IGCSE / O Level Mathematics · Topic 9 Statistics · Chapter 35 · pp. 504–509
Mean, median, mode and range
One number can stand for a whole set of data. That number is an average, and there are three kinds: the mean, the median and the mode. The range tells you how spread out the data is. This chapter does all four for a list, for a frequency table and for grouped data.
Learning objectives
- Find the mean, median and mode of a list of data.
- Find the range and say what it measures.
- Find the mean, median and mode from a frequency table.
- Estimate the mean of grouped data using mid-interval values, and find the modal class.
- Choose which average best describes a set of data.
Book exercises: 35.1 to 35.3 and Student assessment 1 (pp. 504–509).
Three kinds of average
🔑 Key ideas
- Mode: the value that occurs most often.
- Median: the middle value when the data is in order of size.
- Mean: add all the values, then divide by how many values there are.
| Average | How to find it | Watch out |
|---|---|---|
| Mode | Count how often each value appears | There can be two modes, or none if every value appears once |
| Median | Put in order, take the middle. Two middle values: halfway between them | You must sort the data first |
| Mean | Total ÷ number of values | Count the values; do not guess |
💡 With n values, the median is the th value. For 15 values that is the 8th. For 6 values it is the 3.5th, which means halfway between the 3rd and 4th.
📘 Example 1 · Find the mean, median and mode of 1, 0, 2, 4, 1, 2, 1, 1, 2, 5, 5, 0, 1, 2, 3
- Count the values: there are 15.
- Add them: 1 + 0 + 2 + 4 + 1 + 2 + 1 + 1 + 2 + 5 + 5 + 0 + 1 + 2 + 3 = 30.
- Mean = 30 ÷ 15 = 2.
- Sort: 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 4, 5, 5.
- The median is the (15 + 1) ÷ 2 = 8th value: 2.
- The value 1 appears five times, more than any other: mode = 1.
📘 Example 2 · An even number of values: 3, 7, 4, 9, 7, 5
- Sort: 3, 4, 5, 7, 7, 9. There are 6 values.
- The median position is (6 + 1) ÷ 2 = 3.5, so use the 3rd and 4th values: 5 and 7.
- Median = (5 + 7) ÷ 2 = 6.
- Total = 35, so mean = 35 ÷ 6 = 5.83 (3 s.f.).
- 7 appears twice: mode = 7.
📘 Example 3 · The mean of 4 players' masses is 60 kg. With a fifth player the mean is 62 kg. Find the fifth mass
- Total = mean × number. Four players: 60 × 4 = 240 kg.
- Five players: 62 × 5 = 310 kg.
- The fifth player is the difference: 310 − 240 = 70 kg.
- Check: (240 + 70) ÷ 5 = 62. ✔
✏️ Try it
- Find the mean, median and mode of 3, 8, 5, 8, 1.
- Find the median of 2, 9, 4, 7, 6, 10.
- The mean of five numbers is 12. A sixth number, 18, is added. What is the new mean?
Answers
1. Mean 25 ÷ 5 = 5; sorted 1, 3, 5, 8, 8 so median 5; mode 8 · 2. Sorted 2, 4, 6, 7, 9, 10: (6 + 7) ÷ 2 = 6.5 · 3. Old total 60, new total 78, 78 ÷ 6 = 13
Range and spread
🔑 Key ideas
- Spread says how far apart the values are.
- Range = largest value − smallest value. It is one number, not two.
- Two data sets can have the same mean and median but very different spreads.
- The interquartile range is another measure of spread. It is covered in Chapter 37.
Which average should you use?
| Average | Good when | Not good when |
|---|---|---|
| Mean | Values are fairly even; you want to use every value | One value is far bigger or smaller than the rest |
| Median | There is an extreme value (for example one very high salary) | You need the total as well |
| Mode | Data is not numbers (favourite colour) or you want the most popular | Every value appears about equally often |
📘 Example 4 · A javelin thrower's best throws (m): 77, 75, 78, 86, 92, 93, 93, 93, 92, 89. Find the range and mean
- Largest value = 93, smallest value = 75.
