Median of grouped data

Line up fifty students by weight and the middle value is easy to point at — but a grouped frequency table hides individual values inside class intervals like 40–45, so you can no longer just "find the middle one." ICSE Class 10 Statistics solves this with a cumulative frequency (c.f.) column and a linear interpolation formula that estimates where inside the correct class interval — the median class — the middle observation must lie. This article rebuilds that formula from first principles, applies it to a weights-of-students table, and flags the traps that cost the most marks: using the wrong $cf$ , confusing the lower boundary with the class mark, and forgetting the continuity correction for discontinuous classes. The same c.f. column also underlies the less than ogive, the graphical route to the median covered alongside this topic.

The core insight

The median of grouped data is found by locating the median class — the first class whose cumulative frequency reaches or exceeds $N/2$ — and linearly interpolating inside it with $\text{Median} = l + \dfrac{(N/2) - cf}{f}\times i$ .

The core idea

Why a plain "middle value" does not work here. Grouped data only tells you how many observations fall inside each class, not their individual values — so you cannot list all fifty weights in order and pick the 25th and 26th. Instead you locate the class interval that contains the middle observation, then estimate its position inside that interval.

Cumulative frequency finds the right class. Build a running total, the c.f. column, adding each class's frequency $f$ to the sum of all frequencies before it. The median class is the first class whose c.f. reaches or exceeds $\dfrac{N}{2}$ , where $N = \Sigma f$ is the total frequency — this is the class where the middle observation must be sitting.

Interpolation locates the median inside that class. Once the median class is known, the median formula estimates where within it the $\dfrac{N}{2}$ -th value falls, assuming the data is spread evenly across the class:

\text{Median} = l + \frac{\dfrac{N}{2}-cf}{f}\times i

where: $l$ is the lower boundary of the median class, $N$ is the total frequency, $cf$ is the cumulative frequency of the class just before the median class, $f$ is the frequency of the median class itself, and $i$ is the class width (size) of the median class.

Figure

A cumulative frequency staircase rising with class intervals, with a horizontal line at N over 2 crossing into the median class and a dashed vertical line down to the lower boundary l

The reasoning behind the formula

The goal of this derivation is to explain why the interpolation formula $\text{Median} = l + \dfrac{(N/2)-cf}{f}\times i$ correctly estimates the median even though grouped data hides the exact position of every observation inside its class, by treating the median class as if its $f$ observations are spread evenly across its width.

Step 1. Inside the median class, $cf$ observations from earlier classes already lie below its lower boundary $l$ , and the median class itself contributes $f$ more observations before the next class begins. The middle observation, the $\dfrac{N}{2}$ -th one, is therefore the $\left(\dfrac{N}{2}-cf\right)$ -th observation counting from the start of the median class.

Step 2. Assume the $f$ observations of the median class are spread uniformly across its full width $i$ . Under that assumption, each successive observation inside the class advances by an equal fraction of the width, so the position of the $\left(\dfrac{N}{2}-cf\right)$ -th observation, measured from $l$ , is the fraction $\dfrac{(N/2)-cf}{f}$ of the way through the class.

Step 3. Converting that fraction into an actual distance means multiplying by the class width $i$ , giving the offset $\dfrac{(N/2)-cf}{f}\times i$ to add onto the lower boundary $l$ . That produces exactly $\text{Median} = l + \dfrac{(N/2)-cf}{f}\times i$ , the formula used above.

Step 4. The uniform-spread assumption is why this is an estimate, not an exact recovery of the raw data — grouping always loses the precise individual values, and the formula makes the most reasonable guess consistent with the frequencies given.

Worked example

Find the median of the following distribution of weights (in kg) of 50 students.

Weight (kg)	30–35	35–40	40–45	45–50	50–55	55–60
No. of students ( $f$ )	4	8	14	12	8	4

Step 1. Build the cumulative frequency table by running the total of $f$ down the classes.

Class	$f$	c.f.
30–35	4	4
35–40	8	12
40–45	14	26
45–50	12	38
50–55	8	46
55–60	4	50

Step 2. Find $N/2$ . Here $N = 50$ , so $\dfrac{N}{2} = 25$ .

Step 3. Locate the median class. The c.f. first reaches or exceeds 25 in the class 40–45 (c.f. $= 26$ ), so $l = 40$ , $f = 14$ , $cf = 12$ (the c.f. of the class just before, 35–40), and $i = 5$ .

Step 4. Apply the formula. $\text{Median} = 40 + \dfrac{25-12}{14}\times 5 = 40 + \dfrac{13}{14}\times 5 = 40 + 4.64 = 44.64$ . The median weight is approximately 44.64 kg.

Competitive stretch

For 60 students' marks grouped as 0–10, 10–20, 20–30, 30–40, 40–50 with frequencies 3, 12, 27, 14, 4, the same four steps give $N/2 = 30$ , median class 20–30 ( $l = 20$ , $cf = 15$ , $f = 27$ , $i = 10$ ), so $\text{Median} = 20 + \dfrac{30-15}{27}\times 10 = 20 + 5.56 = 25.56$ — approximately 25.56 marks. Notice the median class in this second table has by far the largest frequency, but that is coincidental — the median class is always found from where c.f. crosses $N/2$ , never by looking for the largest $f$ .

Common traps

Using the c.f. of the median class itself instead of the class before it. $cf$ in the formula must be the cumulative frequency of the class preceding the median class.
Confusing $l$ with the class mark (mid-point). $l$ is the lower boundary of the median class, not its centre — that mix-up belongs to the mean formula, not the median formula.
Picking the wrong median class by demanding c.f. equal to $N/2$ exactly. No row may match exactly; always choose the first class where c.f. is greater than or equal to $N/2$ .
Skipping the continuity correction for discontinuous intervals. If classes are stated as 1–10, 11–20 rather than continuous 0.5–10.5 boundaries, the true lower boundary used in the formula differs from the stated lower limit by 0.5.

Ask the tutor: if your distribution has class boundaries that don't line up neatly, or you want to double-check which class is your median class, ask the AI tutor to walk through your specific table step by step.

Why do we need a whole formula instead of just picking the middle student's weight?

Because a grouped table only records how many students fall inside each weight range, not each student's exact weight. You know 14 students weigh between 40 and 45 kg, but not their individual weights, so you cannot simply count to the 25th value. The formula assumes those 14 students are spread evenly across the 40–45 range and estimates where the 25th value must fall inside it — the best guess consistent with the given frequencies.

What if the cumulative frequency lands exactly on N/2?

Then that class is still the median class, since c.f. "reaches" $N/2$ there — you do not move to the next class. Apply the formula normally: with $cf$ taken from the class before it, the fraction $\dfrac{(N/2)-cf}{f}$ simply comes out larger, closer to 1, meaning the median falls near the far end of that class. There is no separate rule for this case; the same four steps apply unchanged.

How is the median different from the mode and mean of grouped data?

All three use the same cumulative-frequency or frequency table but read it differently. The mean uses every class mark weighted by its frequency, so it is pulled by extreme classes. The mode finds the class with the highest frequency and interpolates around its neighbours. The median finds where the middle observation lies via cumulative frequency, ignoring how spread out or skewed the extreme classes are — which is why it is the preferred "typical value" measure for skewed data, such as incomes or exam scores with a few very high or low scorers.