Mean of grouped data

A cricket coach records the runs scored by 40 batsmen in a season, but not as 40 separate numbers — the scorer has already bucketed them into ranges like "20 to 29 runs," "30 to 39 runs," with a count against each range. The individual scores are gone; only the class intervals and their frequencies survive. Can you still find the average runs scored? You can, but not by adding raw numbers and dividing — there are no raw numbers left to add. This is the situation for grouped data, and ICSE Class 10 gives three routes to the same answer: the direct method, the short-cut (assumed mean) method, and the step-deviation method. All three represent each class by its mid-point, weight it by its frequency, and combine the weighted values — they differ only in how much arithmetic they demand. Examiners often name the method they want, so real exam fluency means switching between all three and trusting they land on the same number.

The core insight

The mean of grouped data is the frequency-weighted average of each class's mid-point, computed by the direct, short-cut, or step-deviation formula — all three always give the identical answer.

The core idea

Grouped data hides the individual values inside a range, so before any averaging can happen, each class interval must be collapsed to one representative number: its class mark (mid-value),

x=\frac{\text{lower limit}+\text{upper limit}}{2}.

where: the lower and upper limit are the two boundaries of the class interval. Every data value inside that class is treated, for the purposes of the mean, as if it were exactly $x$ — an approximation, since the true values are spread across the interval, but one that becomes accurate once frequencies are large and classes are reasonably narrow. Once every class has a mark $x$ and a frequency $f$ , three formulae compute the mean:

\text{Direct method: } \bar x=\frac{\Sigma fx}{\Sigma f}

where: $\Sigma fx$ is the sum, across all classes, of (frequency times class mark), and $\Sigma f$ is the total frequency — the total number of data items. This is nothing but the familiar "sum of values divided by count" idea, except each value $x$ is counted $f$ times instead of being listed $f$ times.

\text{Short-cut method: } \bar x = A+\frac{\Sigma fd}{\Sigma f}, \qquad d=x-A

where: $A$ is an assumed mean — any convenient class mark, chosen because it is roughly central — and $d = x - A$ is the deviation of each class mark from $A$ . Instead of multiplying large class marks by frequencies, you multiply small (often single- or double-digit) deviations, which is far friendlier without a calculator.

\text{Step-deviation method: } \bar x = A+\frac{\Sigma ft}{\Sigma f}\times i, \qquad t=\frac{x-A}{i}

where: $i$ is the class size (the common width of every interval, assumed equal across the table) and $t=\dfrac{x-A}{i}$ is the step-deviation — the deviation $d$ further shrunk by dividing out the class size, turning it into a small integer such as $-2, -1, 0, 1, 2$ . Why do three formulae exist for one quantity? They are algebraically the same expression rearranged for arithmetic convenience: the short-cut method avoids multiplying by large $x$ values, and the step-deviation method additionally avoids multiplying by large $d$ values. None of them changes what is computed — they change only how much multiplication you do by hand. Choosing $A$ to be a class mark near the centre of the distribution is good practice: it keeps $d$ and $t$ small and often makes one row exactly zero, a useful checkpoint.

Figure

Three parallel routes from a frequency table to the same mean value, labelled direct method, short-cut method, and step-deviation method

Derivation AWhy the short-cut and step-deviation formulae equal the direct formula

The goal is to show, starting only from the definition $d=x-A$ and $t=\dfrac{x-A}{i}$ , that $A+\dfrac{\Sigma fd}{\Sigma f}$ and $A+\dfrac{\Sigma ft}{\Sigma f}\times i$ are exactly equal to $\dfrac{\Sigma fx}{\Sigma f}$ , so switching methods never changes the answer — it only changes how much arithmetic each row of the table demands.

Step 1. Expand $\Sigma fd$ using $d = x - A$ . For each class, $fd = f(x-A) = fx - fA$ . Summing over all classes, $\Sigma fd = \Sigma fx - \Sigma fA = \Sigma fx - A\Sigma f$ , because $A$ is a constant and can be pulled out of the sum, leaving $\Sigma f$ (the total frequency) multiplying it.

Step 2. Substitute back into the short-cut formula. $\bar x = A + \dfrac{\Sigma fd}{\Sigma f} = A + \dfrac{\Sigma fx - A\Sigma f}{\Sigma f} = A + \dfrac{\Sigma fx}{\Sigma f} - A = \dfrac{\Sigma fx}{\Sigma f}$ . The two occurrences of $A$ cancel exactly, leaving precisely the direct-method formula. This confirms the short-cut method is not an approximation — it is the same sum computed by a different route.

Step 3. Extend the same idea to the step-deviation method. Since $t = \dfrac{x-A}{i} = \dfrac{d}{i}$ , we have $d = it$ , so $\Sigma fd = i\,\Sigma ft$ . Substituting into the short-cut result from Step 2, $\bar x = A + \dfrac{\Sigma fd}{\Sigma f} = A + \dfrac{i\,\Sigma ft}{\Sigma f} = A + \dfrac{\Sigma ft}{\Sigma f}\times i$ , which is exactly the step-deviation formula.

Step 4. Read the result. Every step used only that $A$ and $i$ are constants (the same number for every class), so they can be factored in and out of sums freely. No approximation entered anywhere — the three formulae are algebraically identical expressions, and any correctly filled table must return the same $\bar x$ by all three routes, which is exactly why examiners use this as a self-check.

Worked example

The marks (out of 50) obtained by 40 students in a test are given below. Find the mean marks, and verify the direct method against the step-deviation method with assumed mean $A=25$ and class size $i=10$ .

Marks	0–10	10–20	20–30	30–40	40–50
No. of students ( $f$ )	5	8	15	7	5

Step 1. Find the class marks. Each class mark is the mid-point of its interval: $5, 15, 25, 35, 45$ . Note that $25$ is already one of these marks, which is why it is a good choice for the assumed mean $A$ .

Step 2. Build the direct-method table with columns $f$ , $x$ , $fx$ .

Class	$f$	$x$	$fx$
0–10	5	5	25
10–20	8	15	120
20–30	15	25	375
30–40	7	35	245
40–50	5	45	225
Total	$\Sigma f = 40$		$\Sigma fx = 990$

Applying the direct formula, $\bar x = \dfrac{\Sigma fx}{\Sigma f} = \dfrac{990}{40} = 24.75$ .

Step 3. Build the step-deviation table with $d=x-25$ , $t=\dfrac{x-25}{10}$ , and $ft$ .

Class	$f$	$x$	$d=x-25$	$t=\dfrac{x-25}{10}$	$ft$
0–10	5	5	$-20$	$-2$	$-10$
10–20	8	15	$-10$	$-1$	$-8$
20–30	15	25	$0$	$0$	$0$
30–40	7	35	$10$	$1$	$7$
40–50	5	45	$20$	$2$	$10$
Total	$\Sigma f=40$				$\Sigma ft = -1$

Step 4. Apply the step-deviation formula and compare. $\bar x = A+\dfrac{\Sigma ft}{\Sigma f}\times i = 25+\dfrac{-1}{40}\times 10 = 25-0.25 = 24.75$ . This matches the direct-method value exactly, exactly as the derivation above guarantees. The mean mark is 24.75, confirmed by two independent methods.

Competitive stretch

Suppose the same table instead used assumed mean $A=35$ with $i=10$ . Then $t$ values become $-3,-2,-1,0,1$ for the five classes, giving $ft = -15,-16,-15,0,5$ , so $\Sigma ft = -41$ , and $\bar x = 35 + \dfrac{-41}{40}\times 10 = 35 - 10.25 = 24.75$ — the same answer again, showing that any valid choice of $A$ works, though $A=25$ kept the arithmetic smallest.

Common traps

Using class limits instead of the class mark. Only the mid-value $x$ enters the formula; the lower or upper boundary of the interval is never substituted directly.
Forgetting to adjust discontinuous class intervals. If intervals are stated as $1$ – $10$ , $11$ – $20$ (a gap of $1$ between classes), the true continuous boundaries must be found first (typically by subtracting and adding $0.5$ ) before computing class marks.
Sign errors in $d$ or $t$ . Deviations below the assumed mean are negative; a dropped minus sign anywhere in the table throws off the final sum without any obvious warning.
Dividing by the number of classes instead of $\Sigma f$ . The denominator is always the total frequency, never the number of rows in the frequency table.

Ask the tutor: if you want to practise choosing a good assumed mean or converting a discontinuous class table into a continuous one before finding the mean, ask the AI tutor for a step-by-step walkthrough with your own numbers.

Do I have to use the method the question specifies, or can I always use the direct method?

If the question names "short-cut method" or "step-deviation method," you must show that method's working — examiners award marks for the procedure, not just the final number. If no method is specified, pick whichever is fastest; step-deviation is usually quickest when the class size is round, like 10 or 20. Since all three always agree, using the direct method as a mental check is good practice even if you write up only the required one.

What if the class intervals have unequal widths — can I still use the step-deviation method?

No, not directly. The step-deviation formula relies on dividing every deviation by the same class size $i$ , which only makes sense if every class has identical width. With unequal widths, use the direct or short-cut method instead, both of which work regardless. If asked for step-deviation on unequal-width data, point out the mismatch and fall back to short-cut or direct.

Why does treating every value in a class as its mid-point not distort the mean if the data inside is unevenly spread?

It can introduce a small approximation error, but for grouped data this is accepted as unavoidable — once data is tabulated into intervals, the original individual values are lost and cannot be recovered. The mid-point is the best single guess for a class assuming data is roughly evenly spread within it, and with reasonably narrow class widths and enough classes, the errors from different classes tend to partly cancel, keeping the computed mean close to the true mean of the raw data. This is a standard, examiner-accepted approximation, not a computational mistake.

Mean of grouped data

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads