Every averages and range question from 22 Cambridge IGCSE Mathematics (0580) past papers (May/June 2025 to May/June 2026), syllabus section 9.3. 23 questions covering Core and Extended tiers, ordered from Core to Extended. Each question is labelled with its question paper code (for example 0580/11/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Statistics past paper questions.
0580 Averages and Range past paper questions: Core (Papers 1 and 3)
May/June 2026 · Paper 1 · 0580/11/M/J/26 · Q12
The stem-and-leaf diagram shows the lengths, in centimetres, of 32 wooden planks.
| 23 | 3 4 4 5 6 6 8 9 |
| 24 | 0 0 1 1 1 1 3 6 8 8 9 |
| 25 | 2 3 5 6 7 9 9 9 |
| 26 | 0 0 1 2 4 |
Key: 23 | 3 represents 233 cm
(a) Find the range. (b) Find the mode. (c) Find the median. (d) Find the fraction of all these planks which are less than 236 cm, in its simplest form.
Show solution
- Use the key: 23 | 3 means 233 cm. The smallest length is 233 and the largest is 264.
- Range = largest − smallest = 264 − 233 = 31 cm
- The mode is the value that appears most often. In the 24 row, the leaf 1 appears four times (241).
- Mode = 241 cm
- There are 32 values, so the median is halfway between the 16th and 17th values.
- Count along: the 23 row has 8 values, so the 24 row holds the 9th to 19th values.
- The 16th value is 246 and the 17th value is 248.
- Median = (246 + 248) ÷ 2 = 247 cm
- Planks shorter than 236 cm: 233, 234, 234, 235. That is 4 planks.
- Fraction = 4⁄32. Divide top and bottom by 4: 1⁄8
May/June 2026 · Paper 1 · 0580/12/M/J/26 · Q24
A group of 4 children have a mean age of 11. A fifth child joins the group, and the mean age is now 10. Calculate the age of the fifth child.
Show solution
- Mean = total ÷ number, so total = mean × number.
- Total age of 4 children = 4 × 11 = 44 years.
- Total age of 5 children = 5 × 10 = 50 years.
- Fifth child = 50 − 44 = 6 years
May/June 2026 · Paper 1 · 0580/13/M/J/26 · Q7
Raj records the midday temperature, in °C, for 8 days: 1, 10, 8, −3, −2, 10, 7, 9
(a) Calculate the mean. (b) Find the median. (c) When another temperature is recorded, the range increases by 5. Find the two possible values of this temperature.
Show solution
- Add all eight temperatures: 1 + 10 + 8 + (−3) + (−2) + 10 + 7 + 9 = 40.
- Mean = 40 ÷ 8 = 5 °C
- Put them in order: −3, −2, 1, 7, 8, 9, 10, 10.
- There are 8 values, so the median is halfway between the 4th and 5th: 7 and 8.
- Median = (7 + 8) ÷ 2 = 7.5 °C
- The range now is 10 − (−3) = 13. The new range is 13 + 5 = 18.
- To make the range bigger, the new temperature must become the new lowest or the new highest.
- New lowest: 10 − 18 = −8. New highest: −3 + 18 = 15.
- −8 °C or 15 °C
May/June 2026 · Paper 3 · 0580/32/M/J/26 · Q5
There are 4 adults and 3 children in a room. The height of the shortest adult is 1.59 m. The range of the adults’ heights is 0.25 m. The height of the tallest child is 1.43 m. The range of the children’s heights is 0.38 m.
(a) Work out the range of the heights of the 7 people. (b) Write down the median height of the 7 people.
Show solution
- The tallest person is the tallest adult: 1.59 + 0.25 = 1.84 m.
- The shortest person is the shortest child: 1.43 − 0.38 = 1.05 m.
- Range = tallest − shortest = 1.84 − 1.05 = 0.79 m
- All 3 children are shorter than every adult (the tallest child is 1.43 m and the shortest adult is 1.59 m).
