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When a data set is large or continuous — heights, times, masses — we group it into class intervals. The trade-off is that we no longer know the individual values, only how many fall in each class. That means we cannot find the exact mean, but we can produce an estimated mean using the midpoint of each class as a stand-in for every value in it. This lesson shows how to estimate the mean from grouped data, identify the modal class, locate the median class, and explain why the answer is only an estimate — a point OCR GCSE Mathematics (J560) examiners reward explicitly.
The topic is mostly AO1 (carrying out the ∑fx÷∑f calculation correctly) with an important AO2 element (explaining the meaning of "estimate" and interpreting the modal and median classes in context). OCR command words include "Estimate", "Work out", "Write down" and "Give a reason for your answer". The word "Estimate" is a deliberate signal that grouped-data midpoints are involved.
| Term | Meaning |
|---|---|
| Class interval | A range of values, e.g. 10<t≤20 |
| Midpoint | The value halfway through a class: (lower bound + upper bound) ÷ 2 |
| Estimated mean | The mean found using midpoints, because exact values are unknown |
| Modal class | The class interval with the highest frequency |
| Median class | The class interval that contains the median value |
| ∑fx | The total of (frequency × midpoint) across all classes |
Once data is grouped, the individual values are hidden. If 14 students took between 10 and 20 seconds, we do not know whether one took 10.1 s and another 19.9 s. To make progress we assume every value in a class sits at the class midpoint. This is reasonable on average, but it is an assumption, so the result is an estimate, not the true mean. If the data is not evenly spread within each class, the true mean will differ slightly.
The midpoint of a class is the mean of its two boundaries:
midpoint=2lower bound+upper bound
For example, the midpoint of 10<t≤20 is 210+20=15.
Write down the midpoints of these classes: (a) 0≤m<8; (b) 20<x≤35; (c) 100≤h<110.
Solution:
Common error: using the class width (here 8, 15 and 10) instead of the midpoint. The midpoint is the centre of the interval, not its size.
The table shows the time, t minutes, that 40 students spent on homework one evening. Estimate the mean time.
| Time t (minutes) | Frequency (f) | Midpoint (x) | fx |
|---|---|---|---|
| 0<t≤20 | 5 | 10 | 50 |
| 20<t≤40 | 12 | 30 | 360 |
| 40<t≤60 | 14 | 50 | 700 |
| 60<t≤80 | 7 | 70 | 490 |
| 80<t≤100 | 2 | 90 | 180 |
| Total | 40 | 1780 |
Solution: Estimated mean =∑f∑fx=401780=44.5 minutes.
The shape of the grouped data can be shown with a bar chart of the frequencies:
Estimate the mean mass of 50 apples from the table.
| Mass m (grams) | Frequency | Midpoint | fx |
|---|---|---|---|
| 80≤m<100 | 6 | 90 | 540 |
| 100≤m<120 | 17 | 110 | 1870 |
| 120≤m<140 | 19 | 130 | 2470 |
| 140≤m<160 | 8 | 150 | 1200 |
| Total | 50 | 6080 |
Solution: Estimated mean =506080=121.6 g.
The modal class is simply the class interval with the highest frequency. You quote the whole interval, not a single number.
For the homework data in Worked Example 2, write down the modal class.
Solution: The highest frequency is 14, in the class 40<t≤60. So the modal class is 40<t≤60.
Common error: writing the modal class as "14" (the frequency) or "50" (the midpoint). You must give the full interval 40<t≤60.
The median lies in the class that contains the middle value. With grouped data we usually use position 2n and read down the cumulative frequencies to find which class the value lands in.
For the homework data (n=40), find the median class.
Solution: The median is at position 240=20. Cumulative frequencies are 5, 17, 31, 38, 40. The running total first passes 20 in the third class (it reaches 31 there, having been only 17 before), so the median class is 40<t≤60.
The table shows the heights of 60 plants. Find the modal class and the median class.
| Height h (cm) | 0<h≤10 | 10<h≤20 | 20<h≤30 | 30<h≤40 | 40<h≤50 |
|---|---|---|---|---|---|
| Frequency | 4 | 11 | 21 | 16 | 8 |
Solution: The highest frequency is 21, so the modal class is 20<h≤30. The median is at position 260=30. Cumulative frequencies: 4, 15, 36, 52, 60. The total first passes 30 in the third class, so the median class is 20<h≤30.
Finding the median class tells you which interval the middle value sits in, but sometimes you want a single estimated number. A neat way to picture this is to build a cumulative frequency column — a running total of how many items fall up to the top of each class. The cumulative frequency answers questions such as "how many plants were 30 cm or shorter?" directly, and the median is simply the value where the running total reaches halfway. You will draw these as smooth curves in the Higher cumulative-frequency lesson; here we just use the column to pin down the median class precisely.
The grouped table shows the times, t seconds, taken by 50 swimmers to complete a length.
| Time t (seconds) | Frequency | Cumulative frequency |
|---|---|---|
| 20<t≤25 | 6 | 6 |
| 25<t≤30 | 15 | 21 |
| 30<t≤35 | 19 | 40 |
| 35<t≤40 | 10 | 50 |
(a) Complete the cumulative frequency column (shown above). (b) Use it to state the median class. (c) How many swimmers took 30 seconds or less?
Solution: (a) The cumulative frequencies are formed by adding each frequency to the running total: 6, then 6+15=21, then 21+19=40, then 40+10=50. (b) The median is at position 250=25. The cumulative frequency first reaches or passes 25 in the class 30<t≤35 (the running total jumps from 21 to 40 there), so the median class is 30<t≤35. (c) "30 seconds or less" is read straight from the cumulative frequency at the top of the 25<t≤30 class: 21 swimmers.
Common error: reading the cumulative frequency from the wrong row, or forgetting that the cumulative value sits at the top boundary of its class (the value for "≤30" is the running total after the 25–30 class, not before it).
The table shows the ages of 80 members of a sports club.
| Age a (years) | Frequency | Midpoint | fx |
|---|---|---|---|
| 10≤a<20 | 22 | 15 | 330 |
| 20≤a<30 | 30 | 25 | 750 |
| 30≤a<40 | 18 | 35 | 630 |
| 40≤a<50 | 10 | 45 | 450 |
| Total | 80 | 2160 |
(a) Estimate the mean age. (b) Write down the modal class. (c) Find the median class.
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