Averages from Grouped Data
When data is continuous — measurements like height, weight, or time — exact values cannot be recorded individually, so they are grouped into class intervals. Unlike the discrete frequency table page, the answers here are always estimates. You will learn how to use class midpoints to estimate the mean, identify the modal class (the interval with the highest frequency), and locate the median class (the interval containing the middle value) — all with fully worked examples. Use the Practice Questions below to build your skills from Starter to Master level.
When data is continuous (measurements like height, weight, or time) the exact values are grouped into class intervals. Because we don't know the exact value of each data point, we can only estimate the mean. We also identify the modal class and the median class rather than a single value.
Why do we group data?
Continuous data can take any value in a range; a height might be 1.732 m, or 1.7319 m. Recording every exact value would be impractical. Instead, measurements are placed into class intervals such as \(1.5 \leq h < 2.0\). This makes patterns easier to see, but it means we lose the exact values.
To work around this, we use the midpoint of each class as the best estimate for every value in that group. For a class \(10 \leq t < 20\), the midpoint is \(\dfrac{10+20}{2} = 15\).
Core Ideas: Three Things to Find
For grouped data there are three things the IGCSE exam asks for. Here is what each one means and how to find it.
📐 Estimated Mean
Use the midpoint of each class as the best estimate for all values in that group.
Add a midpoint column \((x)\) and an \(f \times x\) column. Sum both. Divide \(\sum fx\) by \(\sum f\).
Always write "estimate"; the answer is not exact because exact values are unknown.
📊 Modal Class
The class interval with the highest frequency.
No calculation needed; find the row with the biggest \(f\) and write the full class interval as the answer.
Do not write a single number; the modal class is always an interval, e.g. \(10 \leq t < 20\).
🎯 Median Class
The class interval that contains the middle value.
Find the position number, then build a running cumulative frequency total down the table until you pass that position; that class interval is the median class.
For grouped data use \(\frac{n}{2}\) (not \(\frac{n+1}{2}\)) as \(n\) is typically large.
⚙️ Key Formulae: Grouped Frequency Table
Let \(x\) = midpoint of each class interval and \(f\) = its frequency:
\(\sum fx\) = sum of (midpoint \(\times\) frequency) for every class. \(\sum f = n\) = total number of data values.
Worked Examples
💡 Worked Example 1: Estimated Mean, Modal Class, Median Class
The table shows the heights, \(h\) cm, of 40 students measured in PE. Find: (a) an estimate of the mean height, (b) the modal class, (c) the median class.
| Height \(h\) (cm) | Midpoint \(x\) | Frequency \(f\) | \(f \times x\) | Cumulative \(f\) |
|---|---|---|---|---|
| \(150 \leq h < 155\) | 152.5 | 4 | 610 | 4 |
| \(155 \leq h < 160\) | 157.5 | 9 | 1417.5 | 13 |
| \(160 \leq h < 165\) | 162.5 | 14 | 2275 | 27 |
| \(165 \leq h < 170\) | 167.5 | 10 | 1675 | 37 |
| \(170 \leq h \leq 175\) | 172.5 | 3 | 517.5 | 40 |
| Total | \(\sum f = 40\) | \(\sum fx = 6495\) |
\(\displaystyle\bar{x} \approx \frac{\sum fx}{\sum f} = \frac{6495}{40}\)
\(= 162.375 \approx 162.4 \text{ cm (1 d.p.)}\)
Median position \(= \dfrac{40}{2} = 20\)
Cumulative \(f\) reaches 27 at \(160 \leq h < 165\).
💡 Worked Example 2: Finding a Missing Frequency Then Estimating the Mean
The frequency table shows the time, \(t\) minutes, taken by 60 customers to complete a task. One frequency is missing. Find \(p\), then estimate the mean.
| Time \(t\) (min) | Midpoint \(x\) | Frequency \(f\) | \(f \times x\) |
|---|---|---|---|
| \(0 \leq t < 5\) | 2.5 | 8 | 20 |
| \(5 \leq t < 10\) | 7.5 | 20 | 150 |
| \(10 \leq t < 15\) | 12.5 | \(p\) | \(12.5p\) |
| \(15 \leq t < 20\) | 17.5 | 11 | 192.5 |
| \(20 \leq t \leq 25\) | 22.5 | 4 | 90 |
| Total | 60 |
\(\sum fx = 20 + 150 + 212.5 + 192.5 + 90\)
\(= 665\)
\(\displaystyle\bar{x} \approx \frac{665}{60} = 11.08\overline{3}\)
\(\approx 11.1 \text{ min (3 s.f.)}\)
🔑 Key Points
- Always use the midpoint of each class; never the boundary values.
- The mean from grouped data is always an estimate; say so in your answer.
- The modal class is the class interval with the highest \(f\); write the full interval.
- For the median class, find position \(\frac{n}{2}\) and use cumulative frequency to locate which class it falls in.
- Always check \(\sum f = n\) before calculating; missing or wrong frequencies cause every answer to be wrong.
⚠️ Common Mistakes
- Using boundary values instead of midpoints: e.g. using 10 instead of 12.5 for \(10 \leq t < 15\).
- Forgetting "estimate": IGCSE mark schemes deduct marks if you present a grouped mean as exact.
- Writing a single number for modal/median class: always write the full class interval.
- Using \(\frac{n+1}{2}\) for grouped data: use \(\frac{n}{2}\) when \(n\) is large and data is continuous.
- Not checking the total: if there is a missing frequency, find it first before anything else.