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Cambridge 0580 IGCSE  ·  Topic C9.3

C9.3 Averages and Range

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At Core tier, data is always given as a list or an ungrouped frequency table — never grouped into class intervals.

01 — Explanation

How do I find the mean, median and mode from a list?

xˉ=∑xn(mean: sum of values÷number of values)\bar{x} = \frac{\sum x}{n} \quad \text{(mean: sum of values} \div \text{number of values)}

The median is the middle value once the data is ordered (or the average of the two middle values, for an even count). The mode is the most frequent value.

Common mistake

reading the median off the list in its original order. Data must be sorted into ascending order first, before the middle value is picked out — for an even number of values, the median is the mean of the two middle numbers, not either one alone.

Worked example. The number of books read last month by 7 students was 2,2,2,4,5,6,72, 2, 2, 4, 5, 6, 7. Find the mean, median, mode and range.

Ordered already, n=7n=7. Mode =2= 2 (appears 3 times). Median == the 4th value =4= 4.

xˉ=2+2+2+4+5+6+77=287=4range=7−2=5\bar{x} = \frac{2+2+2+4+5+6+7}{7} = \frac{28}{7} = 4 \qquad \text{range} = 7-2 = 5

02 — Explanation

How do I find the mean from a frequency table?

When data is grouped by frequency rather than listed individually, each value must be multiplied by how often it occurs before summing, since simply adding the distinct values would ignore how many times each one actually appeared.

xˉ=∑(x×f)∑f\bar{x} = \frac{\sum (x\times f)}{\sum f}

Worked example. The table shows the number of pets owned by 20 students. Find the mean number of pets.

Number of pets, xx0123
Frequency, ff4853
∑(x×f)=(0×4)+(1×8)+(2×5)+(3×3)=0+8+10+9=27∑f=4+8+5+3=20\sum(x\times f) = (0\times4)+(1\times8)+(2\times5)+(3\times3) = 0+8+10+9=27 \qquad \sum f = 4+8+5+3=20
xˉ=2720=1.35 pets\bar{x} = \frac{27}{20} = 1.35 \text{ pets}

Exam tip

always add an x×fx\times f row (or column) to the table before totalling — trying to do the multiplication mentally for every category is where most arithmetic slips happen in this type of question.

03 — Explanation

How do I find the range?

range=highest value−lowest value\text{range} = \text{highest value} - \text{lowest value}

04 — Explanation

Which average should I use, and when?

Mean uses every value and is best for an overall typical value, but is dragged by extreme outliers. Median is unaffected by outliers, since it only depends on position, not size. Mode is the only average that works for non-numerical (categorical) data, and is useful when the most popular value matters more than a typical one.

Worked example. A small company's salaries (in thousands) are 22,24,25,26,9022, 24, 25, 26, 90. Explain why the median is a better measure of a "typical" salary here than the mean.

xˉ=22+24+25+26+905=1875=37.4median=25\bar{x} = \frac{22+24+25+26+90}{5} = \frac{187}{5}=37.4 \qquad \text{median} = 25

The mean (37.437.4) is pulled upward by the single outlier of 9090, so it doesn't represent what most employees actually earn. The median (2525) is unaffected by that one extreme value and better reflects a typical salary in this data set.