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3.1b · Analysis of Data · Sub-skill

Averages

Mean, median and mode — the three ways to summarise a data set with a single typical value, and how to pick the right one.

Build it up, step by step

Understanding averages

Click each step below to reveal it — work through them in order the first time round.

Step 1 · What “average” means

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An average is a single value that summarises a whole data set. There are three types used in Core Maths: the mean, median and mode. Different situations call for different averages — the exam expects you to know all three and to justify which one is most appropriate.

Step 2 · Calculating each average

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Mean = (sum of all values) ÷ (number of values), written $\bar{x} = \frac{\sum x}{n}$. Median = the middle value once the data is placed in order (average the middle two if $n$ is even). Mode = the value that occurs most often.

Mini example: data set 4, 7, 7, 9, 13 → mean = 40÷5 = 8, median = 7 (middle value), mode = 7 (most frequent).

Step 3 · Choosing the right average

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The mean uses every value, so it's the most representative when data is roughly symmetric — but it's dragged around by extreme values/outliers. The median is unaffected by outliers, so it's better for skewed data (e.g. house prices, incomes). The mode is the only average that works for non-numeric (categorical) data, e.g. ‘most common eye colour’.

Step 4 · Averages from grouped/frequency data

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When data is grouped into classes, you don't know the exact values, so you use the midpoint of each class as an estimate. Estimated mean = $\frac{\sum fx}{\sum f}$, where $f$ is frequency and $x$ is the class midpoint. This is only an estimate, because you're assuming every value in a class sits exactly at the midpoint.

Worked example

A café records the age of 20 customers in a frequency table: 10–19 (freq 4), 20–29 (freq 8), 30–39 (freq 5), 40–49 (freq 3). Estimate the mean age.

Midpoints: 14.5, 24.5, 34.5, 44.5.
Σfx = 4(14.5) + 8(24.5) + 5(34.5) + 3(44.5) = 58 + 196 + 172.5 + 133.5 = 560
Σf = 20
Estimated mean = 560 ÷ 20 = 28 years.

Test yourself

Past-paper style question

The number of goals scored by a football team in its last 9 matches was: 1, 2, 0, 3, 1, 1, 4, 2, 1.

(a) Find the mean, median and mode number of goals.
(b) The team's manager says ‘we typically score about 2 goals a game.’ Using your answers to part (a), comment on whether this claim is reasonable. [5 marks]

Show the answer

Ordered data: 0,1,1,1,1,2,2,3,4.
Mean = (1+2+0+3+1+1+4+2+1) ÷ 9 = 15 ÷ 9 ≈ 1.67.
Median = 5th value = 1.
Mode = 1 (appears 4 times).

(b) The mean (≈1.67) is close to but below 2, and both the median and mode are 1, not 2 — so the manager's claim of ‘about 2 goals a game’ is a slight overstatement; ‘about 1 to 2 goals’ would better reflect the data, especially since two of the three averages equal 1.

Practice

Averages worksheet

Five short questions on averages. Work through them, then reveal the mark scheme to check.

  1. Find the mean, median and mode of: 5, 8, 8, 12, 17.
  2. A data set has mean 20 and 8 values. One value is removed and the new mean is 19. What was the value removed?
  3. Exam scores (out of 10) for a class of 15: 4,5,5,6,6,6,7,7,7,7,8,8,9,9,10. Find the median and mode.
  4. Weekly wages (£) grouped: £200–299 (freq 6), £300–399 (freq 10), £400–499 (freq 4). Estimate the mean wage.
  5. Explain, with a reason, which average (mean or median) would be more appropriate for describing typical house prices in a street where one house is much more expensive than the rest.

Mark scheme

  1. Ordered: 5,8,8,12,17. Mean = 50÷5 = 10. Median = 8 (middle value). Mode = 8 (appears twice).
  2. Original total = 20×8 = 160. New total (7 values) = 19×7 = 133. Value removed = 160−133 = 27.
  3. n = 15, median = 8th value = 7. Mode = 7 (appears 4 times, more than any other value).
  4. Midpoints: 249.5, 349.5, 449.5. Σfx = 6(249.5)+10(349.5)+4(449.5) = 1497+3495+1798 = 6790. Σf = 20. Mean = 6790÷20 = £339.50.
  5. The median — it isn't affected by the one extreme (much higher) value, so it better reflects the ‘typical’ price on the street; the mean would be pulled upward by the expensive outlier and would overstate the typical price.
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