Why bother with grouped data?
Ever wonder how a teacher can turn a long list of test scores into a single, useful number? That’s what mean, median and mode from grouped data help you do. They give a quick snapshot of a whole class without listing every single mark.
💡 In Simple Words: Grouped data is just a way to pack many numbers into tidy intervals, like sorting candies by colour. The mean, median and mode tell you the average colour, the middle colour, and the most common colour, all without counting each candy one by one.
What does "grouped data" mean?
Imagine you have a bunch of exam scores. Instead of writing each score, you bundle them into intervals – say 0‑10, 11‑20, 21‑30, and so on. Each interval is called a class interval. The number of scores that fall inside an interval is the frequency. This compact table is what we call grouped data.
How to find the mean from grouped data
The mean is the “average” value. For grouped data we can’t add up every single score, so we use the midpoint of each class (called the class mark) as a stand‑in for all the scores in that interval.
Step‑by‑step recipe
- Write down the class intervals and their frequencies.
- Find the class mark for each interval: add the lower and upper limits, then divide by 2.
- Multiply each class mark by its frequency (this gives a "weighted" total).
- Add all those products together.
- Divide the sum by the total number of observations (the sum of frequencies).
Worked example
Suppose these are the scores of 30 students:
| Class Interval | Frequency (f) |
|---|---|
| 0‑10 | 2 |
| 11‑20 | 5 |
| 21‑30 | 8 |
| 31‑40 | 9 |
| 41‑50 | 6 |
1. Class marks: (0+10)/2 = 5, (11+20)/2 = 15.5, (21+30)/2 = 25.5, (31+40)/2 = 35.5, (41+50)/2 = 45.5.
2. Multiply by frequencies:
- 5 × 2 = 10
- 15.5 × 5 = 77.5
- 25.5 × 8 = 204
- 35.5 × 9 = 319.5
- 45.5 × 6 = 273
3. Add the products: 10 + 77.5 + 204 + 319.5 + 273 = 884. 4. Total frequency = 30. 5. Mean = 884 ÷ 30 ≈ 29.5.
So the average score is about 29.5.
How to find the median from grouped data
The median is the middle value – the score that splits the data into two equal halves. With grouped data we locate the interval that contains the median position.
Steps
- Compute N, the total number of observations (sum of frequencies).
- Find the median position: N/2.
- Identify the class interval where the cumulative frequency first exceeds N/2.
- Use the formula: Median = L + [(N/2 – CF_prev) / f] × h where:
- L = lower limit of the median class
- CF_prev = cumulative frequency before the median class
- f = frequency of the median class
- h = class width (upper‑lower limit)
Worked example (same data)
Total N = 30, so median position = 30/2 = 15.
Cumulative frequencies: 0‑10 → 2, 11‑20 → 7, 21‑30 → 15, 31‑40 → 24, 41‑50 → 30.
The 15th observation falls in the 21‑30 class (cumulative just reaches 15). So:
- L = 21
- CF_prev = 7 (cumulative before this class)
- f = 8
- h = 10 (interval width)
Median = 21 + [(15 – 7) / 8] × 10 = 21 + (8/8)×10 = 21 + 10 = 31.
The median score is 31.
How to find the mode from grouped data
The mode is the value that appears most often. For grouped data we look at the class with the highest frequency – the modal class. Then we apply a formula to estimate the exact mode.
Steps
- Identify the modal class (the one with the largest frequency).
- Use the formula: Mode = L + [(f₁ – f₀) / (2f₁ – f₀ – f₂)] × h where:
- L = lower limit of the modal class
- f₁ = frequency of the modal class
- f₀ = frequency of the class before the modal class (0 if none)
- f₂ = frequency of the class after the modal class (0 if none)
- h = class width
Worked example (same data)
The highest frequency is 9 in the 31‑40 interval, so that’s the modal class.
- L = 31
- f₁ = 9
- f₀ = 8 (frequency of 21‑30)
- f₂ = 6 (frequency of 41‑50)
- h = 10
Mode = 31 + [(9‑8) / (2×9‑8‑6)] × 10 = 31 + [1 / (18‑14)] × 10 = 31 + (1/4)×10 = 31 + 2.5 = 33.5.
So the most common score is about 33.5.
Quick comparison of formulas
| Measure | Key Formula | What you need |
|---|---|---|
| Mean | Σ(class mark × frequency) ÷ Σfrequency | Class marks, frequencies |
| Median | L + [(N/2 – CF_prev) ÷ f] × h | Total N, cumulative frequencies, class width |
| Mode | L + [(f₁‑f₀) ÷ (2f₁‑f₀‑f₂)] × h | Modal class frequencies, class width |
📝 Likely Exam Questions
- Question: The following grouped data shows the ages of 40 students. Find the mean age.
Age (years): 10‑14 | 15‑19 | 20‑24 | 25‑29 | 30‑34 Frequency: 5 | 12 | 15 | 6 | 2
Answer: Compute class marks (12, 17, 22, 27, 32), multiply by frequencies, sum = 880, total N = 40, mean = 880/40 = 22 years. - Question: Using the same table, determine the median age. Answer: N/2 = 20. Cumulative frequencies: 5, 17, 32... Median class is 20‑24 (cumulative exceeds 20). L=20, CF_prev=17, f=15, h=5. Median = 20 + [(20‑17)/15]×5 = 20 + (3/15)×5 = 21.
- Question: Find the mode of the ages given above. Answer: Modal class = 20‑24 (f₁=15). f₀=12, f₂=6, L=20, h=5. Mode = 20 + [(15‑12)/(2×15‑12‑6)]×5 = 20 + [3/(30‑18)]×5 = 20 + (3/12)×5 = 20 + 1.25 = 21.25 years.
- Question: Explain why we use class marks instead of actual data when calculating the mean for grouped data. Answer: Because individual scores are not listed; the class mark represents the average of all values in that interval, giving a reasonable estimate.