- Range = 93 − 75 = 18 m.
- Total = 77 + 75 + 78 + 86 + 92 + 93 + 93 + 93 + 92 + 89 = 868.
- Mean = 868 ÷ 10 = 86.8 m.
- Sorted: 75, 77, 78, 86, 89, 92, 92, 93, 93, 93. The median is (89 + 92) ÷ 2 = 90.5 m and the mode is 93 m.
📘 Example 5 · Five house prices (thousand baht): 2, 2, 3, 3, 40. Which average is misleading?
- Mean = (2 + 2 + 3 + 3 + 40) ÷ 5 = 50 ÷ 5 = 10.
- Median = the 3rd value = 3.
- Four of the five prices are 3 or less, so 10 describes none of them. The 40 pulled the mean up.
- The mean is misleading here. The median (3) is the better average.
✏️ Try it
- Find the range of 12, 7, 19, 3, 15.
- Set A is 4, 5, 6 and set B is 1, 5, 9. Find the mean and range of each.
- The mean of four numbers is 7. Three of them are 5, 6 and 9. Find the fourth.
Answers
1. 19 − 3 = 16 · 2. Both have mean 5; A has range 2, B has range 8 · 3. Total 7 × 4 = 28; 28 − (5 + 6 + 9) = 8
Frequency tables
🔑 Key ideas
- A frequency table says how many times each value occurs. Do not add the values; use the frequencies.
- Add a row f × x (frequency times value). Its total is the sum of all the data.
- Mean = Σfx ÷ Σf (total of the f × x row ÷ total of the frequencies).
- Mode = the value with the biggest frequency, not the biggest frequency number itself.
- Median: find the (Σf + 1) ÷ 2 position by counting up the frequencies.
| Score | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | Total |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Frequency | 1 | 2 | 3 | 2 | 3 | 5 | 4 | 6 | 4 | 1 | 1 | 32 |
| f × score | 0 | 2 | 6 | 6 | 12 | 25 | 24 | 42 | 32 | 9 | 10 | 168 |
📘 Example 6 · Use the table above to find the mean, median, mode and range
- Σf = 32 students and Σfx = 168 marks (the Total column).
- Mean = 168 ÷ 32 = 5.25.
- Median position = (32 + 1) ÷ 2 = 16.5, so the 16th and 17th values.
- Running totals of the frequencies: 1, 3, 6, 8, 11, 16, 20. The total reaches 16 at score 5, so the 16th value is 5. The 17th is in score 6.
- Median = (5 + 6) ÷ 2 = 5.5.
- The biggest frequency is 6, at score 7. Mode = 7.
- Range = 10 − 0 = 10.
📘 Example 7 · Goals in a match: 0 goals 4 times, 1 goal 6 times, 2 goals 8 times, 3 goals 2 times
- Σf = 4 + 6 + 8 + 2 = 20 matches.
- f × goals: 0, 6, 16, 6. Σfx = 28.
- Mean = 28 ÷ 20 = 1.4.
- Median position = 10.5. Running totals: 4, 10, 18. The 10th is 1 and the 11th is 2.
- Median = (1 + 2) ÷ 2 = 1.5.
- The biggest frequency is 8, at 2 goals. Mode = 2.
💡 The median is the middle value, not the middle of the frequency column. Never take the middle of the row of frequencies.
✏️ Try it
- Goals 0, 1, 2, 3 have frequencies 4, 6, 8, 2. State the mode and range.
- Shoe size 5, 6, 7 has frequencies 4, 10, 6. Find the mean.
- A table has Σf = 25 and Σfx = 100. Find the mean.
