Statistical Measures · Study Notes · Edexcel IGCSE 4MA1 Higher

Mean, median, mode, range, and estimated mean from grouped data. The grouped-data calculation is a guaranteed Higher Tier question.

At a glance

Mean = $\dfrac{\sum x}{n}$ — average.
Median = middle value (sort first!).
Mode = most common value.
Range = max − min.
Modal class = class with highest frequency.
Estimated mean (grouped): use midpoints, $\bar{x} = \dfrac{\sum fx}{\sum f}$ .
Always sort before finding median.
Median least affected by outliers.

What you’ll learn

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

6.1 — Calculate mean, median, mode and range for ungrouped data.
6.1 — Calculate mean, median, modal class and range for grouped data.
6.1 — Estimate the mean from a grouped frequency distribution.
6.1 — Compare distributions using a measure of central tendency and a measure of spread.

The three averages: mean, median, mode

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Three different ways to summarise the centre of a dataset.

Mean — sum divided by count. The arithmetic average.

$\bar{x} = \dfrac{\sum x}{n}$

Median — the MIDDLE value when data is sorted.

For $n$ values, median is at position $(n + 1)/2$ .
If $n$ is even, median = average of the two middle values.

Mode — the most frequently occurring value.

A dataset can have NO mode (all unique), ONE mode, or MULTIPLE modes.

Worked example. Data: $5, 8, 3, 7, 4, 9$ .

Mean: $(5+8+3+7+4+9)/6 = 36/6 = 6$ .
Sorted: $3, 4, 5, 7, 8, 9$ . Median: average of 4th and 3rd, $(5+7)/2 = 6$ .
Mode: no mode (all unique).

Worked example. Data: $4, 7, 2, 4, 9, 4, 6$ .

Mean: $(4+7+2+4+9+4+6)/7 = 36/7 \approx 5.14$ .
Sorted: $2, 4, 4, 4, 6, 7, 9$ . Median (4th): $4$ .
Mode: $4$ (appears 3 times).

When to use which?

Mean: best for symmetric data; pulled by outliers.
Median: robust to outliers (e.g. salaries — Bill Gates skews mean).
Mode: useful for categorical data.

Three averages of one data set: mode is the most frequent, median is the middle, mean is the balance point.

Edexcel tip. Edexcel often asks "give a reason why the median is more appropriate". Answer: "Because the data has outliers/extreme values that distort the mean."

Mean: sum / count.
Median: middle (sort first).
Mode: most common.
Outliers affect mean most.

Range — measure of spread

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Max − min. Single number.

Range = Maximum − Minimum

Worked example. Data $4, 7, 2, 4, 9, 4, 6$ . Max = $9$ , min = $2$ . Range = $7$ .

Limitations. Range only uses two values (max and min) — vulnerable to outliers. For better spread, use interquartile range (IQR — covered in cumulative frequency).

Comparing distributions. A typical exam question:

"Compare the heights of plants in Group A and Group B."
Answer needs: (1) a comparison of CENTRE (mean or median), AND (2) a comparison of SPREAD (range).

Worked example. Group A: mean 12 cm, range 4 cm. Group B: mean 14 cm, range 8 cm.

"Group B has a higher mean height (14 vs 12), so Group B plants are taller on average."
"Group A has a smaller range (4 vs 8), so Group A heights are more consistent."

Edexcel tip. Comparison questions need TWO statements: one about centre, one about spread. Single-statement answers lose half the marks.

Range = max − min.
Single number.
Two-statement comparisons (centre + spread).
Compare in CONTEXT (use 'plants', 'heights', etc., not just numbers).

Mean from a frequency table

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Multiply $f \times x$ , sum, divide by $\sum f$ .

For ungrouped frequency tables (each row is a single value):

$\bar{x} = \dfrac{\sum f x}{\sum f}$

Worked example.

Score $x$	Freq $f$	$fx$
1	3	3
2	5	10
3	8	24
4	4	16
Total	20	53

Mean $= 53/20 = 2.65$ .

Median position. $(n+1)/2 = (20+1)/2 = 10.5$ . So median is between 10th and 11th value.

Cumulative count: 3, 8 (so up to 8 are 2 or less), 16 (up to 16 are 3 or less). The 10th and 11th values are both $3$ . Median = $3$ .

Mode. Highest $f = 8$ → mode is $3$ .

Edexcel tip. Add an $fx$ column to the table — it's the most efficient way and shows method clearly.

