Open a spreadsheet with the marks of 60 students or the daily sales of a shop and you see a wall of numbers. It is very hard to say what the data is telling you. Measures of central tendency fix this: each one gives a single number that stands for the “centre” or the “typical value” of the whole data. The three most important ones are the mean, the median and the mode. In this chapter you will learn what each one means, how to calculate it for raw data and for frequency distributions, how they are related, and which one to trust in which situation. Every idea is followed by fully worked examples, and the chapter ends with 15 practice questions from basic level to GATE DA level.
Consider the five numbers 2, 3, 3, 4, 18. Their mean is 2 + 3 + 3 + 4 + 185 = 305 = 6, their median (middle value) is 3 and their mode (most repeated value) is 3. Which one describes the data best? The single large value 18 drags the mean up, while the median and the mode stay with the bulk of the data. By the end of this chapter you will know exactly why this happens and how to decide.
In this article
The arithmetic mean (or simply the mean or average) of a set of observations is the sum of all the observations divided by the number of observations.
Think of the mean as the fair share. Three friends have ₹20, ₹30 and ₹70. If they pool all the money (₹120) and share it equally, each gets ₹1203 = ₹40. That equal share, 40, is the mean.
Two quantities are needed for the mean. Write the observations as x1, x2, …, xn.
Here is read “x bar” and stands for the mean.
The marks of 8 students in a test are 14, 17, 19, 21, 23, 25, 27, 30. Find the mean mark.
The mean mark is 22.
A shopkeeper’s sales (in ₹) over five days were 120, 150, 135, 160, 145. Find the average daily sale.
The average daily sale is ₹142.
When a value is repeated, adding it again and again is wasteful. If the value x occurs f times, we simply use f · x. The mean then becomes
Here Σf = N is the total frequency (the total number of observations). For a grouped frequency distribution (class intervals), x is taken as the class mark, the middle of the class: class mark = lower limit + upper limit2.
Find the mean of the following distribution.
| x | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|
| f | 4 | 6 | 8 | 5 | 2 |
| x | f | f · x |
|---|---|---|
| 2 | 4 | 8 |
| 4 | 6 | 24 |
| 6 | 8 | 48 |
| 8 | 5 | 40 |
| 10 | 2 | 20 |
| Total | 25 | 140 |
The mean is 5.6.
The marks of 50 students are grouped as shown. Find the mean mark. We will use this same table to find the median and the mode later in the chapter.
| Class | 0–10 | 10–20 | 20–30 | 30–40 | 40–50 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 15 | 16 | 6 |
| Class | Frequency (f) | Class mark (x) | f · x |
|---|---|---|---|
| 0–10 | 5 | 5 | 25 |
| 10–20 | 8 | 15 | 120 |
| 20–30 | 15 | 25 | 375 |
| 30–40 | 16 | 35 | 560 |
| 40–50 | 6 | 45 | 270 |
| Total | 50 | 1350 |
The mean mark is 27.
Sometimes the observations are not equally important. In a course, the final exam may count more than a quiz. We then give every observation a weight w that says how important it is, and use the weighted mean:
If all weights are equal, the weighted mean is the same as the ordinary mean.
A student scores 80 in assignments (weight 20%), 70 in the mid-term (weight 30%) and 90 in the final exam (weight 50%). Find the weighted mean score.
The weighted mean score is 82. (The simple mean of 80, 70, 90 would be 80, which ignores the importance of the final exam.)
A deviation is the difference of an observation from the mean, x − . Some deviations are positive and some are negative, and they cancel exactly:
Why? Σ(x − ) = Σx − n = Σx − Σx = 0, because n = Σx.
Check: for 3, 5, 7, 9, 11 the mean is 7. The deviations are -4, -2, 0, 2, 4, and their sum is (−4) + (−2) + 0 + 2 + 4 = 0.
If a constant k is added to every observation, the mean increases by k. If k is subtracted, the mean decreases by k.
Check: add 5 to 3, 5, 7, 9, 11 to get 8, 10, 12, 14, 16. The new mean is 605 = 12 = 7 + 5.
If every observation is multiplied by k, the mean is multiplied by k. If every observation is divided by k, the mean is divided by k.
Check: multiply 3, 5, 7, 9, 11 by 3 to get 9, 15, 21, 27, 33. The new mean is 1055 = 21 = 3 · 7.
The mean uses every observation, so a single very large or very small value (an outlier) can change it a lot.
Check: the mean of 10, 12, 14, 16, 18 is 705 = 14. Replace 18 by 98 and the data becomes 10, 12, 14, 16, 98. The new mean is 1505 = 30, more than double, although four of the five values did not change.
If the mean of n observations is known, the total is fixed: sum of all observations = n × mean. Subtract the known observations from this total to get the missing one.
The mean of five observations is 24. Four of them are 18, 22, 25, 30. Find the fifth observation.
The fifth observation is 25.
The mean weight of 6 players is 62 kg. The weights of five of them are 58, 60, 65, 61, 66 kg. Find the weight of the sixth player.
