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Cambridge 0580 IGCSE  ·  Topic E9.3

E9.3 Averages and Measures of Spread

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01 — Explanation

Which average should I use — mean, median or mode?

MeasureDefinitionBest used when
Meanxˉ=∑xn\bar{x}=\dfrac{\sum x}{n}Data is fairly symmetric, no extreme outliers
MedianMiddle value once orderedData has outliers or is skewed
ModeMost frequent valueCategorical data, or a "most popular" value is needed

For 4,7,4,9,12,4,84,7,4,9,12,4,8: mean =487=6.86=\dfrac{48}{7}=6.86; ordering gives 4,4,4,7,8,9,124,4,4,7,8,9,12, so the median (4th of 7 values) is 77; the mode is 44 (appears three times); the range is 12−4=812-4=8.

Always sort the data before picking out the median — the middle value of an unsorted list is meaningless.

02 — Explanation

How do I find the quartiles and interquartile range from a raw list?

IQR=Q3−Q1\text{IQR} = Q_3 - Q_1

Once the data is sorted, Q1Q_1 and Q3Q_3 can be found by splitting the ordered list into halves. With an even number of values, split exactly in half and find the median of each half; with an odd number, exclude the middle (overall median) value first, then split the remainder.

Example: 1212 ordered values: 3,5,6,8,9,11,12,14,15,17,19,223,5,6,8,9,11,12,14,15,17,19,22.

Lower half: 3,5,6,8,9,113,5,6,8,9,11 → Q1=6+82=7Q_1=\dfrac{6+8}{2}=7. Upper half: 12,14,15,17,19,2212,14,15,17,19,22 → Q3=15+172=16Q_3=\dfrac{15+17}{2}=16.

IQR=16−7=9\text{IQR} = 16-7=9

Unlike the range, the IQR ignores the most extreme 25%25\% of values at each end, so a single outlier can't distort it.

03 — Explanation

How do I estimate the mean of grouped data?

Once data is grouped into class intervals, individual values are lost, so the mid-interval value stands in for every item in that class:

xˉ≈∑(f⋅x)∑f\bar{x} \approx \frac{\sum(f \cdot x)}{\sum f}

Example: masses of 8080 apples.

Mass (gg)ffMid-interval xxf×xf\times x
100<m≤120100<m\le120121211011013201320
120<m≤140120<m\le140303013013039003900
140<m≤160140<m\le160262615015039003900
160<m≤180160<m\le180121217017020402040
xˉ=1320+3900+3900+204080=1116080=139.5 g\bar{x} = \frac{1320+3900+3900+2040}{80} = \frac{11160}{80} = 139.5\text{ g}

Examiner Tip

Multiply frequency by the mid-interval value, lower bound+upper bound2\frac{\text{lower bound}+\text{upper bound}}{2} — not by the class width, which is a completely different quantity used elsewhere (in histograms). Also, because individual masses are replaced by a mid-interval estimate, the answer must be called an estimate of the mean, never the exact mean.

04 — Explanation

How do I find the modal class?

The modal class is the whole class interval with the highest frequency — never a single number. In the table above, 3030 is the highest frequency, so the modal class is 120<m≤140120<m\le140.

The same "split into halves" method for Q1Q_1 and Q3Q_3 shown earlier works for a list of any size — with an odd number of values like 99, exclude the overall median itself before splitting the remaining 88 values into an upper and lower half of 44 each.

Examiner Tip

"Estimate the mean" and "find the mean" are different instructions — the word "estimate" is the signal that the data has been grouped and the mid-interval method is required, not a hint that any level of inaccuracy is acceptable.