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This lesson covers how to calculate averages and identify key features from grouped frequency tables — a topic that appears regularly on the AQA GCSE Mathematics exam. Because individual data values are lost in grouped data, you cannot find exact averages; instead, you estimate them. Understanding this distinction is crucial for full marks.
When data is presented in a grouped frequency table, you do not know the exact values — only that they fall within certain class intervals. This means:
To estimate the mean from a grouped frequency table:
Estimated mean = sum of (midpoint x frequency) / total frequency
The table shows the times (in minutes) taken by 40 students to complete a puzzle.
| Time (t minutes) | Frequency |
|---|---|
| 0 < t ≤ 5 | 4 |
| 5 < t ≤ 10 | 9 |
| 10 < t ≤ 15 | 14 |
| 15 < t ≤ 20 | 8 |
| 20 < t ≤ 25 | 5 |
Step 1: Find midpoints.
| Time (t minutes) | Midpoint | Frequency | Midpoint x Frequency |
|---|---|---|---|
| 0 < t ≤ 5 | 2.5 | 4 | 10 |
| 5 < t ≤ 10 | 7.5 | 9 | 67.5 |
| 10 < t ≤ 15 | 12.5 | 14 | 175 |
| 15 < t ≤ 20 | 17.5 | 8 | 140 |
| 20 < t ≤ 25 | 22.5 | 5 | 112.5 |
| Total | 40 | 505 |
Step 2: Estimated mean = 505 / 40 = 12.625 minutes
Exam Tip: The midpoint is found by adding the two boundary values and dividing by 2. For the class 5 < t ≤ 10, the midpoint is (5 + 10) / 2 = 7.5. Always show the midpoint column in your working — this is what examiners look for.
The modal class is the class interval with the highest frequency. It is the grouped equivalent of the mode.
From the table above:
The class 10 < t ≤ 15 has the highest frequency (14), so the modal class is 10 < t ≤ 15.
Important: the modal class is the class interval, not just the frequency number. Writing "14" would be incorrect.
Exam Tip: Always give the modal class as the full class interval (e.g. "10 < t ≤ 15"), not just a single number. This is a very common mistake that costs marks.
To find which class contains the median:
Using the same data (total frequency = 40):
The median position is approximately the 20th value (using n/2 = 40/2 = 20).
Build up the cumulative frequency:
| Time (t minutes) | Frequency | Cumulative Frequency |
|---|---|---|
| 0 < t ≤ 5 | 4 | 4 |
| 5 < t ≤ 10 | 9 | 13 |
| 10 < t ≤ 15 | 14 | 27 |
| 15 < t ≤ 20 | 8 | 35 |
| 20 < t ≤ 25 | 5 | 40 |
The 20th value falls in the class 10 < t ≤ 15 (because the cumulative frequency reaches 13 by the end of the second class, and 27 by the end of the third class — the 20th value is somewhere in between).
The distances (in kilometres) cycled by 50 people in a week are shown below.
| Distance (d km) | Frequency |
|---|---|
| 0 < d ≤ 10 | 8 |
| 10 < d ≤ 20 | 15 |
| 20 < d ≤ 30 | 12 |
| 30 < d ≤ 40 | 10 |
| 40 < d ≤ 50 | 5 |
(a) Write down the modal class.
The highest frequency is 15, which is in the class 10 < d ≤ 20.
(b) Find the class containing the median.
Total frequency = 50. Median position = 50/2 = 25th value.
| Distance (d km) | Frequency | Cumulative Frequency |
|---|---|---|
| 0 < d ≤ 10 | 8 | 8 |
| 10 < d ≤ 20 | 15 | 23 |
| 20 < d ≤ 30 | 12 | 35 |
| 30 < d ≤ 40 | 10 | 45 |
| 40 < d ≤ 50 | 5 | 50 |
The 25th value falls in the class 20 < d ≤ 30 (cumulative frequency is 23 after the second class, so the 25th value is in the next class).
(c) Calculate an estimate for the mean distance.
| Distance (d km) | Midpoint | Frequency | Midpoint x Frequency |
|---|---|---|---|
| 0 < d ≤ 10 | 5 | 8 | 40 |
| 10 < d ≤ 20 | 15 | 15 | 225 |
| 20 < d ≤ 30 | 25 | 12 | 300 |
| 30 < d ≤ 40 | 35 | 10 | 350 |
| 40 < d ≤ 50 | 45 | 5 | 225 |
| Total | 50 | 1140 |
Estimated mean = 1140 / 50 = 22.8 km
graph LR
A[Grouped Frequency Table] --> B[Find Midpoints]
B --> C[Multiply Midpoint x Frequency]
C --> D[Sum all Midpoint x Frequency]
D --> E[Divide by Total Frequency]
E --> F[Estimated Mean]
Exam Tip: The word "estimate" in the question is your clue that the data is grouped. Always use midpoints. If the question says "calculate an estimate for the mean", you will lose marks if you do not clearly show midpoints and the (midpoint x frequency) column.
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