- In order from shortest: child, child, child, then the 4 adults.
- With 7 people the median is the 4th value, which is the shortest adult: 1.59 m
Feb/Mar 2026 · Paper 3 · 0580/32/F/M/26 · Q13
A set of five different whole numbers has the smallest number −3, a range of 14, a median of 8 and a mean of 6. Work out the four other numbers. Write your numbers in order, starting with the smallest.
Show solution
- The smallest number is −3 and the range is 14, so the largest = −3 + 14 = 11.
- There are 5 numbers, so the median is the middle (3rd) one: 8.
- The mean is 6, so the total = 5 × 6 = 30.
- The two missing numbers add to 30 − (−3 + 8 + 11) = 30 − 16 = 14.
- One must be between −3 and 8, and the other between 8 and 11, all different. Try 10 + 4 = 14.
- −3, 4, 8, 10, 11 (−3, 5, 8, 9, 11 also works)
Oct/Nov 2025 · Paper 1 · 0580/11/O/N/25 · Q10
The bar chart shows the number of books sold in a shop each month from March to July: March 3, April 9, May 7, June 11, July 8.
(a) 4 books were sold in August. Complete the bar chart. (b) Calculate the mean number of books sold in the six months.
Show solution
- Draw a bar for August of height 4, the same width as the other bars with the same gaps.
- Add all six months: 3 + 9 + 7 + 11 + 8 + 4 = 42 books.
- Mean = total ÷ number of months = 42 ÷ 6 = 7
Oct/Nov 2025 · Paper 1 · 0580/12/O/N/25 · Q11
The stem-and-leaf diagram shows the number of customers entering a shop each day for 20 days.
| 3 | 0 1 5 9 |
| 4 | 1 4 6 8 9 |
| 5 | 1 3 4 4 4 7 8 |
| 6 | 3 6 6 |
| 7 | 2 |
Key: 3 | 0 represents 30 customers
(a) Find the range. (b) Find the mode.
(c) Work out the percentage of these 20 days when more than 55 customers enter the shop.
(d) (i) Find the median. (ii) On the next day 48 customers enter the shop. Will the median increase or decrease? Give a reason for your answer.
Show solution
- Use the key: 3 | 0 means 30. The smallest value is 30 and the largest is 72.
- Range = 72 − 30 = 42
- The mode appears most often. In the 5 row the leaf 4 appears three times.
- Mode = 54
- Days with more than 55 customers: 57, 58, 63, 66, 66, 72. That is 6 days.
- Percentage = 6⁄20 × 100 = 30%
- There are 20 values, so the median is halfway between the 10th and 11th.
- Count along: the 3 row has 4 values and the 4 row has 5, so the 10th value is the first in the 5 row (51) and the 11th is 53.
- Median = (51 + 53) ÷ 2 = 52
- 48 is below the median, so adding it pulls the middle value down.
- With 21 values the median is now the 11th value, which is 51.
- Decrease, because 48 is below the median
Oct/Nov 2025 · Paper 1 · 0580/13/O/N/25 · Q8
These are the masses, in kg, of each of 11 bags: 23, 16, 8, 10, 27, 19, 4, 17, 13, 4, 14
(a) Complete the stem-and-leaf diagram. Key: 2 | 7 represents 27 kg. (b) Find the median.
Show solution
- Split each mass into tens (stem) and units (leaf). Single-digit masses go on the 0 stem.
- Put the leaves for each stem in order: 0 | 4 4 8, 1 | 0 3 4 6 7 9, 2 | 3 7.
- 0 | 4 4 8; 1 | 0 3 4 6 7 9; 2 | 3 7
- There are 11 values, so the median is the middle one: the (11 + 1) ÷ 2 = 6th value.
- Count along: 4, 4, 8, 10, 13, 14, … The 6th value is 14.
- Median = 14 kg
Oct/Nov 2025 · Paper 3 · 0580/31/O/N/25 · Q6
Find (a) the mode (b) the range of these numbers: 17, 4, 25, 18, 14, 6, 3, 18, 12
Show solution
- Put the numbers in order: 3, 4, 6, 12, 14, 17, 18, 18, 25.
- The mode is the value that appears most often: 18 appears twice.
- Mode = 18
- Range = largest − smallest = 25 − 3 = 22
Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q10
The bar chart shows the frequency of each score on a spinner: score 1 (7), 2 (5), 3 (8), 4 (9), 5 (6). Calculate the mean score.
Show solution
- Mean = total of all scores ÷ number of spins.
- Total of scores = 1 × 7 + 2 × 5 + 3 × 8 + 4 × 9 + 5 × 6 = 7 + 10 + 24 + 36 + 30 = 107.
- Number of spins = 7 + 5 + 8 + 9 + 6 = 35.
- Mean = 107 ÷ 35 = 3.057… = 3.06
- Tip: Do not divide by 5 (the number of different scores). Divide by the total frequency.
Oct/Nov 2025 · Paper 3 · 0580/32/O/N/25 · Q20
A town has two cinemas. Movie Scene has mean audience 83 and range 52. Flix has mean audience 105 and range 25.
(a) Which cinema has higher audience numbers on average? Give a reason. (b) Which cinema has greater variation in its audience numbers? Give a reason.
Show solution
- The mean is a type of average. Flix has the higher mean (105 > 83).
- Flix, because its mean is higher
- The range measures how spread out the numbers are. Movie Scene has the larger range (52 > 25).
- Movie Scene, because its range is larger
Oct/Nov 2025 · Paper 3 · 0580/33/O/N/25 · Q10
(a) Jason records his score in each of 10 games of cricket: 16, 7, 21, 8, 20, 9, 11, 7, 14, 3. (i) Find the mode. (ii) Work out the range.
(b) Ed plays 5 games. His mean score is 28. After one more game his mean score is now 26. Work out his score in the sixth game.
Show solution
- The mode is the score that appears most often. 7 appears twice; every other score appears once.
- Mode = 7
- Range = highest − lowest = 21 − 3 = 18
- Mean = total ÷ number of games, so total = mean × number of games.
- Total after 5 games = 5 × 28 = 140.
- Total after 6 games = 6 × 26 = 156.
- The sixth game added the difference: 156 − 140 = 16
May/June 2025 · Paper 1 · 0580/12/M/J/25 · Q11
0, 2, 2, 3, 4, 7
For these six numbers (a) write down the mode (b) work out the range (c) work out the median (d) work out the mean.
Show solution
- The mode is the value that appears most often. 2 appears twice.
- Mode = 2
- Range = largest − smallest = 7 − 0 = 7
- The numbers are already in order. With 6 numbers, the median is halfway between the 3rd and 4th: 2 and 3.
- Median = (2 + 3) ÷ 2 = 2.5
- Total = 0 + 2 + 2 + 3 + 4 + 7 = 18.
- Mean = total ÷ how many numbers = 18 ÷ 6 = 3
May/June 2025 · Paper 1 · 0580/13/M/J/25 · Q6
These are the lengths of time, in minutes, of seven phone calls. 10, 22, 5, 7, 35, 8, 75
(a)(i) Find the median. (a)(ii) Find the range.
(b) The longest phone call is 75 minutes. Write 75 minutes in hours.
Show solution
- Put the values in order: 5, 7, 8, 10, 22, 35, 75.
- There are 7 values, so the median is the middle one, the 4th.
- Median = 10 min
- Range = largest − smallest = 75 − 5 = 70 min
- There are 60 minutes in an hour, so divide by 60.
- 75 ÷ 60 = 1.25 h
0580 Averages and Range past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 2 · 0580/23/M/J/26 · Q4
These are the wages of 8 office workers: $220, $210, $850, $265, $245, $230, $245, $235
(a) Work out the median. (b) Give a reason why the mean may not be a suitable average to represent the data.
