1
(b) numbers written in order (2 4 5 7 8 9 9 12) or 7 and 8 identified as the middle pairM1
(b) 7.5A1
(c) 10B1
Answer: (a) 9 (b) 7.5 (c) 10
Examiner tip: For the median, put the numbers in order first; with an even number of values the median is halfway between the middle two.

2Grade 4–5Non-calculator3 marks

The mean of five numbers is 8.

Four of the numbers are 6, 11, 4 and 9.

Work out the fifth number. [3]

$5 \times 8$ (= 40) for the total of all five numbersM1
$40 - (6 + 11 + 4 + 9)$ or $40 - 30$M1
10A1
Answer: 10
Examiner tip: Multiply the mean by how many numbers there are to get the total – do not just average the four numbers you are given.

3Grade 4–5Calculator4 marks

The table shows the number of goals scored by a football team in each of 25 matches.

Number of goals	Frequency
0	5
1	8
2	6
3	4
4	2

(a) Find the median number of goals. [1]

(b) Work out the mean number of goals per match. [3]

(a) 1B1
(b) products $0\times5,\ 1\times8,\ 2\times6,\ 3\times4,\ 4\times2$ (0, 8, 12, 12, 8) – at least 3 correctM1
(b) (dep) $\Sigma fx \div \Sigma f$, e.g. $40 \div 25$M1
(b) 1.6A1
Answer: (a) 1 (b) 1.6
Examiner tip: The mean is $\Sigma fx \div 25$, not $\Sigma fx \div 5$ – divide by the total frequency, not the number of rows.

4Grade 4–5Calculator6 marks

The table shows information about the times, $t$ minutes, that 30 people waited at a doctor's surgery.

Time ($t$ minutes)	Frequency
$0 \lt t \le 10$	4
$10 \lt t \le 20$	9
$20 \lt t \le 30$	12
$30 \lt t \le 40$	5

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

(b) Work out an estimate for the mean time. [4]

(a) $20 \lt t \le 30$ (accept 20–30 or '20 to 30'; do not accept 12, the frequency)B1
(b) products $f \times x$ using consistent values of $x$ within each class (including the ends)M1
(dep) use of all correct midpoints 5, 15, 25, 35 (products 20, 135, 300, 175)M1
(dep on first M1) $\Sigma fx \div \Sigma f$, e.g. $630 \div 30$M1
(b) 21 (minutes) (accept 21.0)A1
(c) the exact times are not known, so the midpoint of each class is used to represent every value in that class (accept e.g. 'we don't know the actual times, only the class they are in', 'midpoints are used' or 'the data is grouped so the exact values are not known'; do not accept 'the answer is rounded' or 'the data might be wrong')C1
Answer: (a) $20 \lt t \le 30$ (b) 21 minutes (c) midpoints are used, the actual times are unknown
Examiner tip: Use midpoints, and divide by the total frequency (30) – dividing by 4 (the number of classes) loses the last two marks.

5Grade 6–7Calculator3 marks

Class A has 12 students. Their mean test score is 64.

Class B has 18 students. Their mean test score is 71.

Work out the mean test score of all 30 students. [3]

$12 \times 64$ (= 768) or $18 \times 71$ (= 1278)M1
$(768 + 1278) \div 30$ (= $2046 \div 30$)M1
68.2A1
Answer: 68.2
Examiner tip: The combined mean is not $(64 + 71) \div 2 = 67.5$ – the classes are different sizes, so find each total first.

6Grade 6–7Non-calculator3 marks

Four positive whole numbers have

a mode of 5
a median of 6
a mean of 7.

Find the four numbers. [3]

two of the numbers are 5 (for the mode), e.g. 5, 5, ..., ...B1
total of the four numbers is $4 \times 7 = 28$, or the third number found as 7 from $(5 + x) \div 2 = 6$M1
5, 5, 7, 11A1
Answer: 5, 5, 7, 11
Examiner tip: Use each fact in turn: mode → two 5s, median → the middle pair adds to 12, mean → the total is 28.

7Grade 6–7HigherNon-calculator5 marks

Here are five expressions.

$x + 2$ $2x - 1$ $3x$ $x + 8$ $3x + 1$

The mean of the five expressions is 12.

(a) Work out the value of $x$. [3]

(b) Work out the median of the five values. [2]

(a) $10x + 10$ for the sum of the expressionsM1
(a) $10x + 10 = 60$ or $\frac{10x + 10}{5} = 12$M1
(a) $x = 5$A1
(b) substitutes their $x$ to find the five values (7, 9, 15, 13, 16)M1
(b) 13 (ft their $x$ if M1 awarded)A1
Answer: (a) $x = 5$ (b) 13
Examiner tip: In (b), the median is the middle VALUE once the numbers are in order (7, 9, 13, 15, 16) – not the middle expression as written.

