Central tendency of ungrouped data

A cricket coach wants one number that sums up a batter's last eleven innings. A teacher wants one number that sums up a class's test scores. Neither wants to read out eleven separate values every time — they want a single, typical figure that represents the whole set. That is exactly what a measure of central tendency does, and for raw, un-tabulated data there are three standard ways to get it: the mean, the median, and the mode. This article stays entirely with ungrouped data — individual values listed one by one, not bundled into class intervals — which is prerequisite recall for ICSE Class 10 Statistics before grouped-data techniques (mean of grouped data, median class, modal class) are introduced. Getting comfortable here, including why each measure behaves the way it does, makes every later grouped-data method feel like a natural extension rather than a new idea.

The core insight

The mean is the arithmetic average of all values, the median is the middle value of the data once arranged in order (with a different rule for odd and even counts), and the mode is the most frequently occurring value — three different "typical values" that agree for symmetric data but diverge for skewed or outlier-heavy data.

The core idea

Mean. For $n$ raw values $x_1, x_2, \ldots, x_n$ , the mean adds everything up and divides by the count.

\bar x = \frac{\Sigma x}{n} = \frac{x_1+x_2+\cdots+x_n}{n}

where: $\bar x$ is the mean, $\Sigma x$ is the sum of all values, and $n$ is the number of values. Every single value pulls on $\bar x$ , which is both its strength (it uses all the information) and its weakness (one extreme outlier can drag it far from where most of the data actually sits).

Median. Before finding the median, the data must first be arranged in ascending or descending order — this ordered list is called an array. Once arrayed, the position of the middle term depends on whether $n$ is odd or even.

n \text{ odd: Median} = \left(\frac{n+1}{2}\right)\text{th term} \qquad n \text{ even: Median} = \frac{\left(\frac{n}{2}\right)\text{th term} + \left(\frac{n}{2}+1\right)\text{th term}}{2}

where: $n$ is the number of values in the array. For odd $n$ there is a single middle term; for even $n$ there is no single middle term, so the two central values are averaged. Because the median depends only on position in the array rather than the exact size of every value, one wildly large or small entry barely moves it.

Mode. The mode is simply the value that occurs most often. A data set can be unimodal (one mode), bimodal or multimodal (two or more tied values), or have no mode at all if every value occurs equally often. Unlike the mean and median, the mode is also the only one of the three that makes sense for non-numeric categories, such as the most commonly worn shoe size in a shop.

Figure

A number line showing eleven ordered data points with the mean marked as a balance point above the line and the median marked as the middle point below the line

Derivation AThe mean is the balance point of the data

The goal here is to show, with nothing beyond algebra, why the mean $\bar x$ is often described as the point where the data "balances" — the sum of every value's distance above the mean exactly cancels the sum of every value's distance below it, which is a stronger and more useful fact than the bare formula alone.

Step 1. Write each value's deviation from the mean as $d_i = x_i - \bar x$ for $i = 1, 2, \ldots, n$ . Summing all $n$ deviations gives $\Sigma d = \Sigma(x_i - \bar x) = \Sigma x_i - n\bar x$ .

Step 2. Substitute the definition of the mean, $\bar x = \Sigma x / n$ , so $n\bar x = \Sigma x$ . Then $\Sigma d = \Sigma x - \Sigma x = 0$ .

Step 3. Read the result physically. The sum of all deviations from the mean is always exactly zero, for any data set whatsoever. Values above the mean and values below the mean cancel out perfectly — this is precisely why $\bar x$ is called the "balance point": it is the single value around which the data's ups and downs sum to nothing.

Step 4. This is also why the mean is sensitive to outliers. A single very large value creates a large positive deviation, and because all deviations must still sum to zero, every other value's deviation (and hence $\bar x$ itself) is pulled to compensate — unlike the median, which does not use the magnitude of any value at all, only its rank.

Worked example

Find the mean, median, and mode of the following set of 11 test scores: $12, 15, 12, 18, 20, 15, 15, 22, 18, 12, 15$ .

Step 1. Find the mean. $\Sigma x = 12+15+12+18+20+15+15+22+18+12+15 = 174$ , and $n = 11$ , so $\bar x = \dfrac{174}{11} = 15.82$ (to 2 d.p.).

Step 2. Find the median. Arrange the data into an array: $12, 12, 12, 15, 15, 15, 15, 18, 18, 20, 22$ . Here $n = 11$ is odd, so the median is the $\left(\dfrac{11+1}{2}\right)$ th $= 6$ th term. Counting through the array, the 6th term is $15$ .

Step 3. Find the mode. Counting frequencies: $12$ appears 3 times, $15$ appears 4 times, $18$ appears 2 times, $20$ and $22$ appear once each. The value $15$ occurs most often.

Result: mean $\approx 15.82$ , median $= 15$ , mode $= 15$ .

Competitive stretch

Suppose the highest score of $22$ were mistyped as $52$ . The array becomes $12,12,12,15,15,15,15,18,18,20,52$ ; the sum becomes $174 - 22 + 52 = 204$ , so $\bar x = 204/11 = 18.55$ — a jump of nearly $3$ marks from one bad entry. The median, however, is still the 6th term of the array, which is still $15$ , completely unaffected. This single swapped digit is a clean illustration of why the median is called "resistant to outliers" while the mean is not.

Common traps

Finding the median without arranging the data first. Picking the "middle" entry of an unordered list gives a meaningless answer — the array step is not optional.
Using the wrong position rule for odd versus even $n$ . For odd $n$ there is one middle term; for even $n$ the two middle terms must be averaged, not one of them picked at random.
Miscounting the position number. Confusing "value at position $\dfrac{n+1}{2}$ " with "value at position $\dfrac{n}{2}$ " is a common slip — for small data sets, count through the array explicitly to check.
Assuming every data set has exactly one mode. Some sets are bimodal or have no repeated value at all — state this honestly instead of forcing a single answer.

Ask the tutor: if you want to practise spotting which measure best represents a skewed data set, or want the median walked through step by step for an even-sized array, ask the AI tutor to work an example with you.

Why do we have to arrange data before finding the median but not before finding the mean?

The mean only cares about the total and the count — addition does not care what order the numbers are added in, so $\Sigma x$ comes out the same regardless of arrangement. The median, however, is defined by position: it is "the middle value," and "middle" only means something once the data is sorted from smallest to largest. Skipping the arrangement step and picking whatever value happens to sit in the middle of the raw list, as it was originally recorded, gives a number with no statistical meaning at all.

If mean, median, and mode all come out different, which one should I actually report?

It depends on what you are trying to communicate and how skewed the data is. Report the mean when every value matters and there are no extreme outliers, since it uses all the information. Report the median when a few very large or very small values would otherwise distort the picture, such as with incomes or house prices. Report the mode when you need the single most common or most popular value, especially for categorical data where "mean" and "median" do not even apply, such as the most common blood group in a class.

Can a data set have no mode at all, and does that mean it has no central tendency?

Yes, a data set can have no mode if every value occurs exactly once, with no repeats. That is a normal, honest outcome — it just means the mode is not a useful summary for that particular data set. It does not mean the data set has no central tendency at all, because the mean and the median can still always be calculated (the mean from any list of numbers, the median from any array) and either of those can still represent the data's typical value.

Central tendency of ungrouped data

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads