Calculate the mean, median, mode and range; find the mean from a frequency table and an estimated mean from grouped data; and compare distributions using an average and the range (and quartiles at Higher tier).

What this dot point is asking

OCR references S4 and S5 cover the averages (mean, median, mode) and the range as a measure of spread, including finding the mean from a frequency table and an estimated mean from grouped data, and comparing distributions. Averages summarise data in a single value, and the range measures how spread out it is. This content appears on every tier, with the frequency-table mean and the grouped-data estimate being reliable mid-tariff questions.

The four summary measures

Each average summarises the data differently.

So for $3, 5, 5, 8, 9$: the mean is $\tfrac{30}{5} = 6$, the median is $5$ (third of five), the mode is $5$, and the range is $9 - 3 = 6$. The mean uses every value but is affected by extreme values (outliers); the median is more robust to outliers; the mode is the only average usable for qualitative data.

Choosing which average to quote is itself an exam skill. If a data set has an extreme outlier, such as one very high salary among modest ones, the mean is pulled upward and the median gives a fairer picture of the typical value. For a data set with no clear repeated value, the mode may not exist or may not be useful. A frequent question gives a data set, asks for all three averages, and then asks which best represents the data and why, which rewards a reasoned AO2 answer rather than just a calculation.

The mean from a frequency table

A frequency table groups repeated values.

So for shoe sizes $5$ (frequency $3$), $6$ (frequency $5$) and $7$ (frequency $2$), the total is $(5 \times 3) + (6 \times 5) + (7 \times 2) = 15 + 30 + 14 = 59$, over $3 + 5 + 2 = 10$ items, giving a mean of $5.9$. Adding an "$fx$" column keeps the calculation organised, and dividing by the total frequency (not the number of rows) is the crucial step.

The estimated mean from grouped data

When data is grouped into intervals, the exact values are lost.

For grouped data, use the midpoint of each class as the representative value, then apply the same $\dfrac{\sum fx}{\sum f}$ formula. The result is an estimate, because the midpoint assumes the data is evenly spread within each class. So for a class $10 \le x < 20$ with frequency $8$, use the midpoint $15$ and the product $15 \times 8 = 120$. OCR always calls this an "estimate" of the mean, and stating that it is an estimate is part of a complete answer.

Comparing distributions

A good comparison uses both location and spread.

To compare two data sets, compare an average (which is typically larger, showing location) and the range (which is more spread out, showing consistency). So "Class A has a higher mean, so scored better on average, but a larger range, so was less consistent" is a full comparison. At Higher tier, the interquartile range (the spread of the middle $50\%$ of the data) is a more robust measure of spread than the range, because it ignores outliers.