Learn how to find and compare the mean, median and mode from lists, frequency tables and grouped data, work backwards from a known mean, combine groups, and choose the fairest measure of the centre.
When a data set is large, it is hard to understand by reading every value. A measure of central tendency gives one number that describes where the data are centred. The three most common are the mean, the median and the mode. They often give different answers, because each one answers a different question. This guide explains how to find each measure from a list, from a frequency table and from grouped data, how to work backwards from a known mean, how to combine groups, and how to decide which measure is the fairest summary. It has 25 worked examples checked by computer, diagrams, three calculators, a practice drill, exam questions with answers, and a 20-question quiz.
Central tendency means a typical or central value of a data set. Instead of listing every observation, we use one representative number.
| Measure | Basic idea | How to find it |
|---|---|---|
| Mean | The equal-share value | Add all the values and divide by how many there are |
| Median | The middle value | Put the values in order and take the middle one |
| Mode | The most common value | Find the value that occurs most often |
In everyday speech, “the average” usually means the mean, but news reports and advertisements sometimes use it for the median or the mode without saying so. In mathematics you should always name the measure you are using. The mean of a data set is often written x̄, read “x bar”.
In symbols, x̄ = Σx ÷ n, where Σx (“sigma x”) means the sum of the values and n is the number of values.
(a) Find the mean of 6, 8, 10, 12 and 14. The sum is 6 + 8 + 10 + 12 + 14 = 50 and there are 5 values, so the mean is 50 ÷ 5 = 10.
(b) Find the mean of 7, 9, 4 and 12. The sum is 32 and there are 4 values, so the mean is 8.
(c) Find the mean of 2, 3, 5 and 9. The sum is 19, so the mean is 19 ÷ 4 = 4.75. The mean does not have to be one of the original values, or even a whole number.
Imagine towers of blocks with 4, 6, 8, 10 and 12 blocks. There are 40 blocks altogether. If you share them equally among the five towers, each tower gets 8 blocks. That is the mean. Another way to see it: if the values sit on a number line, the mean is the point where the line would balance, because the distances to the left of the mean cancel the distances to the right (Figure 1).
Five containers hold 4, 6, 8, 10 and 12 litres. Total = 40 litres. Shared equally, each gets 40 ÷ 5 = 8 litres, so the mean is 8. Check: the differences from 8 are −4, −2, 0, +2, +4, and their sum is 0. The differences from the mean always add up to zero.
The median is the middle value when the data are arranged in order. The most important rule is: sort the data first. Unsorted data have no “middle”.
Data: 9, 3, 7, 5, 11. Ordered: 3, 5, 7, 9, 11. There are 5 values, so the 3rd value is the middle one. Median = 7.
When there are two middle values, the median is the mean of those two.
Data: 4, 12, 7, 9, 15, 5. Ordered: 4, 5, 7, 9, 12, 15. The middle values are 7 and 9, so the median is (7 + 9) ÷ 2 = 8. Notice that 8 is not in the list.
For n ordered values, if n is odd the median is at position (n + 1) ÷ 2. If n is even, the middle values are at positions n ÷ 2 and (n ÷ 2) + 1.
9 values: position (9 + 1) ÷ 2 = 5th. 101 values: position (101 + 1) ÷ 2 = 51st. 10 values: positions 5th and 6th. 20 values: positions 10th and 11th.
What did the writer forget to do first?
The fix: the data must be put in order first: 2, 4, 5, 7, 9. The 3rd value is then 5, so the median is 5. The 3rd value of the unsorted list has no special meaning.
The mode is the value that occurs most often. Finding it needs counting, not calculating, so it also works for categories.
(a) Data: 2, 4, 4, 5, 7, 4, 9. The value 4 occurs three times, and every other value occurs less often. Mode = 4.
(b) Data: 2, 2, 4, 5, 5, 7, 8. Both 2 and 5 occur twice. The data are bimodal and the modes are 2 and 5.
(c) Data: 1, 2, 3, 4. Every value occurs once, so there is no mode.
If more than two values share the highest frequency, the data are sometimes called multimodal. Some teachers prefer to say there is no mode when every value occurs equally often. Follow the convention used in your own course.
Is the mode the count or the value?
The fix: the mode is the value that occurs most often, not how many times it occurs. The value 2 occurs 3 times, so the mode is 2. The number 3 is the frequency of the mode.
Data: 2, 3, 3, 4, 8. Mean = 20 ÷ 5 = 4. Median = the 3rd value = 3. Mode = 3, because it occurs twice. The three answers differ because they measure different things.
| Feature | Mean | Median | Mode |
|---|---|---|---|
| Uses every value? | Yes | No (only the middle) | No (only the most common) |
| Needs ordered data? | No | Yes | No |
| Always exists? | Yes, for numbers | Yes, for ordered data | Not always, and there may be several |
| Affected by extreme values? | Strongly | Little | Usually not |
| Works for categories? | No | Only if the categories can be put in order | Yes |
The mean uses every value, so one unusually large or small value pulls it towards itself. The median depends only on the middle, so it hardly moves.
Set A: 10, 11, 12, 13, 14 has mean 12 and median 12. Replace 14 with 100 to get 10, 11, 12, 13, 100. The sum is 146, so the mean is 146 ÷ 5 = 29.2, but the median is still 12. Four of the five values are close to 12, so 29.2 is not a fair picture of a typical value.
Five people in a small business earn Rs 30 000, Rs 32 000, Rs 35 000, Rs 36 000 and Rs 200 000 a month. The total is Rs 333 000, so the mean is Rs 66 600. The median is the 3rd value, Rs 35 000. Four of the five earn between Rs 30 000 and Rs 36 000, so the median describes a typical wage better. An owner who wants to say that “the average wage is high” might choose the mean.
A data set with a long tail on the right (a few very large values) is called right-skewed, and the mean is usually pulled above the median. With a long tail on the left, it is left-skewed, and the mean is usually pulled below the median. Incomes and house prices are often right-skewed. For symmetrical data with one peak, the mean, median and mode are close to each other or equal, but their being close does not prove the data are symmetrical.
A batter scores 12, 45, 0, 78, 23, 45 and 31 in seven innings. Ordered: 0, 12, 23, 31, 45, 45, 78. The sum is 234, so the mean is 234 ÷ 7 = 33.43 (2 decimal places). The median is 31, the mode is 45, and the range is 78 − 0 = 78.
When values repeat, a frequency table is shorter than a list. Here f is the frequency and x is the value.
Multiply each value by its frequency to get fx, then add the fx column and divide by the total frequency. For the median, add up the frequencies cumulatively and find which value the middle position falls in. The mode is the value with the highest frequency.
| Value x | Frequency f | fx |
|---|---|---|
| 2 | 3 | 6 |
| 4 | 2 | 8 |
| 6 | 4 | 24 |
| 8 | 1 | 8 |
| Total | 10 | 46 |
Mean = 46 ÷ 10 = 4.6. For the median, write the values out: 2, 2, 2, 4, 4, 6, 6, 6, 6, 8. There are 10 values, so use the 5th and 6th, which are 4 and 6. Median = (4 + 6) ÷ 2 = 5. The highest frequency is 4, belonging to the value 6, so the mode is 6.
| Value | Frequency | Cumulative frequency |
|---|---|---|
| 1 | 2 | 2 |
| 2 | 3 | 5 |
| 3 | 4 | 9 |
| 4 | 1 | 10 |
The total frequency is 10, so the middle positions are 5 and 6. The cumulative frequencies show that the 5th value is 2 (positions 3 to 5 are 2) and the 6th value is 3 (positions 6 to 9 are 3). Median = (2 + 3) ÷ 2 = 2.5. The mean is (2 + 6 + 12 + 4) ÷ 10 = 2.4 and the mode is 3.
What has been ignored?
The fix: the frequencies were ignored. The value 3 occurs four times but was counted once. The correct mean is Σfx ÷ Σf = 24 ÷ 10 = 2.4. Never divide by the number of rows. Divide by the total frequency.
Since Mean = Sum ÷ Number, we also have Sum = Mean × Number of values. This one idea solves many exam questions.
The mean of five values is 18. Four values are 12, 16, 20 and 21. The required total is 18 × 5 = 90. The known total is 69. The missing value is 90 − 69 = 21.
Five values have mean 12. A sixth value, 18, is added. Original total = 60. New total = 78. New mean = 78 ÷ 6 = 13.
Eight values have mean 15. One value, 22, is removed. Original total = 120. New total = 98. Seven values remain, so the new mean is 98 ÷ 7 = 14.
The mean of 5 numbers is 8. Three of them have a mean of 6. Find the mean of the other two. Total of all five = 5 × 8 = 40. Total of the three = 3 × 6 = 18. Total of the other two = 40 − 18 = 22. Their mean is 22 ÷ 2 = 11.
Group A has 20 values with mean 12. Group B has 30 values with mean 18. Totals: 20 × 12 = 240 and 30 × 18 = 540. Combined mean = 780 ÷ 50 = 15.6.
One group has 10 values with mean 20. Another has 90 values with mean 30. The simple average of the means is (20 + 30) ÷ 2 = 25, which wrongly treats the groups as equal in size. The correct combined mean is (10 × 20 + 90 × 30) ÷ 100 = 29.
When is this correct?
The fix: averaging two means is correct only when both groups have the same size. If class A has 10 students and class B has 30, the combined mean is (10 × 70 + 30 × 80) ÷ 40 = 77.5, closer to the larger group’s mean. Always combine the totals.
A weighted mean gives each value a weight showing how important it is.
(a) Scores 70, 80 and 90 have weights 2, 3 and 5. Weighted total = 140 + 240 + 450 = 830. Total weight = 10. Weighted mean = 83.
(b) A course gives a mark of 72 a weight of 3 and a mark of 85 a weight of 2. Weighted mean = (216 + 170) ÷ 5 = 386 ÷ 5 = 77.2. The simple mean, 78.5, would give both marks equal importance.
When data are grouped into classes, such as 10 ≤ t < 20 (“10 up to but not including 20”), we no longer know the exact values. We can only estimate the mean and the median. For the mode we can name the modal class, the class with the highest frequency.
The journey times, in minutes, of 30 students are grouped below.
| Time t (minutes) | Frequency f | Midpoint | f × midpoint | Cumulative frequency |
|---|---|---|---|---|
| 0 ≤ t < 10 | 4 | 5 | 20 | 4 |
| 10 ≤ t < 20 | 9 | 15 | 135 | 13 |
| 20 ≤ t < 30 | 12 | 25 | 300 | 25 |
| 30 ≤ t < 40 | 5 | 35 | 175 | 30 |
| Total | 30 | 630 |
Estimated mean: 630 ÷ 30 = 21 minutes.
Modal class: the highest frequency is 12, so the modal class is 20 ≤ t < 30.
Estimated median: N ÷ 2 = 15, and the cumulative frequencies 4, 13, 25, 30 show that the 15th value is in the class 20 ≤ t < 30. Here L = 20, CF = 13, f = 12 and h = 10. Median ≈ 20 + (15 − 13) ÷ 12 × 10 = 20 + 1.67 = 21.7 minutes (1 decimal place).
Estimated mode: f₁ = 12, f₀ = 9, f₂ = 5. Mode ≈ 20 + (12 − 9) ÷ (24 − 9 − 5) × 10 = 20 + 3 = 23 minutes.
The median can also be read from a cumulative frequency graph. Plot each cumulative frequency at the upper boundary of its class, join the points, draw a line across from N ÷ 2 = 15, and read the value below where it meets the curve (Figure 5). The graph gives about 21.7, matching the calculation. These are estimates, because we never knew the exact times inside each class.
Some data are categories, such as transport type or favourite fruit. Nominal categories have no natural order, so the mean and median make no sense. The mode still tells you the most common category.
Transport choices: bus, car, bus, bicycle, bus, train, car. Bus occurs 3 times, car 2 times, and the others once. The mode is bus.
If the same number is added to every value, the mean, median and mode all go up by that number, and the range does not change. If every value is multiplied by the same number, the mean, median and mode are all multiplied by it as well.
Data: 4, 6, 8, 8, 10 has mean 7.2, median 8 and mode 8. Add 5 to every value: 9, 11, 13, 13, 15 has mean 12.2, median 13 and mode 13.
Data: 2, 3, 3, 5, 7 has mean 4, median 3 and mode 3. Multiply every value by 2: 4, 6, 6, 10, 14 has mean 8, median 6 and mode 6.
This is useful when converting units. If the mean of some lengths is 12 cm, the mean in millimetres is 120 mm.
The range is a measure of spread, not of centre: Range = Maximum − Minimum. Two data sets can have the same mean and be very different.
For 5, 6, 6, 7, 16: the mean is 8, the median is 6, the mode is 6, and the range is 16 − 5 = 11.
Team A scores 40, 50 and 60. Team B scores 10, 50 and 90. Both have mean 50. The range for A is 20 and for B is 80, so A is much more consistent. A mean alone would hide that.
A mean must lie between the smallest and largest values (it can equal one of them only if all values are equal). The median and the mode always lie within the same range. If your answer falls outside it, there is an error.
All the values lie between 20 and 50, but a student gets a mean of 57. That cannot be right. Recheck the addition and the number of values you divided by.
| Situation | Useful measure | Why |
|---|---|---|
| Numbers with no extreme values | Mean | Uses every value |
| Numbers with extreme values or a skewed shape (wages, house prices) | Median | Not pulled by the extremes |
| The most popular size, colour or choice | Mode | Shows the most common item |
| Categories with no order | Mode | Mean and median are meaningless |
| Values with different importance | Weighted mean | Weights show importance |
A shoe shop wants to know which size to stock the most of. The mode (the most common size) answers that. A town planner studying household income might report the median, because a few very high incomes would drag the mean upwards. A teacher working out a final grade from tests with different weights uses a weighted mean.
Whenever someone quotes an “average”, ask which one it is, how many values it is based on, and how spread out the data are. The same data can be described honestly with the median and misleadingly with the mean, or the other way round.
Type a list of numbers separated by commas or spaces. You will see the ordered list, the mean, the median with its working, the mode and the range. Try 4, 12, 7, 9, 15, 5 (an even number of values), then 10, 11, 12, 13, 100 to see an extreme value pull the mean.
Type one row per line as value, frequency. You get the mean (Σfx ÷ Σf), the median from the cumulative frequencies, and the mode.
Type one class per line as lower boundary, upper boundary, frequency. The classes must join up. The tool estimates the mean and the median, names the modal class and, when it can, estimates the mode.
Tap the correct answer for each question. Your best score is saved on this device.
1. Ordered: 3, 4, 5, 6, 9, 9. The sum is 36, so the mean is 36 ÷ 6 = 6. The median is (5 + 6) ÷ 2 = 5.5. The mode is 9.
2. Original total = 4 × 15 = 60. New total = 85. New mean = 85 ÷ 5 = 17.
3. Σfx = 2 + 10 + 9 = 21 and Σf = 10. Mean = 2.1.
1. Ordered: 2, 2, 2, 5, 6, 8, 9, 10. The sum is 44, so the mean is 44 ÷ 8 = 5.5. The median is (5 + 6) ÷ 2 = 5.5. The mode is 2. The range is 10 − 2 = 8.
2. Total of five = 40. Total of three = 18. Total of the other two = 22. Their mean is 11.
3. Σf = 24 and Σfx = 0 + 8 + 12 + 9 + 8 = 37. (a) Mean = 37 ÷ 24 = 1.54. (b) Use the 12th and 13th values. The cumulative frequencies are 5, 13, 19, 22, 24, so both are 1 and the median is 1. (c) The highest frequency is 8, so the mode is 1.
4. Totals: 12 × 60 = 720 and 8 × 75 = 600. Combined mean = 1320 ÷ 20 = 66.
5. (a) The highest frequency is 14, so the modal class is 160 ≤ h < 170. (b) Midpoints 155, 165, 175, 185 give Σfx = 930 + 2310 + 1400 + 370 = 5010. Mean ≈ 5010 ÷ 30 = 167. (c) N ÷ 2 = 15. The cumulative frequencies are 6, 20, 28, 30, so the median class is 160 ≤ h < 170. Median ≈ 160 + (15 − 6) ÷ 14 × 10 = 160 + 6.43 = 166.4.
6. Mean = 333 000 ÷ 5 = Rs 66 600. Median = Rs 35 000. The median is more typical, because four of the five earn between Rs 30 000 and Rs 36 000 and the one very high wage pulls the mean up.
7. Old total = 6 × 9 = 54. New total = 7 × 10 = 70. The seventh number is 70 − 54 = 16.
| Task | Formula or rule |
|---|---|
| Mean | Σx ÷ n |
| Mean from a frequency table | Σfx ÷ Σf |
| Total from a known mean | Mean × number of values |
| Median, n odd | Value at position (n + 1) ÷ 2 |
| Median, n even | Mean of the values at positions n ÷ 2 and (n ÷ 2) + 1 |
| Mode | Most frequent value or category |
| Range | Maximum − minimum |
| Weighted mean | Σ(weight × value) ÷ Σ(weights) |
| Midpoint of a class | (lower boundary + upper boundary) ÷ 2 |
| Estimated mean, grouped data | Σ(f × midpoint) ÷ Σf |
| Estimated median, grouped data | L + [(N ÷ 2 − CF) ÷ f] × h |
The “equal share” idea is the best way in. Give learners towers of linking cubes of different heights and ask them to level the towers by moving cubes: the final height is the mean. For the median, ask learners to stand in a line in order of height and find the middle person. For the mode, make a quick class survey (for example, how learners travel to school) and ask which is the most common. Then collect a real set of measurements, such as arm spans or journey times, and ask learners to predict whether the mean or the median will be higher before they calculate. The Data Calculator, Frequency Table Calculator and Grouped Data Calculator on this page let learners check their own working, including the working for a median with an even number of values. Encourage learners to say in a sentence what each number means.
This article is original EDUSAMBAM writing. Examination requirements differ between boards, so compare your practice with the specification of your own board.
20 questions covering mean, median, mode, frequency tables, missing values, grouped data and interpretation. Answer every question, then submit to see your score instantly.