Median of Grouped Data Class 10: Formula & Examples

Quick answer: The median of grouped data is the value of the middle-most observation, found with

Median = l + ((n/2 − cf) / f) × h

where l = lower limit of the median class, n = total number of observations, cf = cumulative frequency of the class before the median class, f = frequency of the median class, and h = class size.

The median splits the data into two equal halves — half the values lie below it and half above. For grouped data you can’t just count to the middle, so you build a cumulative frequency column, find n/2, locate the median class, and apply the formula.

This guide defines cumulative frequency, shows how to find the median class, and works a full example that gives Median = 25.33, using the same data set as the mean and mode guides. At the end we connect all three with the empirical relationship.

What is the median, and why cumulative frequency?

The median is the middle value when data is arranged in order. For ungrouped data with n values: if n is odd, the median is the ((n+1)/2)ᵗʰ value; if n is even, it’s the average of the (n/2)ᵗʰ and ((n/2)+1)ᵗʰ values.

For grouped data the middle observation falls somewhere inside a class interval, so we need a way to see how the counts build up. That’s what cumulative frequency does.

Cumulative frequency (cf) of a class is the running total of frequencies up to and including that class — the number of observations below its upper limit.

Our data set (shared across the Statistics guides):

Each cf is the previous cf plus the current frequency: 7, then 7+10 = 17, then 17+15 = 32, and so on.

examfront’s Topic Practice starts you on cumulative-frequency drills before the full median formula, so the foundation is solid before the harder problems arrive.

How to find the median class

The median class is the class that contains the middle observation. To find it:

1. Compute n/2. Here n = 50, so n/2 = 25.
2. Read down the cf column and pick the first class whose cf is ≥ n/2.

The cumulative frequencies are 7, 17, 32, 40, 50. The first one that reaches or passes 25 is 32, which belongs to the class 20–30. So the median class is 20–30.

examfront’s Practice Companion checks your cumulative frequency column as you build it, so you never pick the wrong median class from a single addition slip.

🎬 Quick challenge: Can you spot the median class from the cumulative frequencies? 👀

The median formula, with every variable defined

Median = l + ((n/2 − cf) / f) × h

• Meaning in words: start at the lower limit of the median class, then move into it by the fraction of the class you still need to cross to reach the middle observation.
• Variables:
  • l = lower limit of the median class
  • n = total number of observations (Σf)
  • cf = cumulative frequency of the class preceding the median class
  • f = frequency of the median class
  • h = class size
• Condition: class intervals must be continuous before applying the formula.

Worked example — finding the median step by step

Step 1 — Build the cf column and find n/2. Done above: n/2 = 25.

Step 2 — Identify the median class. First cf ≥ 25 is 32 → median class 20–30.

Step 3 — Read off the values.

• l = 20 (lower limit of 20–30)
• cf = 17 (cumulative frequency of 10–20, the class before)
• f = 15 (frequency of the median class)
• h = 10 (class size)

Step 4 — Substitute and simplify.

Median = 20 + ((25 − 17) / 15) × 10 = 20 + (8 / 15) × 10 = 20 + 5.33 = 25.33

So the median is 25.33: about half the observations are below 25.33 and about half are above it.

What if the data is “less than” type or has gaps?

Two common variations show up in exams:

• “Less than” / “more than” tables already give you cumulative frequencies. Convert them back to ordinary class frequencies (by subtracting consecutive cumulative values), then proceed as above.
• Discontinuous classes like 118–126, 127–135 must be made continuous first (117.5–126.5, 126.5–135.5, …) by adjusting each boundary by 0.5, because the formula assumes continuous classes.

The heavier versions — finding a missing frequency from a given median, or median from a “less than” ogive — build directly on these steps.

Those missing-frequency and ogive-based median problems are where marks are won or lost. Work the full graded set inside examfront and get personalised help exactly where your cumulative-frequency reasoning slips.

The empirical relationship (linking all three measures)

Once you have any two of the three measures, the third follows approximately from the empirical relationship:

3 Median = Mode + 2 Mean

Check it on our shared data, where mean = 25.8 and mode = 24.17:

3 × Median ≈ Mode + 2 × Mean = 24.17 + 2(25.8) = 75.77, so Median ≈ 75.77 / 3 ≈ 25.26 — very close to the calculated 25.33. The small gap is expected, because the relationship is empirical (approximate), not exact.

A key strength of the median: it is not affected by extreme values, because it depends only on the position of the middle observation. That’s why it’s the preferred “typical” value for skewed data like wages.

🎬 Quick challenge: Can you calculate the median using the correct n/2 value? 📊

Key Takeaways

• Median of grouped data: Median = l + ((n/2 − cf)/f) × h.
• Cumulative frequency = running total of frequencies; build this column first.
• Median class = first class whose cumulative frequency is ≥ n/2.
• Make class intervals continuous before applying the formula.
• Empirical relationship: 3 Median = Mode + 2 Mean (approximate); the median is not affected by extreme values.

Quick Facts

• Formula: Median = l + ((n/2 − cf)/f) × h.
• Use n/2 (not (n+1)/2) to locate the median class in grouped data.
• For our data, median class = 20–30, median = 25.33.
• cf in the formula is the cumulative frequency of the class before the median class.
• Empirical relationship: 3 Median = Mode + 2 Mean.

Common Mistakes

1. Using (n+1)/2 instead of n/2 — for grouped data you locate the median class with n/2.
2. Taking cf of the median class itself — cf must be the cumulative frequency of the preceding class.
3. Not making classes continuous — 118–126 style classes must become 117.5–126.5 first.
4. Forgetting to convert “less than” tables back to ordinary frequencies before finding f.
5. Adding frequencies wrongly in the cf column — a single slip moves the median class and ruins the answer.

FAQ

What is the formula for the median of grouped data in Class 10? Median = l + ((n/2 − cf) / f) × h, where l is the lower limit of the median class, n the total observations, cf the cumulative frequency of the class before the median class, f the frequency of the median class, and h the class size.

What is cumulative frequency? The cumulative frequency of a class is the running total of frequencies up to and including that class — how many observations fall below its upper limit. It is essential for finding the median.

How do you find the median class? Compute n/2, then pick the first class whose cumulative frequency is ≥ n/2. That class is the median class.

What is the empirical relationship between mean, median and mode? The approximate relationship 3 Median = Mode + 2 Mean lets you estimate any one measure from the other two. With mean = 25.8 and mode = 24.17, Median ≈ (24.17 + 51.6)/3 ≈ 25.26.

Why is the median not affected by extreme values? Because it depends only on the position of the middle observation, not on the size of the largest or smallest values, so outliers don’t pull it up or down.

Related Concepts

• Measures of central tendency — how the median compares with mean and mode.
• Mean of grouped data — the average and its three methods.
• Mode of grouped data — the most frequent value.
• Cumulative frequency curves (ogives) — the graphical way to read the median.

Ready to test yourself on median problems? Start with examfront’s Statistics quiz, and let Progress Tracking flag whether it’s the cumulative frequency or the formula that trips you up.