Mode of grouped data

A shoe shop does not care about the average shoe size it sells — it cares which size sells the most, because that is the size to stock in bulk. That single most-frequent value is the mode. For raw data you just count repeats, but real survey data almost always arrives already sorted into class intervals — marks bucketed 0 to 10, 10 to 20, and so on — and once the individual values are hidden inside a class, "count the repeats" no longer works. You only know how many observations fell in each class, not what those observations actually were. This is where the mode of grouped data formula comes in. It locates the class with the highest frequency — the modal class — and then uses that class's frequency together with its two immediate neighbours to estimate exactly where inside that class the true peak of the distribution most likely sits. This article builds that formula, applies it step by step to a worked example, and flags the mix-ups that cost easy marks in exams.

The core insight

The mode of grouped data is estimated from the modal class (highest frequency) and its two neighbouring class frequencies using Mode=l+(f1−f02f1−f0−f2)×h\text{Mode} = l + \left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h.

The core idea

In a grouped frequency distribution, the first job is always to find the modal class — the class interval whose frequency is the largest of all. Everything else in the formula is read off relative to that one class. Once the modal class is identified, five quantities are pulled from the table:

  • ll = lower limit of the modal class
  • f1f_1 = frequency of the modal class
  • f0f_0 = frequency of the class immediately before the modal class
  • f2f_2 = frequency of the class immediately after the modal class
  • hh = class width (size) of the modal class

These feed into the mode formula:

Mode=l+(f1−f02f1−f0−f2)×h\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h

where: ll is the modal class's lower limit, f1f_1, f0f_0, f2f_2 are the frequencies of the modal class and its immediate left and right neighbours, and hh is the modal class width. The formula assumes all class widths are equal, which is the standard CBSE setup. The correction term (f1−f02f1−f0−f2)×h\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h nudges the estimate away from the bare lower limit ll toward wherever the peak most likely sits inside the modal class. If the class before the modal class is nearly as crowded as the modal class itself (f0f_0 close to f1f_1), the true peak is probably near the start of the modal class, so the shift from ll stays small. If both neighbours are comparatively sparse, the modal class's own frequency dominates the picture and the shift is larger, pushing the mode further into (or past) the class mid-point. Both neighbours enter the formula because a single class's frequency alone cannot tell you which side of it the data leans toward — only comparing it against both flanks can. Like mean and median, the mode summarises an entire distribution with one representative number, but unlike the mean it ignores extreme classes with tiny frequencies far from the modal class, so an odd outlier class does not drag the mode away from where most of the data actually is.

Figure
A grouped frequency histogram with the tallest bar marked modal class and its lower limit and neighbouring bars labelled
Modal class sits between two neighbours that shape the mode class interval frequency 0-10 10-20 (f0) 20-30 (f1) modal class 30-40 (f2) 40-50 l = 20
Derivation ABuilding the mode formula for a modal class

Starting from the histogram picture of a grouped frequency distribution, derive why the estimated mode inside the modal class equals its lower limit plus a correction fraction of the class width, and show that this fraction depends only on how the modal class frequency compares with its two immediate neighbours.

Step 1. Draw the histogram bars for the modal class and its two neighbours on either side. The modal class bar is the tallest, standing at height f1f_1; the bar just before it stands at height f0f_0 and the bar just after at height f2f_2, with f1>f0f_1 > f_0 and f1>f2f_1 > f_2 since it is the modal class.

Step 2. Join the top-right corner of the bar before the modal class to the top-left corner of the modal class bar, and join the top-left corner of the bar after the modal class to the top-right corner of the modal class bar. These two joining lines cross at a single point inside the modal class bar, and that crossing point is taken as an estimate of where the true peak of the underlying (continuous) distribution lies.

Step 3. By similar triangles formed on either side of the vertical through the crossing point, the horizontal distance of the crossing point from the modal class's left edge works out to be proportional to f1−f0f_1 - f_0, while the total horizontal span used for that proportion is (f1−f0)+(f1−f2)=2f1−f0−f2(f_1-f_0)+(f_1-f_2) = 2f_1-f_0-f_2. So the fraction of the class width hh that the peak sits away from ll is f1−f02f1−f0−f2\frac{f_1-f_0}{2f_1-f_0-f_2}.