Show solution
- Put the wages in order: 210, 220, 230, 235, 245, 245, 265, 850.
- There are 8 values, so the median is halfway between the 4th and 5th: 235 and 245.
- Median = (235 + 245) ÷ 2 = $240
- One wage is very different from the rest (an outlier).
- $850 is much larger than the other wages, so it pulls the mean up, which makes the mean bigger than what most workers earn.
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q2
A librarian records the number of books borrowed by each of 40 students.
| number of books | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| frequency | 5 | 14 | 10 | 8 | 1 | 2 |
(a) Find the mode. (b) Find the median. (c) Calculate the mean.
Show solution
- The mode is the value with the highest frequency. The highest frequency is 14, for 1 book.
- Mode = 1 book
- There are 40 values, so the median is half-way between the 20th and 21st values.
- Running totals: 0 books → 5, 1 book → 5 + 14 = 19, 2 books → 19 + 10 = 29.
- The 20th and 21st values are both in the “2 books” group.
- Median = 2 books
- Multiply each number of books by its frequency and add.
- 0 × 5 + 1 × 14 + 2 × 10 + 3 × 8 + 4 × 1 + 5 × 2 = 0 + 14 + 20 + 24 + 4 + 10 = 72 books.
- Mean = total ÷ number of students = 72 ÷ 40 = 1.8 books
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q9
The table shows the marks of 100 students in a test.
| mark | 0 < x ≤ 20 | 20 < x ≤ 40 | 40 < x ≤ 60 | 60 < x ≤ 80 | 80 < x ≤ 100 |
|---|---|---|---|---|---|
| frequency | 4 | 20 | 31 | 34 | 11 |
(a) Write down the modal class. (b) Calculate an estimate of the mean.
Show solution
- The modal class is the class with the highest frequency. The highest frequency is 34.
- 60 < x ≤ 80
- Use the midpoint of each class as an estimate: 10, 30, 50, 70, 90.
- Multiply each midpoint by its frequency: 4 × 10 = 40, 20 × 30 = 600, 31 × 50 = 1550, 34 × 70 = 2380, 11 × 90 = 990.
- Total = 40 + 600 + 1550 + 2380 + 990 = 5560.
- Mean = 5560 ÷ 100 = 55.6
Feb/Mar 2026 · Paper 2 · 0580/22/F/M/26 · Q6
Nisha plays basketball. For the number of points she scored in her last 4 games: the mean is 10, the median is 13 and the range is 12. Find the number of points she scored in each of her last 4 games.
Show solution
- Call the four scores a ≤ b ≤ c ≤ d (in order).
- Mean 10 means the total is 4 × 10 = 40: a + b + c + d = 40.
- The median is the average of the middle two: (b + c) ÷ 2 = 13, so b + c = 26.
- The range is 12: d − a = 12, so d = a + 12.
- Substitute into the total: a + 26 + (a + 12) = 40, so 2a + 38 = 40, 2a = 2 and a = 1.
- d = 1 + 12 = 13. Since c ≤ d = 13 and b + c = 26 with b ≤ c, both b and c must be 13.
- 1, 13, 13, 13
- Check: mean (1 + 13 + 13 + 13) ÷ 4 = 10 ✓, median 13 ✓, range 13 − 1 = 12 ✓
Oct/Nov 2025 · Paper 2 · 0580/21/O/N/25 · Q13
These are Rahul’s 10 test scores: 9, 8, 9, 10, 7, x, 9, 9, x, 7. The mean of these scores is 8. Find the interquartile range. You must show all your working.
Show solution
- The mean is 8 and there are 10 scores, so the total is 8 × 10 = 80.
- Add the known scores: 9 + 8 + 9 + 10 + 7 + 9 + 9 + 7 = 68.
- 68 + 2x = 80, so 2x = 12 and x = 6.