Add $fx$ column.
Mean = $\sum fx / \sum f$ .
Median: cumulative count to position $(n+1)/2$ .
Mode: row with highest $f$ .

Grouped data — estimated mean and modal class

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Use midpoints. Result is an estimate.

For grouped frequency tables, exact values are unknown — only intervals are given. We estimate:

Estimated mean. Use class midpoints as $x$ in $\bar{x} = \sum fx / \sum f$ .

Worked example.

Heights (cm)	Freq $f$	Midpoint $x$	$fx$
$150 \le h < 160$	5	155	775
$160 \le h < 170$	12	165	1980
$170 \le h < 180$	8	175	1400
Total	25		4155

Estimated mean $= 4155 / 25 = 166.2$ cm.

Estimated mean uses the midpoint of each class; the modal class is the interval with the tallest bar.

Modal class — the class with the HIGHEST frequency. Here: $160 \le h < 170$ .

(NOT 165, NOT "the modal frequency". State the full INTERVAL.)

Median class. The class containing the median position $(n+1)/2 = 13$ . Cumulative: 5, 17 (after 2nd class). 13 falls in 2nd class. Median class: $160 \le h < 170$ .

Why "estimated"? Because we don't know whether someone in the $160$ – $170$ class is at $160$ cm or $169.9$ cm — we use midpoint $165$ as a representative.

Edexcel tip. Mark scheme breakdown for estimated mean (5 marks):

M1: midpoints identified
M1: $fx$ products
M1: $\sum fx / \sum f$ method
A1: correct numerical answer
B1 (sometimes): rounded sensibly with units

Use midpoints.
$\bar{x} = \sum fx / \sum f$ .
Modal class = highest $f$ interval.
Median class = interval containing middle position.
Always say 'estimated' mean.

See the full worked example for statistical measures →

Quick recap

Mean = $\sum x / n$ . Median = middle (sort!). Mode = most common.
Range = max − min.
Frequency table: $\bar{x} = \sum fx / \sum f$ .
Grouped data: use midpoints; result is ESTIMATED mean.
Modal class: highest-frequency interval (state full range).
Compare distributions with one centre statement + one spread statement.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Mean $= \dfrac{\sum x}{n}$ or $\dfrac{\sum fx}{\sum f}$ .
Median position = $(n + 1)/2$ .
Range = max − min.
Modal class — full interval, not just midpoint.
Estimated mean — say 'estimated' for grouped data.

How it’s examined

Estimated mean from grouped data is a near-guaranteed Paper 2H question (4-5 marks). Comparison questions need TWO statements (centre + spread). Examiner reports flag: (1) midpoints wrongly calculated, (2) division by sum of $x$ instead of $\sum f$ , (3) modal class given as midpoint.

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. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on statistical measures , written exactly the way a tutor would explain them at the board.

1Mean from a list
Foundation• mean
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Question
Find the mean of $5, 8, 3, 7, 4, 9$ .
Step-by-step solution
1. Step 1
  Sum: $5 + 8 + 3 + 7 + 4 + 9 = 36$ .
2. Step 2
  $\bar{x} = 36 / 6 = 6$ .
Answer
$\bar{x} = 6$
2Median
Foundation• median
▼
Question
Find the median of $7, 3, 9, 5, 11, 4, 8$ .
Step-by-step solution
1. Step 1
  Sort ascending: $3, 4, 5, 7, 8, 9, 11$ .
2. Step 2
  $n = 7$ . Median is at position $(n+1)/2 = 4$ . Median = $7$ .
Answer
Median = $7$
3Mode and range
Foundation• mode, range
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Question
Find the mode and range of $4, 7, 2, 4, 9, 4, 6$ .
Step-by-step solution
1. Step 1
  Mode = most frequent value = $4$ (appears 3 times).
2. Step 2
  Range = max − min = $9 - 2 = 7$ .
Answer
Mode = $4$ , range = $7$ .
4Estimated mean from grouped data
Higher• Adapted from 4MA1/2H May/Jun 2024 Q15• estimated mean, grouped data
▼
Question
Heights (cm): $150 \le h < 160$ (5 people), $160 \le h < 170$ (12), $170 \le h < 180$ (8). Estimate the mean.
Step-by-step solution
1. Step 1
  Use midpoint $\times$ frequency. Midpoints: $155, 165, 175$ .
2. Step 2
  $\sum f x = 5(155) + 12(165) + 8(175) = 775 + 1980 + 1400 = 4155$ .
3. Step 3
  $\sum f = 25$ . Estimated mean $= 4155 / 25 = 166.2$ cm.
Answer
Estimated mean = $166.2$ cm
Examiner tip
Always say 'estimated mean' for grouped data — actual values are unknown. Mark scheme: M1 for midpoints, M1 for $\sum fx / \sum f$ , A1 for value.
5Modal class
Higher• modal class
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Question
From the heights table above, find the modal class.
Step-by-step solution
1. Step 1
  Highest frequency = 12 → class $160 \le h < 170$ .
Answer
$160 \le h < 170$
6Median of the first ten square numbers [1 mark]
Getting started• Paper 1H/2H style — 1 mark• median, averages
▼
Question
Write down the median of the first ten square numbers. [1]
Step-by-step solution
1. Step 1
  The first ten square numbers, already in order, are $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ .
2. Step 2
  B1 There are $10$ values, so the median is halfway between the $5$ th and $6$ th values.
  $\frac{25 + 36}{2} = 30.5$
Answer
$30.5$
Examiner tip
Examiner tip: With an even number of values, the median is the mean of the middle two. It does not have to be one of the values — or even a whole number.