8Grade 6–7HigherCalculator4 marks

Kai has taken 10 tests. His mean score is 62.

He takes some more tests and scores 80 in each one.

His mean score for all of the tests is now 68.

How many more tests did Kai take? [4]

$10 \times 62$ (= 620) for the total of the first 10 scoresM1
forms an equation, e.g. $\frac{620 + 80n}{10 + n} = 68$, or a trial that finds a new total and divides by the new number of testsM1
rearranges to $620 + 80n = 680 + 68n$ or $12n = 60$M1
5A1
Answer: 5 more tests
Examiner tip: Each new test changes BOTH the total score and the number of tests – write the new mean as (new total) ÷ (new number of tests).

9Grade 1–3FoundationNon-calculator4 marks

The bar chart shows the number of books each member of a book club read last month.

(a) Write down the mode. [1]

(b) Work out the range. [1]

(a) 2 (do not accept 7, the frequency)B1
(b) 4 (from $5-1$; do not accept 6, the range of the frequencies $7-1$)B1
(c) total of 20 members found ($3+7+4+5+1$) and attempt to locate the 10th and 11th values, e.g. cumulative totals 3, 10, 14 or the values listed in orderM1
(c) 2.5 (10th value is 2 and 11th value is 3)A1
Answer: (a) 2 (b) 4 (c) 2.5
Examiner tip: Read the averages from the horizontal axis (number of books), not the bar heights; the median is the middle member when all 20 are lined up in order.

10Grade 1–3FoundationNon-calculator4 marks

Ellie records how many minutes late her train is on each of 7 days.

4 3 6 2 40 3 5

(a) Work out the mean number of minutes late. [2]

(b) Ellie says, ‘The mean is the best average to show how late my train usually is.’

Give a reason why Ellie is wrong. [1]

Work out the range of the other six values. [1]

(a) $(4+3+6+2+40+3+5) \div 7$ or $63 \div 7$M1
(a) 9A1
(b) reason referring to the outlier, e.g. '40 is an outlier / extreme value that makes the mean too big' or 'the mean (9) is bigger than 6 of the 7 values, so the median (4) is better' (accept 'the median is not affected by the 40'; do not accept 'the mode is better' without a reason)C1
(c) 4 (from $6-2$)B1
Answer: (a) 9 minutes (b) e.g. the 40 is an outlier, so the mean is higher than almost every value; the median (4) is more typical (c) 4 minutes
Examiner tip: When asked which average is best, name the outlier and say what it does to the mean; a bare 'the median is better' scores nothing.

11Grade 4–5Non-calculator4 marks

The table shows information about the heights, $h$ cm, of 50 seedlings.

Height ($h$ cm)	Frequency
$4 \lt h \le 8$	6
$8 \lt h \le 12$	11
$12 \lt h \le 16$	15
$16 \lt h \le 20$	12
$20 \lt h \le 24$	6

(a) Write down the class interval that contains the median. [1]

(b) Sam says, ‘The modal class is 15.’

Explain why Sam is wrong. [1]

Is Priya correct? You must give a reason. [2]

(a) $12 \lt h \le 16$ (accept 12–16; the 25th and 26th heights are both in this class since cumulative frequencies are 6, 17, 32)B1
(b) 15 is the frequency, not a class; the modal class is $12 \lt h \le 16$ (accept 'he has given the frequency instead of the class interval')C1
(c) the smallest height is more than 4 cm and the largest is at most 24 cm (or greatest possible range $24-4=20$ seen)M1
(c) No, with a correct conclusion that the range must be less than (or at most) 20 cm, so it cannot be 22 cmC1
Answer: (a) $12 \lt h \le 16$ (b) 15 is a frequency; the modal class is $12 \lt h \le 16$ (c) No: all heights lie between 4 and 24 cm, so the range is less than 20 cm
Examiner tip: For grouped data the range can't be found exactly, but the class boundaries at each end give a maximum; and always write a class interval, not a frequency, for 'modal class'.

12Grade 4–5Non-calculator5 marks

Jamal grows tomato plants using two different fertilisers, A and B.

He measures the heights, in cm, of seven plants given each fertiliser.