Answers
1. Mode 2 (frequency 8), range 3 − 0 = 3 · 2. Σf = 20, Σfx = 20 + 60 + 42 = 122, mean 6.1 · 3. 100 ÷ 25 = 4
The mean for grouped data
🔑 Key ideas
- In a grouped table you know which class each value is in, but not the exact value.
- So the mean is only an estimate: pretend every value in a class is at the mid-interval value (the middle of the class).
- Midpoint = (lower end + upper end) ÷ 2.
- The modal class is the class with the greatest frequency (when the class widths are equal).
| Score S | Frequency f | Mid-interval x | f × x |
|---|---|---|---|
| 0 ≤ S ≤ 19 | 2 | 9.5 | 19 |
| 20 ≤ S ≤ 39 | 4 | 29.5 | 118 |
| 40 ≤ S ≤ 59 | 14 | 49.5 | 693 |
| 60 ≤ S ≤ 79 | 16 | 69.5 | 1112 |
| 80 ≤ S ≤ 99 | 4 | 89.5 | 358 |
| Total | 40 | 2300 |
📘 Example 8 · The history test table above. Estimate the mean and find the modal class
- Midpoints: (0 + 19) ÷ 2 = 9.5, then 29.5, 49.5, 69.5, 89.5.
- f × x: 2 × 9.5 = 19, 4 × 29.5 = 118, 14 × 49.5 = 693, 16 × 69.5 = 1112, 4 × 89.5 = 358.
- Add: 19 + 118 + 693 + 1112 + 358 = 2300. Σf = 2 + 4 + 14 + 16 + 4 = 40.
- Estimated mean = 2300 ÷ 40 = 57.5.
- The greatest frequency is 16, so the modal class is 60 ≤ S ≤ 79.
📘 Example 9 · Finger lengths: 5.0− (3 students), 5.5− (8), 6.0− (10), 6.5− (7), 7.0−7.5 (2). Note: 5.0− means 5.0 ≤ L < 5.5
- Each class is 0.5 wide. Midpoints: 5.25, 5.75, 6.25, 6.75, 7.25.
- f × x: 15.75, 46, 62.5, 47.25, 14.5.
- Σfx = 186, Σf = 30.
- Estimated mean = 186 ÷ 30 = 6.2 cm.
- Greatest frequency 10, so the modal class is 6.0 ≤ L < 6.5.
🧪 Try it: averages calculator
✏️ Try it
- Classes 0 ≤ x < 10, 10 ≤ x < 20, 20 ≤ x < 30 have frequencies 5, 15, 10. Estimate the mean.
- What is the modal class in question 1?
- Classes 0−4, 5−9, 10−14 (whole numbers) have frequencies 2, 6, 2. Estimate the mean.
Answers
1. Midpoints 5, 15, 25: 25 + 225 + 250 = 500; 500 ÷ 30 = 16.7 (3 s.f.) · 2. 10 ≤ x < 20 (frequency 15) · 3. Midpoints 2, 7, 12: 4 + 42 + 24 = 70; 70 ÷ 10 = 7
Chapter summary
- Mode: most frequent value. Median: middle value of the sorted data, at position (n + 1) ÷ 2. Mean: total ÷ number of values.
- Range = largest − smallest. It measures spread.
- Total = mean × number of values. Use it to find a missing value or a new mean.
- Frequency table: mean = Σfx ÷ Σf. The mode is the value with the biggest frequency.
- Grouped data: use mid-interval values, so the mean is only an estimate. The modal class has the greatest frequency.
- An extreme value pulls the mean but barely moves the median.
📄 Practice questions
The book's Exercises 35.1 to 35.3 and Student assessment 1 are in the printable assignment for this chapter.
Mini-quiz
Source: Cambridge IGCSE Mathematics Core and Extended (4th edition, Pimentel and Wall), Chapter 35 (Mean, median, mode and range). Explanations, the Try it answers, the tool and the quiz are Guru Jazzy's own. This page is not made or endorsed by Cambridge or Hodder Education.