Common mistake: Giving $25$ (the $5$ th value on its own), or finding the median of $1$ to $10$ ( $5.5$ ) and squaring it.
7Mean of temperatures including negatives [2 marks]
Getting started• Paper 1H/2H style — 2 marks• mean, negative numbers
▼
Question
Here are the night-time temperatures, in °C, in a village on eight nights.

$-4, \; 3, \; -1, \; 6, \; -7, \; 0, \; 2, \; -3$

Work out the mean temperature. [2]
Step-by-step solution
1. Step 1
  M1 Add the eight temperatures, taking care with the signs.
  $-4 + 3 - 1 + 6 - 7 + 0 + 2 - 3 = -4$
2. Step 2
  A1 Divide by $8$ . The night with $0$ °C still counts as one of the eight values.
  $\frac{-4}{8} = -0.5$
Answer
$-0.5$ °C
Examiner tip
Examiner tip: Add the positives ( $11$ ) and the negatives ( $-15$ ) separately, then combine: $11 - 15 = -4$ . A negative mean is fine for temperatures.

Common mistake: Dividing by $7$ because the $0$ "doesn't count", or dropping the minus sign and giving $0.5$ .
8Modal class and the class containing the median [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• grouped data, median, modal class
▼
Question
The table gives information about the masses, $m$ kg, of $50$ parcels.

$0 < m \leqslant 2$ : frequency $7$

$2 < m \leqslant 4$ : frequency $15$

$4 < m \leqslant 6$ : frequency $11$

$6 < m \leqslant 8$ : frequency $12$

$8 < m \leqslant 10$ : frequency $5$

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

(b) Find the class interval that contains the median. [2]
Step-by-step solution
1. Step 1
  B1 (a) The highest frequency is $15$ .
  $2 < m \leqslant 4$
2. Step 2
  M1 (b) With $50$ parcels, the median is halfway between the $25$ th and $26$ th masses. Running totals: $7$ , $22$ , $33$ , $45$ , $50$ .
3. Step 3
  A1 The $23$ rd to $33$ rd masses are in the third class, so the $25$ th and $26$ th are both there.
  $4 < m \leqslant 6$
Answer
(a) $2 < m \leqslant 4$ (b) $4 < m \leqslant 6$
Examiner tip
Examiner tip: The median class is found from running totals, not from the size of the frequencies. It need not be the modal class.

Common mistake: Assuming the median is in the modal class and giving $2 < m \leqslant 4$ — only $22$ parcels have a mass of $4$ kg or less, fewer than half of $50$ . Giving the frequency $15$ as the modal class is also wrong: the answer is a class interval.
9Correcting a mean after a recording error [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• mean, problem solving
▼
Question
The mean of $12$ scores was worked out as $58$ .

It was then found that one score had been recorded as $47$ when it should have been $74$ .

Work out the correct mean of the $12$ scores. [3]
Step-by-step solution
1. Step 1
  M1 The total used for the wrong mean.
  $12 \times 58 = 696$
2. Step 2
  M1 The correct total is $74 - 47 = 27$ more.
  $696 + 27 = 723$
3. Step 3
  A1
  $\frac{723}{12} = 60.25$
Answer
$60.25$
Examiner tip
Examiner tip: Correcting one value changes the total by the difference, and the mean by that difference divided by the number of values: $58 + \frac{27}{12} = 60.25$ is a quick check.

