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How do you find and compare the mean, median, mode and range, including from grouped data?

Find the mean, median, mode and range of a data set, estimate the mean from grouped data, find the modal class, and use averages and range to compare two distributions.

A CCEA GCSE Mathematics answer on averages and spread, covering the mean, median, mode and range, estimating the mean from grouped data, finding the modal class, and comparing two distributions using an average and a measure of spread.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. The three averages and the range
  3. Finding the median carefully
  4. Averages from grouped data
  5. Comparing distributions
  6. Why this matters

What this dot point is asking

Averages summarise a data set with a single typical value, and spread measures how varied the data is. In the CCEA Handling Data strand you must find the mean, median, mode and range, estimate the mean from grouped data, identify the modal class, and use an average together with a measure of spread to compare two distributions. The comparison questions carry the AO2 and AO3 reasoning marks, so a good answer states both an average and a spread and interprets them in context.

The three averages and the range

Each average summarises the data differently, and each has its uses.

The mean uses every value but is pulled by extreme values (outliers). The median is unaffected by outliers, which makes it the better average for skewed data. The mode is the only average that works for non-numerical data.

Finding the median carefully

To find the median you must first order the data. For an odd number of values the median is the single middle one; for an even number it is the mean of the two middle values. With nn values, the median is at position n+12\tfrac{n + 1}{2}, which helps locate it in a long list or a frequency table.

Averages from grouped data

When data is grouped into classes, you do not know the exact values, so you estimate the mean using the midpoint of each class as a representative value.

Comparing distributions

A full comparison uses two things: an average and a measure of spread. Compare the means or medians to say which group is higher on average, then compare the ranges (or interquartile ranges) to say which group is more consistent. The interpretation must be in context, for example "Class A scored higher on average and was also more consistent because its range was smaller."

Choosing which average to compare matters. If one data set contains an obvious outlier, comparing medians is fairer than comparing means, because the mean of that set is distorted. The mode is the right choice only when you care about the most frequent value, such as the most common shoe size a shop should stock. A reliable structure for a comparison answer is two sentences: one about the average ("on average, A is higher/lower than B") and one about the spread ("A is more/less consistent than B because its range or IQR is smaller/larger"). Marks are lost when a candidate quotes the numbers but never says what they mean for the two groups being compared, so always finish with the interpretation.

Why this matters

Averages and spread are the everyday language of statistics, used in reports, league tables and research, and the comparison questions are exactly the AO2 and AO3 reasoning CCEA rewards. The grouped-mean method links to histograms and frequency tables, and the idea of spread leads on to the interquartile range, cumulative frequency and box plots. Choosing the most suitable average for the data is a judgement the exam tests.

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA 20203 marksFind the mean, median and range of 7,3,9,3,87, 3, 9, 3, 8. (Non-calculator.)
Show worked answer →

Mean: add the values and divide by how many: 7+3+9+3+85=305=6\dfrac{7 + 3 + 9 + 3 + 8}{5} = \dfrac{30}{5} = 6.

Median: order the data 3,3,7,8,93, 3, 7, 8, 9 and take the middle value: 77.

Range: largest minus smallest: 93=69 - 3 = 6.

One mark each for the mean, the median (ordering first), and the range. Forgetting to order the data before reading the median is the most common slip.

CCEA 20224 marksEstimate the mean of the grouped data: 0<t100 < t \le 10 (frequency 4), 10<t2010 < t \le 20 (frequency 6), 20<t3020 < t \le 30 (frequency 10). (Calculator.)
Show worked answer →

Use the midpoint of each class as the estimate for that group: 5,15,255, 15, 25.

Multiply each midpoint by its frequency: 5×4=205 \times 4 = 20, 15×6=9015 \times 6 = 90, 25×10=25025 \times 10 = 250.

Total of these is 20+90+250=36020 + 90 + 250 = 360, and the total frequency is 4+6+10=204 + 6 + 10 = 20.

Estimated mean =36020=18= \dfrac{360}{20} = 18. Marks are for the midpoints, the products, the total, and the division. It is an estimate because exact values within each class are unknown.

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