- Put all 10 scores in order: 6, 6, 7, 7, 8, 9, 9, 9, 9, 10.
- Split into a lower half (6, 6, 7, 7, 8) and an upper half (9, 9, 9, 9, 10).
- Lower quartile = middle of the lower half = 7. Upper quartile = middle of the upper half = 9.
- IQR = 9 − 7 = 2
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q16
The table shows the heights of 50 plants.
| height (h cm) | 5 < h ≤ 10 | 10 < h ≤ 12 | 12 < h ≤ 20 |
|---|---|---|---|
| frequency | 3 | 24 | 23 |
Calculate an estimate of the mean height.
Show solution
- The exact heights are not known, so use the midpoint of each class.
- Midpoints: (5 + 10) ÷ 2 = 7.5, (10 + 12) ÷ 2 = 11, (12 + 20) ÷ 2 = 16.
- Multiply each midpoint by its frequency: 3 × 7.5 = 22.5, 24 × 11 = 264, 23 × 16 = 368.
- Add them: 22.5 + 264 + 368 = 654.5.
- Divide by the total frequency, 50: 654.5 ÷ 50 = 13.09 cm
Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q6
A dentist has appointments for 10 patients. The stem-and-leaf diagram shows the waiting times of 8 of them: 1, 4, 10, 12, 19, 21, 25, 25 minutes. The times for patients P and Q are not shown. The mean waiting time of all 10 patients is 16 minutes. The range of waiting times is 26 minutes. Patient P waits longer than patient Q. Find the waiting time for each of patient P and patient Q.
Show solution
- Total of all 10 times = mean × number = 16 × 10 = 160 minutes.
- The 8 known times add up to 1 + 4 + 10 + 12 + 19 + 21 + 25 + 25 = 117.
- So P + Q = 160 − 117 = 43.
- If both P and Q were inside the known values, the range would be 25 − 1 = 24. The range is 26, so one of them must be the new largest (or smallest) time.
- The smallest time cannot go below 1 and still have P + Q = 43 with the largest at 25, so the largest time = 1 + 26 = 27. P waits longer, so P = 27 min
- Q = 43 − 27 = 16 min
- Check: 16 is between 1 and 27, so the range stays 26 ✓
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q2
The stem-and-leaf diagram shows the age of each of 16 adults.
| 3 | 2 3 3 5 6 7 |
| 4 | 0 1 5 5 6 8 9 |
| 5 | 1 1 1 |
Key: 3 | 2 represents age 32 years
(a) Find the mode. (b) Find the median. (c) Find the percentage of the 16 adults with an age of less than 38 years.
Show solution
- Write the ages in order: 32, 33, 33, 35, 36, 37, 40, 41, 45, 45, 46, 48, 49, 51, 51, 51.
- The mode is the age that appears most often. 51 appears three times.
- Mode = 51 years
- There are 16 ages, so the median is half-way between the 8th and 9th values.
- 8th = 41 and 9th = 45. Median = (41 + 45) ÷ 2 = 43 years
- Ages less than 38: 32, 33, 33, 35, 36, 37. That is 6 adults.
- Percentage = 6⁄16 × 100 = 37.5%
May/June 2025 · Paper 2 · 0580/21/M/J/25 · Q9
The table shows some information about the marks scored by a group of students in a test.
| Test mark | 4 | 5 | 8 |
| Frequency | 2 | 4 | n |
The mean mark is 6. Work out the value of n.
Show solution
- Mean = total of all marks ÷ number of students.
- Total of marks = 4 × 2 + 5 × 4 + 8 × n = 8 + 20 + 8n = 28 + 8n.
- Number of students = 2 + 4 + n = 6 + n.
- Set up the equation: 28 + 8n⁄6 + n = 6.
- Multiply both sides by (6 + n): 28 + 8n = 36 + 6n.
- Subtract 6n and 28 from both sides: 2n = 8.
- n = 4
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