Common mistake: Adding $27$ to the mean instead of the total, giving $85$ .
10Unknown frequency from an estimated mean [5 marks]
Stretch• Paper 1H/2H style — 5 marks• estimated mean, grouped data, algebra
▼
Question
The table gives information about the times, $t$ minutes, that some people waited at a clinic.

$0 < t \leqslant 10$ : frequency $5$

$10 < t \leqslant 20$ : frequency $x$

$20 < t \leqslant 30$ : frequency $8$

$30 < t \leqslant 40$ : frequency $2$

An estimate for the mean waiting time is $17.5$ minutes.

Work out the value of $x$ . Show clear algebraic working. [5]
Step-by-step solution
1. Step 1
  M1 Use the midpoints $5$ , $15$ , $25$ and $35$ to write $\sum fx$ .
  $5 \times 5 + 15x + 25 \times 8 + 35 \times 2 = 295 + 15x$
2. Step 2
  M1 The total frequency.
  $5 + x + 8 + 2 = 15 + x$
3. Step 3
  M1 Form an equation.
  $\frac{295 + 15x}{15 + x} = 17.5$
4. Step 4
  M1 Multiply both sides by $(15 + x)$ and collect terms.
  $295 + 15x = 262.5 + 17.5x \;\Rightarrow\; 32.5 = 2.5x$
5. Step 5
  A1
  $x = 13$
Answer
$x = 13$
Examiner tip
Examiner tip: Check by substituting back: $\sum f = 28$ and $\sum fx = 490$ , and $\frac{490}{28} = 17.5$ .

Common mistake: Multiplying out without brackets, writing $295 + 15x = 17.5 \times 15 + x$ . The whole of $15 + x$ must be multiplied by $17.5$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
The range of a set of test marks is $23$ . The lowest mark is $48$ .

Work out the highest mark. [1]
Model answer
$48 + 23 = 71$

Answer: $71$
Why this scores
Mark scheme:
- B1 for $71$ cao
Why this scores full marks: Range = highest − lowest, so the highest mark is the lowest mark plus the range. Subtracting ( $48 - 23 = 25$ ) gives a mark below the lowest, which cannot be right.
Question 2
Paper 1H/2H style — 2 marks2 marks
A shop sold $35$ pairs of shoes in one day. The sizes sold are shown below.
- size $4$ : frequency $3$
- size $5$ : frequency $7$
- size $6$ : frequency $12$
- size $7$ : frequency $9$
- size $8$ : frequency $4$
(a) Write down the mode. [1]

(b) Work out the range. [1]
Model answer
(a) The highest frequency is $12$ , for size $6$ .

Answer: size $6$

(b) $8 - 4 = 4$

Answer: $4$
Why this scores
Mark scheme:
- B1 for $6$ cao
- B1 for $4$ cao
Why this scores full marks: Both answers are sizes, not frequencies. Weaker scripts give $12$ as the mode and $12 - 3 = 9$ as the range, using the frequency column.
Question 3
Paper 1H/2H style — 2 marks2 marks
(a) A class records each student's favourite fruit. Which average should be used to describe these data? Give a reason for your answer. [1]

(b) The prices of six houses in a village are £185 000, £192 000, £198 000, £205 000, £210 000 and £1 400 000. Which average, the mean or the median, better describes a typical house price? Give a reason for your answer. [1]
Model answer
(a) The mode, because the data are not numbers, so the mean and the median cannot be worked out.

(b) The median, because the £1 400 000 house is an extreme value that makes the mean much higher than the price of five of the six houses.
Why this scores
Mark scheme:
- B1 for mode with a reason referring to the data not being numerical (categories) oe
- B1 for median with a reason referring to the extreme value oe
Why this scores full marks: Each choice comes with a reason about these particular data. "The mode is easiest" or "the median is most accurate" would not score.
Question 4
Paper 1H/2H style — 3 marks3 marks
The mean of the four numbers

$2x, \; x + 3, \; x + 6, \; 4x - 1$

is $10$ .

Work out the value of $x$ . [3]
Model answer
The four numbers add up to $4 \times 10 = 40$ .

$2x + x + 3 + x + 6 + 4x - 1 = 40$

$8x + 8 = 40$

$x = 4$

Check: the numbers are $8, 7, 10, 15$ , total $40$ , mean $10$ .

