Mean of data

Five students write a test and score 8, 9, 6, 7, 10 marks. What single number best represents how the class did? Add all five marks and share the total equally among the five students — that shared-out value is the mean, the number every observation would have if the total were spread evenly. It is the most-used summary in all of statistics: report cards, cricket averages, weather bulletins and exam toppers' lists all lean on it. This section builds the mean from the ground up: the formula, the picture of the mean as a balance point on a dot plot, and the rules that predict how the mean shifts when data is added, removed, or scaled. These shift-rules are not just theory — they let you check an answer "feels right" without re-adding a whole list of numbers, a genuinely useful exam shortcut once a data set grows long. This topic sits in the Class 8 chapter "Tales by Dots and Lines," which introduces the mean through dot plots before connecting the picture to the familiar arithmetic formula.

The core insight

The mean of a data set is the sum of all observations divided by their count, and it behaves as the balance point of the data because the total distance of values above it always equals the total distance of values below it.

The core idea

For $n$ observations $x_1, x_2, \ldots, x_n$ , the mean is defined as:

\text{Mean} = \dfrac{x_1+x_2+\cdots+x_n}{n} = \dfrac{\text{sum of observations}}{\text{number of observations}}

where: each $x_i$ is one data value and $n$ is how many values are in the list.

The balance-point picture. Plot every value as a dot on a number line and the mean is the point where the line balances like a seesaw: the total distance of the dots to the right of the mean equals the total distance of the dots to the left. A single dot far from the rest — an outlier — can pull the mean noticeably, just as one heavy weight far from a seesaw's pivot outweighs several light weights close to it.

Figure

A dot plot of five test scores with the mean marked as a balance point on a number line

Effect of changing the data. Three rules let you predict a new mean without recomputing the whole sum.

Add a value greater than the current mean and the mean rises; add one less than the mean and it falls.
Remove a value greater than the current mean and the mean falls; remove one less than the mean and it rises.
Scale every value by the same factor $k$ and the mean scales by $k$ too; add the same constant $c$ to every value and the mean increases by $c$ . This shortcut needs every value to change the same way — it breaks if only some values are altered.

Derivation AWhy the balance-point rule holds

Step 1. The claim that "distances above the mean balance distances below it" is not a coincidence of the marks 8, 9, 6, 7, 10 — it is a direct algebraic consequence of the mean's own definition, and this derivation shows why it must be true for any list of numbers, giving the tool a genuinely universal grip rather than one that happens to work on convenient examples.

Step 2. Start from the definition. $\text{Mean} = \dfrac{x_1+x_2+\cdots+x_n}{n}$ , which rearranges to $n \times \text{Mean} = x_1+x_2+\cdots+x_n$ , i.e. the sum of all $n$ observations equals $n$ copies of the mean added together.

Step 3. Now look at the deviation of each value from the mean, $d_i = x_i - \text{Mean}$ , and add up all $n$ deviations: $d_1+d_2+\cdots+d_n = (x_1+x_2+\cdots+x_n) - n\times\text{Mean}$ .

Step 4. Substitute the result from Step 2 into Step 3: the sum of observations is exactly $n\times\text{Mean}$ , so $d_1+d_2+\cdots+d_n = n\times\text{Mean} - n\times\text{Mean} = 0$ . The deviations always add to zero — positive deviations (values above the mean) exactly cancel negative deviations (values below it), which is precisely the "balance" seen on the dot plot.

Worked example

Five students score 8, 9, 6, 7, 10 marks in a class test.

Step 1. Add the observations: $8+9+6+7+10 = 40$ .

Step 2. Divide by the count: $\text{Mean} = \dfrac{40}{5} = 8$ .

Step 3. Check the balance-point idea using deviations from 8: $0, +1, -2, -1, +2$ . Sum of positive deviations $=1+2=3$ ; sum of negative deviations (as size) $=2+1=3$ — equal, confirming 8 is the balance point, matching the derivation above.

Step 4. A sixth student joins and scores 14. Since $14 > 8$ (the current mean), the new mean must rise above 8. New sum $=40+14=54$ , new count $=6$ , so new mean $=\dfrac{54}{6}=9$ — higher than 8, exactly as the add-a-value rule predicted, without re-checking every deviation from scratch.

Competitive stretch

A shopkeeper's weekly sales (in ₹ hundreds) are 20, 22, 18, 25, 20, 21, 22, giving mean $\dfrac{148}{7}\approx 21.14$ . Removing the 18 (an off day) drops $n$ to 6 and raises the sum-per-count: new mean $=\dfrac{130}{6}\approx 21.67$ , up from 21.14 as expected since 18 was below the mean. If sales are reported in rupees instead of hundreds, every value scales by 100, so the mean simply scales too: $21.14\times100=2114$ — no re-addition needed. Notice how each answer was found using only the shift-rules from the core idea, without ever re-adding all seven original numbers — that speed is the whole point of learning the rules rather than only the formula.

Common traps

Forgetting to update $n$ . Adding or removing a value changes the sum and the count — students often update only the sum and divide by the old $n$ .
Guessing the direction of change. Always compare the new or removed value to the current mean, not to the other data values, to know whether the mean will rise or fall.
Treating the mean as always "typical." One outlier can drag the mean far from where most of the data actually sits.
Partial scaling. The "mean scales by $k$ " shortcut only holds if every observation is scaled — it fails the moment only some values are changed.
Adding constant shifts and scaling together. If a data set is both scaled by $k$ and shifted by $c$ (say every value is doubled, then increased by 5), apply the rules in the same order as the change was made to the data — scale the old mean by $k$ first, then add $c$ .

Ask the tutor: Want to try predicting how the mean shifts for a new data set of your own before checking the arithmetic?

Why does the mean sometimes not match any actual data value, like 21.14 for whole-number sales?

Because the mean is a computed share-out, not a value picked from the list. Dividing a total by a count can land anywhere, including between two data values or with a decimal, as with the shopkeeper's sales averaging ₹2114.14 in hundreds. That is expected and correct — the mean answers "what would each observation be if the total were shared equally," and an equal share rarely lands exactly on one of the original numbers.

If I add a value equal to the current mean, does the mean change at all?

No. Adding a value exactly equal to the current mean leaves the mean unchanged, because that value sits exactly at the balance point already — it adds zero deviation on either side. You can check this with the sum-of-deviations idea from the derivation above: a new value with deviation zero does not disturb a total of deviations that was already zero.

Why does removing an outlier always move the mean, but not always by much?

Removing any value changes both the sum and the count, so the mean almost always moves at least slightly — the only exception is removing a value exactly equal to the current mean. How much it moves depends on how far the removed value was from the mean and how large the data set is: removing one extreme value from five numbers (as in the worked example) shifts the mean noticeably, but removing the same-sized outlier from five hundred numbers barely moves it, because the total is being shared among far more remaining observations. This is also why frequency tables, which summarise larger data sets, use the same mean formula with sums weighted by how often each value occurs.

Mean of data

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads