Mode of grouped data

Ask a shopkeeper which shoe size sells the most and you get a single number — the size that occurs most often. That number is the mode, the third of the three measures of central tendency alongside mean and median. For raw, ungrouped data the mode is easy to spot: just look for the value that repeats the most. But once data is bundled into class intervals — ages grouped in tens, marks grouped in bands of twenty — no single value repeats at all, because every entry inside a class has been merged into a range. You can no longer point at "the most common number." Instead, you first locate the modal class, the interval with the highest frequency, and then use an interpolation formula to estimate exactly where inside that class the true peak most likely sits. This article builds that formula from the ground up, applies it to a fully worked frequency table, and flags the mix-ups — modal class versus median class, boundary versus class mark, swapped neighbour frequencies — that cost ICSE students marks. By the end you should be able to pick out a modal class on sight and compute the mode of any equal-width grouped distribution with confidence.

The core insight

The mode of grouped data is found by identifying the modal class (the interval with the highest frequency) and interpolating within it using $\text{Mode} = l + \dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$ , which leans the estimate toward whichever neighbouring class is less crowded.

The core idea

Why grouping changes the problem. With raw data — say the list of ages $12, 15, 15, 18, 15, 20$ — the mode is simply the most repeated value, here $15$ . Grouped data throws that method away: once you group ages into "10–20," "20–30," and so on, you know how many people fall in each band but not their individual ages, so no value can literally "repeat" anymore. The best you can do is find the band with the most people and then interpolate a single representative value inside it.

Step 1 — find the modal class. Scan the frequency column and pick the class interval with the single highest frequency. That interval, and only that interval, is the modal class — nothing else about it (its position in the table, its width, its class mark) matters for this choice.

Step 2 — interpolate inside it. Once the modal class is fixed, the exact mode is estimated with:

\text{Mode} = l + \frac{f_1-f_0}{2f_1-f_0-f_2}\times i

where: $l$ is the lower boundary of the modal class, $f_1$ is the frequency of the modal class, $f_0$ is the frequency of the class immediately before the modal class, $f_2$ is the frequency of the class immediately after it, and $i$ is the class size (width), assumed equal across all intervals. The formula's job is to decide where inside the modal-class interval the true peak most likely lies, rather than just declaring the whole interval or its midpoint as "the mode." It does this by comparing two things: how much the frequency jumped going into the modal class ( $f_1-f_0$ ) against the combined rise-and-fall across both of its neighbours ( $2f_1-f_0-f_2$ ). If the class before is much smaller than the class after ( $f_0$ small, $f_2$ large), the numerator stays large relative to the denominator and the mode is pulled further from $l$ , toward the crowded neighbour — and vice versa. The graphical picture below shows the same idea as a geometric construction.

Derivation AMode Formula from Histogram Geometry

Goal: show why interpolating with $\dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$ correctly locates the mode inside the modal-class bar of a histogram, using nothing beyond the geometry of two intersecting straight lines.

Step 1. Draw the histogram of the grouped frequency distribution and focus on the modal-class bar, whose top-left corner sits at height $f_0$ (matching the top of the preceding bar's shared side, extended) and whose top-right corner sits at a similarly matched height with $f_2$ . Concretely, join the top-left corner of the modal-class bar to the top-right corner of the preceding bar with a straight line, and join the top-right corner of the modal-class bar to the top-left corner of the following bar with another straight line. These two diagonals cross at a single point inside the modal-class bar; call its horizontal distance from $l$ the value $x$ .

Step 2. Set up the two diagonal lines algebraically, measuring horizontal distance $t$ from $l$ (so $t$ runs from $0$ to $i$ across the modal class). The first diagonal falls from height $f_1 - f_0$ at $t=0$ (the excess of the modal class over its predecessor) down to height $0$ at $t = i$ ; its height at position $t$ is $(f_1-f_0)\left(1 - \dfrac{t}{i}\right)$ . The second diagonal rises from height $0$ at $t=0$ to height $f_1-f_2$ at $t=i$ (the excess of the modal class over its successor); its height at $t$ is $(f_1-f_2)\dfrac{t}{i}$ .