Answer: $x = 4$
Why this scores
Mark scheme:
- M1 for $2x + x + 3 + x + 6 + 4x - 1$ ( $= 8x + 8$ )
- M1 for $8x + 8 = 40$ or $\frac{8x + 8}{4} = 10$ oe
- A1 for $4$ cao
Why this scores full marks: The candidate collects like terms carefully, sets the total equal to $40$ and checks by working out the four numbers.
Question 5
Paper 1H/2H style — 3 marks3 marks
Here are the numbers of visitors to a small museum on $15$ days.

$23, \; 31, \; 18, \; 27, \; 35, \; 29, \; 22, \; 40, \; 26, \; 33, \; 19, \; 28, \; 37, \; 24, \; 30$

Find the interquartile range. [3]
Model answer
In order:

$18, \; 19, \; 22, \; 23, \; 24, \; 26, \; 27, \; 28, \; 29, \; 30, \; 31, \; 33, \; 35, \; 37, \; 40$

Lower quartile is the $\frac{15 + 1}{4} = 4$ th value $= 23$

Upper quartile is the $12$ th value $= 33$

$33 - 23 = 10$

Answer: $10$
Why this scores
Mark scheme:
- M1 for the data in order
- M1 for $23$ and $33$ identified as the quartiles
- A1 for $10$ cao
Why this scores full marks: The data are ordered first and the quartile positions are stated, so the method is clear. Taking the $4$ th and $12$ th values from the unordered list gives $27$ and $28$ — a common and costly slip.
Question 6
Paper 1H/2H style — 3 marks3 marks
A set of exam marks has mean $52$ , median $50$ and range $36$ .

Every student is given $5$ extra marks.

Write down the new mean, the new median and the new range. [3]
Model answer
Every mark goes up by $5$ , so every average goes up by $5$ . The highest and lowest marks both go up by $5$ , so their difference does not change.

New mean $= 57$

New median $= 55$

New range $= 36$
Why this scores
Mark scheme:
- B1 for $57$
- B1 for $55$
- B1 for $36$
Why this scores full marks: The candidate sees that adding the same amount to every value shifts the averages but not the spread. Weaker scripts add $5$ to the range as well.
Question 7
Paper 1H/2H style — 4 marks4 marks
Aisha's mean score in her first $7$ tests was $64$ .

After $3$ more tests, her mean score in all $10$ tests was $68$ .

Work out her mean score in the last $3$ tests. Give your answer correct to $1$ decimal place. [4]
Model answer
Total for $7$ tests: $7 \times 64 = 448$

Total for $10$ tests: $10 \times 68 = 680$

Total for the last $3$ tests: $680 - 448 = 232$

$\frac{232}{3} = 77.333\ldots$

Answer: $77.3$
Why this scores
Mark scheme:
- M1 for $7 \times 64$ ( $= 448$ )
- M1 for $10 \times 68$ ( $= 680$ )
- M1 for $\frac{680 - 448}{3}$
- A1 for awrt $77.3$
Why this scores full marks: Totals are found before anything is subtracted, and the total of the last three tests is divided by $3$ , not by $10$ . The answer is given to $1$ decimal place as asked.
Question 8
Paper 1H/2H style — 4 marks4 marks
The table shows the number of hours of homework done last week by each of $19$ students in class A.
- $1$ hour: frequency $2$
- $2$ hours: frequency $4$
- $3$ hours: frequency $5$
- $4$ hours: frequency $4$
- $5$ hours: frequency $3$
- $6$ hours: frequency $1$
The median for class A is $3$ hours.

(a) Work out the interquartile range for class A. [2]

For class B, the median is $4$ hours and the interquartile range is $3$ hours.

(b) Compare the hours of homework done by class A and class B. [2]
Model answer
(a) Running totals: $2, 6, 11, 15, 18, 19$

Lower quartile is the $\frac{19 + 1}{4} = 5$ th value $= 2$

Upper quartile is the $15$ th value $= 4$

$4 - 2 = 2$

Answer: $2$ hours

(b) Class A's median ( $3$ hours) is lower than class B's ( $4$ hours), so on average class A did less homework.