The sixth player weighs 62 kg.
If we know the size and mean of each group, we can find the mean of all the groups together without the original data. The idea: the total of a group is (size) × (mean), so we add the totals and divide by the total size.
For k groups: c = Σ niiΣ ni. Note that the combined mean is not the simple average of the group means unless the groups have equal sizes.
Section A has 40 students with mean marks 60, and Section B has 60 students with mean marks 70. Find the mean marks of all 100 students.
The combined mean is 66. (The simple average of 60 and 70 is 65, which is wrong because Section B is larger.)
Three groups have 10, 15 and 25 observations with means 20, 30 and 40 respectively. Find the combined mean.
The combined mean is 33.
The median is the middle value of the data when the observations are arranged in order. Half of the observations are smaller than or equal to it, and half are larger than or equal to it.
The median can only be found after the observations are arranged in ascending order (smallest to largest) or descending order. Skipping this step is the most common mistake. Once the data is ordered, the answer depends on whether n is odd or even.
If n is odd there is exactly one middle value.
Find the median of 7, 3, 12, 5, 9, 15, 11.
The median is 9.
If n is even there are two middle values. The median is their average.
Find the median of 18, 25, 12, 30, 22, 16.
The median is 20. Notice that 20 is not one of the observations; with an even number of values the median can fall between two of them.
Writing out every repeated value would take very long. Instead we use the cumulative frequency (cf), the running total of the frequencies.
The number of goals scored by a team in 50 matches is given below. Find the median.
| Goals (x) | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Matches (f) | 4 | 7 | 10 | 14 | 8 | 7 |
| x | f | cf |
|---|---|---|
| 1 | 4 | 4 |
| 2 | 7 | 11 |
| 3 | 10 | 21 |
| 4 | 14 | 35 |
| 5 | 8 | 43 |
| 6 | 7 | 50 |
The median number of goals is 4.
For grouped data we first find the median class, the class in which the N2th observation lies. Then we assume the observations in that class are spread evenly and place the median inside it:
The median class is the first class whose cumulative frequency is greater than or equal to N2. The classes must be continuous (for example 0–10, 10–20, …). If they are written as 0–9, 10–19, …, first convert them to class boundaries (−0.5–9.5, 9.5–19.5, …).
Find the median of the 50 marks in the table of Solved Example 4.
| Class | Frequency (f) | Cumulative frequency (cf) |
|---|---|---|
| 0–10 | 5 | 5 |
| 10–20 | 8 | 13 |
| 20–30 | 15 | 28 |
| 30–40 | 16 | 44 |
| 40–50 | 6 | 50 |
The median mark is 28.
Find the median of the following distribution.
| Class | 10–20 | 20–30 | 30–40 | 40–50 | 50–60 |
|---|---|---|---|---|---|
| Frequency | 4 | 8 | 20 | 12 | 6 |
| Class | Frequency (f) | Cumulative frequency (cf) |
|---|---|---|
| 10–20 | 4 | 4 |
| 20–30 | 8 | 12 |
| 30–40 | 20 | 32 |
| 40–50 | 12 | 44 |
| 50–60 | 6 | 50 |
The median is 36.5.
Repeated values must be kept, each one counted as a separate observation. Never remove duplicates before finding the median.
Find the median of 5, 7, 7, 7, 9, 10, 10, 12.
The median is 8. (If we had wrongly removed the repeated 7s and 10, we would have got the median of 5, 7, 9, 10, 12, which is 9.)
The median depends only on the middle of the ordered data, so it is hardly affected by extreme values. Go back to 10, 12, 14, 16, 18: the median is 14 and the mean is 14. Replace 18 by 98 to get 10, 12, 14, 16, 98. The median is still the middle (3rd) value, 14, while the mean jumped to 30. This is why the median is preferred for incomes, house prices and other data with a few very large values.
Write the ordered data with the unknown in its correct position, write the median in terms of the unknown, and solve. Afterwards check that the answer keeps the ordering you assumed.
The observations 4, 7, 9, x, 15, 18 are arranged in ascending order and their median is 11. Find x.
x = 13
The observations 12, 15, x + 2, 22, 25 are in ascending order and their median is 19. Find x.
x = 17
The mode is the observation that occurs most often, that is, the value with the highest frequency. In a grouped distribution, the class with the highest frequency is called the modal class.
The mode answers questions such as “which shoe size sells the most?” or “which score did most students get?”. It is the only one of the three measures that can also be used for non-numerical data, such as the most popular colour.
Count how many times each value appears (a tally helps) and pick the value with the largest count.
Find the mode of 3, 5, 7, 5, 8, 5, 9, 7.
| Value | 3 | 5 | 7 | 8 | 9 |
|---|---|---|---|---|---|
| Times it occurs | 1 | 3 | 2 | 1 | 1 |
The mode is 5.
The mode is simply the value with the largest frequency. No calculation is needed.
A shop sold the following number of pairs of shoes in one week. Find the modal shoe size.
| Shoe size | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|
| Pairs sold | 6 | 14 | 25 | 12 | 3 |
The modal shoe size is 7.
First find the modal class (the class with the highest frequency). The mode lies inside it, and its exact position depends on the frequencies of the neighbouring classes:
Intuition: if the class after the modal class is more crowded than the class before it, the mode is pulled towards the upper side, and the other way round.
Find the mode of the 50 marks in the table of Solved Example 4.
The mode is about 30.91.
Find the mode of the distribution in Solved Example 14.
The mode is 36.
| Type | Meaning | Example | Mode(s) |
|---|---|---|---|
| Unimodal | One value occurs most often | 2, 3, 3, 3, 5, 7 | 3 |
| Bimodal | Two values tie for the highest frequency | 2, 3, 3, 5, 7, 7, 9 | 3 and 7 |
| Multimodal | More than two values tie for the highest frequency | 1, 1, 2, 2, 3, 3, 4 | 1, 2 and 3 |
If every value occurs the same number of times, no value stands out and the data has no mode. Example: 4, 6, 8, 9, 12, where each value occurs once. (The word “no mode” is correct here; it is wrong to say the mode is 0.)
| Advantages | Limitations |
|---|---|
|
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The mode depends only on how often a value occurs, not on how large or small the other values are. In 3, 5, 5, 6, 7, 5, 100 the mode is 5. If 100 is changed to 1000, or to −50, the mode is still 5. So, like the median, the mode is not affected by extreme values.
For data whose shape is moderately skewed and has a single peak, the three measures are connected by a handy approximate rule discovered by the statistician Karl Pearson:
The same rule can be written in other useful forms:
| To find | Formula |
|---|---|
| Mode | Mode = 3 · Median − 2 · Mean |
| Median | Median = Mode + 2 · Mean3 |
| Mean | Mean = 3 · Median − Mode2 |
| Gap form | Mean − Mode = 3 (Mean − Median) |
So the distance from the mode to the median is about twice the distance from the median to the mean. The diagram below shows the order of the three measures in different shapes.
Solid line = mean, long dashes = median, dots = mode. The mean is pulled the most towards the long tail, the mode the least, and the median stays in between.
“Empirical” means based on observation, not on proof. The one-third rule was found by looking at many data sets; it was not derived from a theorem that holds for every data set. That is why we write ≈ (approximately equal) and not =.
| Use it when | Do not use it when |
|---|---|
|
|
Take the small, strongly skewed data 1, 2, 2, 3, 10. Mean = 185 = 3.6, median = 2, and the true mode = 2. The rule would give 3 · 2 − 2 · 3.6 = 6 − 7.2 = −1.2, which is not even a value in the data. For well-behaved, moderately skewed data the rule is close; for data like this it fails.
For the 50 marks in Solved Example 4 we found mean = 27, median = 28 and (by the modal-class formula) mode ≈ 30.91. The empirical rule gives
which is very close to 30.91. The data is unimodal and moderately skewed (mean < median < mode, a slight tail on the left), exactly where the rule works well.
Substitute the given mean and median into Mode = 3 · Median − 2 · Mean. Always state clearly that the answer is an estimate.
For a moderately skewed distribution the mean is 42 and the median is 40. Estimate the mode.
The estimated mode is 36.
For a moderately skewed distribution the mean is 18 and the mode is 12. Estimate the median.
The estimated median is 16.
For a moderately skewed distribution the median is 30 and the mode is 36. Estimate the mean.
The estimated mean is 27.
| Feature | Mean | Median | Mode |
|---|---|---|---|
| Definition | Sum of observationsNumber of observations | Middle value of the ordered data | Most frequent value |
| How it is calculated | = Σxn (or ΣfxΣf) | Order the data and pick the middle position (or use the grouped formula) | Count frequencies (or use the modal-class formula) |
| Uses every value? | Yes | No, only the position | No, only the frequencies |
| Effect of extreme values | Strongly affected | Hardly affected | Not affected |
| Always exists and is unique? | Yes | Yes | No: may be absent or have several values |
| Suitable data | Numerical, fairly symmetric data without outliers | Numerical, skewed data or data with outliers; ordered data | Any data, including categories and colours |
| Advantages | Easy, uses all data, good for further calculations | Robust to outliers, can be found with open-ended classes | Simplest, works for non-numerical data |
| Disadvantages | Misleading when outliers are present | Needs ordering, ignores the size of other values | May be missing or not unique, ignores most data |
| Real-life use | Average marks, average rainfall, batting average | Household income, house prices, waiting times | Most common shoe size, most popular product, most frequent score |
| Shape of the data | Order of the measures | Why |
|---|---|---|
| Symmetric | Mean = Median = Mode | Both sides of the peak are mirror images |
| Positively skewed (long tail to the right) | Mode < Median < Mean | Large values drag the mean to the right |
| Negatively skewed (long tail to the left) | Mean < Median < Mode | Small values drag the mean to the left |
Refer to the diagram in the previous section: the mean always moves farthest towards the tail, and the median always stays between the mean and the mode.