Fertiliser A: 14 18 11 16 20 13 17

Fertiliser B: 15 22 9 19 24 12 21

(a) Work out the median and the range for each fertiliser. [3]

(b) Compare the heights of the plants given fertiliser A with the heights of the plants given fertiliser B. [2]

(a) at least one set of heights written in order (11 13 14 16 17 18 20 or 9 12 15 19 21 22 24) or the 4th value of an ordered list identifiedM1
(a) medians: A = 16 and B = 19A1
(a) ranges: A = 9 and B = 15B1
(b) correct comparison of averages in context, e.g. 'plants given fertiliser B were taller on average (higher median)' (ft their medians; accept comparison of means, A 15.6, B 17.4)C1
(b) correct comparison of spread in context, e.g. 'heights with fertiliser A were more consistent / less spread out (smaller range)' (ft their ranges; do not accept 'the range of A is smaller' with no interpretation)C1
Answer: (a) A: median 16 cm, range 9 cm; B: median 19 cm, range 15 cm (b) B plants were taller on average; A plants' heights were more consistent
Examiner tip: A comparison needs two statements: one about the average and one about the spread, each saying what it means for the plants, not just which number is bigger.

13Grade 6–7HigherCalculator5 marks

A company surveys a sample of its staff. Each person is asked how many cups of coffee they drink in a working day.

Number of cups	Frequency
0	6
1	9
2	$x$
3	5
4	2

The mean number of cups is 1.6

(a) Work out the value of $x$. [4]

The company has 240 staff.

(b) Use the sample to estimate the total number of cups of coffee drunk by all 240 staff in one working day. [1]

(a) $\Sigma fx$ in terms of $x$, e.g. $0 + 9 + 2x + 15 + 8$ (= $32 + 2x$), or $\Sigma f = 22 + x$M1
(a) forms a correct equation, e.g. $\dfrac{32 + 2x}{22 + x} = 1.6$ oeM1
(a) correct rearrangement to a linear form, e.g. $32 + 2x = 35.2 + 1.6x$ or $0.4x = 3.2$M1
(a) $x = 8$ (correct answer from trial and improvement with $\frac{48}{30} = 1.6$ shown scores 4)A1
(b) 384 (from $1.6 \times 240$)B1
Answer: (a) $x = 8$ (b) 384 cups
Examiner tip: The unknown frequency changes both the total of $fx$ and the total frequency, so put $x$ in the denominator as well as the numerator.

14Grade 6–7HigherCalculator4 marks

A weather station records the temperature at noon, in °C, on 10 days.

For these temperatures, the mean is 15 °C and the range is 8 °C.

Each temperature is converted to °F using the formula

$F = 1.8C + 32$

(a) Work out the mean of the 10 temperatures in °F. [2]

(b) Work out the range of the 10 temperatures in °F. [2]

(a) $1.8 \times 15 + 32$ (or each value multiplied by 1.8 and 32 added, so the mean does the same)M1
(a) 59 (°F)A1
(b) $1.8 \times 8$, or $(1.8H + 32) - (1.8L + 32)$ oe showing the 32 cancels, or a correct example with two values 8 apartM1
(b) 14.4 (°F) (46.4, from $1.8 \times 8 + 32$, scores M0 A0)A1
Answer: (a) 59 °F (b) 14.4 °F
Examiner tip: Adding 32 to every value moves the mean but does not change the gap between the largest and smallest values, so the range is only multiplied by 1.8.

15Grade 8–9HigherCalculator6 marks

The table shows information about the masses, $m$ kg, of 40 parcels.

Mass ($m$ kg)	Frequency
$0 \lt m \le 2$	7
$2 \lt m \le 4$	$a$
$4 \lt m \le 6$	$b$
$6 \lt m \le 10$	4

An estimate for the mean mass of the parcels is 4 kg.

(a) Work out the value of $a$ and the value of $b$. [4]

(b) Olu says, ‘The median mass of the parcels is more than 4 kg.’

Is Olu correct? You must show how you get your answer. [2]

(a) midpoints 1, 3, 5, 8 used, e.g. $7 \times 1 + 3a + 5b + 4 \times 8$M1
(a) forms an equation for $\Sigma fx$, e.g. $7 + 3a + 5b + 32 = 160$ (from $4 \times 40$) oe, and $a + b = 29$M1
(a) correct method to eliminate one variable, e.g. $3a + 5(29 - a) = 121$ or $2b = 121 - 87$M1
(a) $a = 12$ and $b = 17$A1
(b) cumulative frequencies, e.g. 7, 19, 36 (ft their $a$, $b$), or shows that the 20th (and 21st) parcel lies in the class $4 \lt m \le 6$M1
(b) Yes, with a correct reason, e.g. 'only 19 parcels are 4 kg or less, so the 20th and 21st are in $4 \lt m \le 6$, so the median is more than 4 kg' (ft their $a$, $b$)C1
Answer: (a) $a = 12$, $b = 17$ (b) Yes: only 19 parcels weigh 4 kg or less, so the median (between the 20th and 21st) is in $4 \lt m \le 6$
Examiner tip: The last class is wider, so its midpoint is 8, not 7; use the total frequency (40) to get a second equation in $a$ and $b$.

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