Class A's interquartile range ( $2$ hours) is smaller than class B's ( $3$ hours), so the hours of homework in class A were more consistent.
Why this scores
Mark scheme:
- M1 for identifying the $5$ th and $15$ th values or LQ $= 2$ and UQ $= 4$
- A1 for $2$ cao
- B1 for comparing the medians in context, e.g. class A did less homework on average
- B1 for comparing the IQRs in context, e.g. class A was more consistent ft their (a)
Why this scores full marks: The quartile positions come from $n = 19$ , and the comparison has one sentence about the average and one about the spread, each naming both classes and saying what it means for homework.
Question 9
Paper 1H/2H style — 5 marks5 marks
The speed limit on a road is $50$ km/h. The table gives information about the speeds, $s$ km/h, of $60$ cars on the road.
- $30 < s \leqslant 40$ : frequency $8$
- $40 < s \leqslant 45$ : frequency $14$
- $45 < s \leqslant 50$ : frequency $17$
- $50 < s \leqslant 55$ : frequency $13$
- $55 < s \leqslant 70$ : frequency $8$
(a) Work out an estimate for the mean speed. [3]

(b) Find the class interval that contains the median. [1]

A newspaper says, "On average, cars on this road break the speed limit."

(c) Is the newspaper correct? Give a reason for your answer. [1]
Model answer
(a) Midpoints: $35, \; 42.5, \; 47.5, \; 52.5, \; 62.5$

$35 \times 8 + 42.5 \times 14 + 47.5 \times 17 + 52.5 \times 13 + 62.5 \times 8 = 280 + 595 + 807.5 + 682.5 + 500 = 2865$

$\frac{2865}{60} = 47.75$

Answer: $47.75$ km/h

(b) The median is between the $30$ th and $31$ st speeds. Running totals: $8, 22, 39, \ldots$

Answer: $45 < s \leqslant 50$

(c) No. The estimated mean ( $47.75$ km/h) is below $50$ km/h, and the median is in the class $45 < s \leqslant 50$ , so it is not more than $50$ km/h.
Why this scores
Mark scheme:
- M1 for finding $fx$ for at least $4$ classes using midpoints
- M1 (dep) for $\frac{2865}{60}$ oe
- A1 for $47.75$ oe (accept awrt $47.8$ )
- B1 for $45 < s \leqslant 50$
- B1 for No, with a reason based on the mean or the median being below $50$ km/h ft
Why this scores full marks: The unequal class widths are handled by finding each midpoint separately, and the conclusion in (c) quotes an average and compares it with the limit of $50$ km/h.
Question 10
Paper 1H/2H style — 5 marks5 marks
The mean age of a group of $n$ people is $20$ years.

$4$ more people join the group. Their mean age is $32$ years.

The mean age of everyone in the group is now $23$ years.

Work out the value of $n$ . Show clear algebraic working. [5]
Model answer
Total age of the first group: $20n$

Total age of the $4$ new people: $4 \times 32 = 128$

There are now $n + 4$ people, so

$\frac{20n + 128}{n + 4} = 23$

$20n + 128 = 23n + 92$

$36 = 3n$

$n = 12$

Check: $\frac{240 + 128}{16} = \frac{368}{16} = 23$

Answer: $n = 12$
Why this scores
Mark scheme:
- M1 for $20n$ as the total age of the first group
- M1 for $20n + 128$ as the total of everyone
- M1 for $\frac{20n + 128}{n + 4} = 23$ or $20n + 128 = 23(n + 4)$ oe
- M1 for correctly expanding and collecting terms to reach $3n = 36$ oe
- A1 for $12$ cao
Why this scores full marks: The equation is built from totals, and $23(n + 4)$ is expanded with the bracket. Trial and improvement would not satisfy "Show clear algebraic working".

4MA1 Statistical Measures

At a glance

What you’ll learn

The three averages: mean, median, mode

Range — measure of spread

Mean from a frequency table

Grouped data — estimated mean and modal class

Quick recap

Memorise this

How it’s examined

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

1Mean from a list

2Median

3Mode and range

4Estimated mean from grouped data

6Median of the first ten square numbers [1 mark]

7Mean of temperatures including negatives [2 marks]

8Modal class and the class containing the median [3 marks]

9Correcting a mean after a recording error [3 marks]

10Unknown frequency from an estimated mean [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

4MA1 Statistical Measures

At a glance

What you’ll learn

The three averages: mean, median, mode

Range — measure of spread

Mean from a frequency table

Grouped data — estimated mean and modal class

Quick recap

Memorise this

How it’s examined

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

1Mean from a list

2Median

3Mode and range

4Estimated mean from grouped data

5Modal class

6Median of the first ten square numbers [1 mark]

7Mean of temperatures including negatives [2 marks]

8Modal class and the class containing the median [3 marks]

9Correcting a mean after a recording error [3 marks]

10Unknown frequency from an estimated mean [5 marks]

Model AnswersFull mark-scheme-style answers