Averages from Grouped Data

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 $\sum fx \div \sum f$∑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.

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Key Vocabulary

Term	Meaning
Class interval	A range of values, e.g. $10 < t \le 20$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
$\sum fx$∑fx	The total of (frequency × midpoint) across all classes

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Why the Mean is Only Estimated

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.

Finding the Midpoint

The midpoint of a class is the mean of its two boundaries:

$\text{midpoint} = \dfrac{\text{lower bound} + \text{upper bound}}{2}$midpoint=2lower bound+upper bound​

For example, the midpoint of $10 < t \le 20$10<t≤20 is $\dfrac{10 + 20}{2} = 15$210+20​=15.

Worked Example 1

Write down the midpoints of these classes: (a) $0 \le m < 8$0≤m<8; (b) $20 < x \le 35$20<x≤35; (c) $100 \le h < 110$100≤h<110.

Solution:

• (a) $\dfrac{0 + 8}{2} = 4$20+8​=4.
• (b) $\dfrac{20 + 35}{2} = 27.5$220+35​=27.5.
• (c) $\dfrac{100 + 110}{2} = 105$2100+110​=105.

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.

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Estimating the Mean: the Method

1. Find the midpoint $x$x of each class.
2. Multiply each midpoint by its frequency to fill an $fx$fx column.
3. Add the $fx$fx column to get $\sum fx$∑fx.
4. Divide by the total frequency: $\bar{x} \approx \dfrac{\sum fx}{\sum f}$xˉ≈∑f∑fx​.

Worked Example 2

The table shows the time, $t$t minutes, that 40 students spent on homework one evening. Estimate the mean time.

Time $t$t (minutes)	Frequency $(f)$(f)	Midpoint $(x)$(x)	$fx$fx
$0 < t \le 20$0<t≤20	5	10	50
$20 < t \le 40$20<t≤40	12	30	360
$40 < t \le 60$40<t≤60	14	50	700
$60 < t \le 80$60<t≤80	7	70	490
$80 < t \le 100$80<t≤100	2	90	180
Total	40		1780

Solution: Estimated mean $= \dfrac{\sum fx}{\sum f} = \dfrac{1780}{40} = 44.5$=∑f∑fx​=401780​=44.5 minutes.

The shape of the grouped data can be shown with a bar chart of the frequencies:

Worked Example 3

Estimate the mean mass of 50 apples from the table.

Mass $m$m (grams)	Frequency	Midpoint	$fx$fx
$80 \le m < 100$80≤m<100	6	90	540
$100 \le m < 120$100≤m<120	17	110	1870
$120 \le m < 140$120≤m<140	19	130	2470
$140 \le m < 160$140≤m<160	8	150	1200
Total	50		6080

Solution: Estimated mean $= \dfrac{6080}{50} = 121.6$=506080​=121.6 g.

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The Modal Class

The modal class is simply the class interval with the highest frequency. You quote the whole interval, not a single number.

Worked Example 4

For the homework data in Worked Example 2, write down the modal class.

Solution: The highest frequency is 14, in the class $40 < t \le 60$40<t≤60. So the modal class is $40 < t \le 60$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 \le 60$40<t≤60.

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The Median Class

The median lies in the class that contains the middle value. With grouped data we usually use position $\dfrac{n}{2}$2n​ and read down the cumulative frequencies to find which class the value lands in.

Worked Example 5

For the homework data ($n = 40$n=40), find the median class.

Solution: The median is at position $\dfrac{40}{2} = 20$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 \le 60$40<t≤60.

Worked Example 6

The table shows the heights of 60 plants. Find the modal class and the median class.

Height $h$h (cm)	$0 < h \le 10$0<h≤10	$10 < h \le 20$10<h≤20	$20 < h \le 30$20<h≤30	$30 < h \le 40$30<h≤40	$40 < h \le 50$40<h≤50
Frequency	4	11	21	16	8

Solution: The highest frequency is 21, so the modal class is $20 < h \le 30$20<h≤30. The median is at position $\dfrac{60}{2} = 30$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 \le 30$20<h≤30.

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From Median Class to an Estimated Median

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.

Worked Example 6b

The grouped table shows the times, $t$t seconds, taken by 50 swimmers to complete a length.

Time $t$t (seconds)	Frequency	Cumulative frequency
$20 < t \le 25$20<t≤25	6	6
$25 < t \le 30$25<t≤30	15	21
$30 < t \le 35$30<t≤35	19	40
$35 < t \le 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$6, then $6 + 15 = 21$6+15=21, then $21 + 19 = 40$21+19=40, then $40 + 10 = 50$40+10=50. (b) The median is at position $\dfrac{50}{2} = 25$250​=25. The cumulative frequency first reaches or passes 25 in the class $30 < t \le 35$30<t≤35 (the running total jumps from 21 to 40 there), so the median class is $30 < t \le 35$30<t≤35. (c) "30 seconds or less" is read straight from the cumulative frequency at the top of the $25 < t \le 30$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 "$\le 30$≤30" is the running total after the $25$25–$30$30 class, not before it).

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Putting It Together

Worked Example 7

The table shows the ages of 80 members of a sports club.

Age $a$a (years)	Frequency	Midpoint	$fx$fx
$10 \le a < 20$10≤a<20	22	15	330
$20 \le a < 30$20≤a<30	30	25	750
$30 \le a < 40$30≤a<40	18	35	630
$40 \le a < 50$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.