Step 4. Multiplying that fraction by the class width hh and adding it to the lower limit ll gives the estimated mode, Mode=l+(f1−f02f1−f0−f2)×h\text{Mode} = l + \left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h, which is exactly the CBSE mode formula — and it confirms the earlier intuition that a modal class flanked by a near-equal neighbour on one side pulls the peak toward that side, while a sharply higher modal class pulls the estimate further inside.

Worked example

Find the mode of the following frequency distribution.

Class0-1010-2020-3030-4040-50
Frequency5820126

Step 1. Scan the frequency row for the largest value: 2020, occurring in the class 2020-3030. So 2020-3030 is the modal class.

Step 2. Read off the terms directly from the table: l=20l = 20 (lower limit of the modal class), f1=20f_1 = 20 (modal class frequency), f0=8f_0 = 8 (frequency of the class before, 1010-2020), f2=12f_2 = 12 (frequency of the class after, 3030-4040), and h=10h = 10 (class width).

Step 3. Substitute into the formula:

Mode=20+(20−82(20)−8−12)×10=20+(1240−20)×10=20+(1220)×10\text{Mode} = 20 + \left(\frac{20-8}{2(20)-8-12}\right)\times 10 = 20 + \left(\frac{12}{40-20}\right)\times 10 = 20 + \left(\frac{12}{20}\right)\times 10

Step 4. Simplify the fraction before multiplying: 1220=0.6\frac{12}{20} = 0.6, so Mode=20+0.6×10=20+6\text{Mode} = 20 + 0.6\times 10 = 20 + 6. The mode of the distribution is 26.

Competitive stretch

Change only the frequency of the class after the modal class from 1212 to 88 in the same table, keeping everything else the same. Now f2=8f_2 = 8, so 2f1−f0−f2=40−8−8=242f_1 - f_0 - f_2 = 40-8-8=24, giving Mode=20+(1224)×10=20+5=25\text{Mode} = 20 + \left(\frac{12}{24}\right)\times 10 = 20+5 = 25. Notice the mode moved closer to ll even though f0f_0 did not change — because lowering f2f_2 raised the denominator, showing the mode depends on both neighbours together, not just the one immediately before the modal class.

Common traps

  • Picking the modal class by eyeballing which interval "looks middle" instead of checking which one actually has the highest frequency in the table.
  • Swapping f0f_0 and f2f_2: f0f_0 is always the class before the modal class and f2f_2 is always the class after it, and the formula is not symmetric in them.
  • Using the class mark (mid-point) instead of the lower limit ll — the mid-point belongs to the mean formula, not the mode formula.
  • Forgetting to simplify the fraction before multiplying by hh, which leads to arithmetic slips especially when 2f1−f0−f22f_1-f_0-f_2 shares a common factor with f1−f0f_1-f_0.

Ask the tutor: Not sure which class is the modal class in your own dataset, or want your mode calculation checked step by step? Ask the AI tutor to work through it with you.

What if the modal class is the very first or very last class in the table

Then there is no class before (or after) it, so f0f_0 (or f2f_2) has nothing to be read from the table. Most CBSE problems are designed to avoid this edge case entirely by putting the modal class safely in the middle of the distribution, so in practice you rarely need to decide what to do here — but if it does come up, the missing neighbouring frequency is conventionally taken as 00.

Why does the mode formula need both f0 and f2 instead of just f1

Because f1f_1 alone only tells you the modal class is the tallest bar — it says nothing about which side the underlying peak leans toward. Comparing f1f_1 against f0f_0 tells you how the modal class compares with what came before it, and comparing against f2f_2 tells you how it compares with what comes after; together they pin down where inside the class the peak most plausibly sits.

Is the mode from this formula the exact mode of the raw data, or just an estimate

It is an estimate. Because grouping hides the individual values inside each class, there is no way to recover the exact most-frequent raw value from a frequency table alone. The formula gives the best estimate obtainable from the class structure, assuming the data is roughly evenly spread within each class — which is why it is called the mode of grouped data rather than simply "the mode."

Before this

Where this leads