Step 3. The intersection point is where these two heights are equal: $(f_1-f_0)\left(1-\dfrac{t}{i}\right) = (f_1-f_2)\dfrac{t}{i}$ . Expanding gives $(f_1-f_0) - (f_1-f_0)\dfrac{t}{i} = (f_1-f_2)\dfrac{t}{i}$ , so $(f_1-f_0) = \dfrac{t}{i}\left[(f_1-f_2)+(f_1-f_0)\right] = \dfrac{t}{i}\left(2f_1-f_0-f_2\right)$ .

Step 4. Solve for $t$ : $t = \dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$ . Since $t$ was measured as horizontal distance from $l$ , the actual mode is $l + t$ , which is exactly $\text{Mode} = l + \dfrac{f_1-f_0}{2f_1-f_0-f_2}\times i$ — the formula stated above, now justified by the geometry of the two intersecting diagonals rather than assumed.

Worked example

Find the mode of the following distribution of ages (in years) of 100 people.

Age (years)	10–20	20–30	30–40	40–50	50–60
No. of people ( $f$ )	12	30	34	18	6

Step 1. Scan the frequency row for the highest value: $34$ , occurring in the class 30–40. This is the modal class.

Step 2. Read off the values needed by the formula. $l = 30$ (lower boundary of the modal class), $f_1 = 34$ (frequency of the modal class), $f_0 = 30$ (frequency of the preceding class, 20–30), $f_2 = 18$ (frequency of the following class, 40–50), and $i = 10$ (common class width).

Step 3. Substitute into the formula and simplify the fraction first.

\text{Mode} = 30 + \frac{34-30}{2(34)-30-18}\times 10 = 30 + \frac{4}{68-48}\times 10 = 30 + \frac{4}{20}\times 10

Step 4. Complete the arithmetic: $\dfrac{4}{20}\times 10 = 2$ , so the mode is $30 + 2 = 32$ . The mode of this distribution is 32 years.

Competitive stretch

Suppose instead the modal class had been the last interval, 50–60, with no class beyond it. There is no $f_2$ to read off the table — but the correct move is not to leave it blank or skip the term. Treat the missing neighbour as having frequency $0$ , so $f_2 = 0$ in the formula, and proceed exactly as before. The same rule applies symmetrically if the modal class is the first interval: take $f_0 = 0$ .

Common traps

Confusing the modal class with the median class. The modal class is chosen purely by highest frequency; the median class is chosen using cumulative frequency reaching $N/2$ . They are frequently different classes in the same table.
Using the class mark instead of the lower boundary $l$ . Just as with the median formula, $l$ is the lower limit of the modal class, not its midpoint — plugging in the class mark shifts the whole answer.
Swapping $f_0$ and $f_2$ . $f_0$ is always the frequency of the class before the modal class; $f_2$ is always the frequency of the class after it. Reversing them changes both the sign pattern and the final value.
Leaving $f_0$ or $f_2$ undefined at the table's edge. If the modal class is the first or last interval, the missing neighbour is taken as frequency $0$ , never omitted from the formula.

Ask the tutor: if you're stuck deciding which frequency is $f_0$ and which is $f_2$ , or want to practise a combined mean-median-mode problem on one data set, ask the AI tutor to walk through it with you.

How is the modal class different from the class with the highest class mark

The modal class has nothing to do with the size of the numbers in the interval — it is chosen only by which class has the largest frequency, that is, the most data points falling inside it. A class with a high class mark, like 90–100 in a marks distribution, could easily have very few students in it and would not be the modal class at all. Always scan the frequency column, not the interval labels, to find it.

Can the mode formula give an answer outside the modal class interval

It should not, if the data and the modal class are chosen correctly, because the fraction $\dfrac{f_1-f_0}{2f_1-f_0-f_2}$ always works out to lie between 0 and 1 when $f_1$ is genuinely the largest frequency and its neighbours are non-negative. If your calculated mode falls below $l$ or above $l+i$ , re-check that you picked the correct modal class and did not swap $f_0$ and $f_2$ .

Why do mean, median, and mode of the same data usually come out different

Each measure answers a different question: the mean balances all the values by total sum, the median splits the data exactly in half by position, and the mode picks out the single most frequent region. Unless the distribution is perfectly symmetric, these three questions have three different answers, which is exactly why ICSE problems often ask you to find and compare all three for one data set.

Mode of grouped data

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads