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How do you find quartiles and use them to detect outliers?

Finding quartiles from a list and from cumulative frequency, the interquartile range, percentiles and identifying outliers.

A focused answer to AQA GCSE Statistics on quartiles and the interquartile range, covering how to find quartiles from a list and from a cumulative frequency curve, calculate the interquartile range, and use the 1.5 times IQR rule to identify outliers.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. Finding quartiles from a list
  3. The interquartile range
  4. Quartiles from cumulative frequency
  5. Identifying outliers

What this dot point is asking

AQA wants you to find quartiles from an ordered list and from a cumulative frequency curve, calculate the interquartile range, work with percentiles, and use the 1.5Γ—IQR1.5 \times \text{IQR} rule to decide whether a value is an outlier. This material underpins box plots, so accurate quartiles here feed straight into the diagram questions.

Finding quartiles from a list

For a list of nn values that has been sorted, AQA uses these positions: Q1Q_1 at n+14\frac{n+1}{4}, the median at n+12\frac{n+1}{2}, and Q3Q_3 at 3(n+1)4\frac{3(n+1)}{4}. When a position is a whole number, read off that value directly. When it falls between two positions, interpolate: a position of 3.53.5 means halfway between the 33rd and 44th ordered values. Note the small but important difference: for a raw list you use n+14\frac{n+1}{4}, but for a cumulative frequency graph of grouped data you use n4\frac{n}{4}, because a continuous curve treats the data as one large block rather than discrete items.

The interquartile range

Because it discards the bottom and top quarters, the interquartile range is the natural partner to the median for skewed data. It also drives the outlier test below and sets the box length on a box plot.

Quartiles from cumulative frequency

For grouped data plotted as a cumulative frequency curve, read Q1Q_1 at n4\frac{n}{4}, the median at n2\frac{n}{2} and Q3Q_3 at 3n4\frac{3n}{4} on the vertical axis, then go across to the curve and down to the horizontal axis to estimate each value. These are estimates because the original values inside each class are unknown. The points are read where the curve crosses, so a smooth, accurately plotted ogive is essential for marks.

Quartiles also feed directly into other diagrams and statistics. The five-number summary (minimum, Q1Q_1, median, Q3Q_3, maximum) is exactly what a box plot displays, so accurate quartiles here translate into a correct box plot in the next topic. The interquartile range is the box length, the median position inside the box hints at skew, and the outlier boundaries decide where the whiskers stop. Because the quartiles ignore the most extreme quarter at each end, they give a robust picture of the data that is not thrown off by a single unusual value, which is why they are the standard summary for skewed real-world data such as incomes or waiting times.

Identifying outliers

The standard AQA outlier rule flags any value that sits well outside the central half of the data. The factor 1.51.5 is a convention: it sets the boundary far enough from the box that ordinary values are kept but genuinely unusual ones are flagged. A value beyond the boundary is sometimes called a "mild" outlier, and the rule is applied at both ends, so a data set can have low outliers, high outliers, both or none.

Exam-style practice questions

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

AQA 20185 marksThe masses (kg) of 1111 parcels are 2,3,3,5,6,7,8,9,11,12,302, 3, 3, 5, 6, 7, 8, 9, 11, 12, 30. (a) Find the median and the lower and upper quartiles. (b) Using the 1.5Γ—IQR1.5 \times \text{IQR} rule, determine whether the parcel of mass 3030 kg is an outlier.
Show worked answer β†’

(a) With n=11n = 11 ordered values: median at position 11+12=6\frac{11+1}{2} = 6, so Q2=7Q_2 = 7. Lower quartile at position 11+14=3\frac{11+1}{4} = 3, so Q1=3Q_1 = 3. Upper quartile at position 3(11+1)4=9\frac{3(11+1)}{4} = 9, so Q3=11Q_3 = 11.

(b) IQR=11βˆ’3=8\text{IQR} = 11 - 3 = 8. Upper boundary =Q3+1.5Γ—IQR=11+12=23= Q_3 + 1.5 \times \text{IQR} = 11 + 12 = 23. Since 30>2330 > 23, the 3030 kg parcel is an outlier.

Markers reward correct quartile positions, the interquartile range, the boundary calculation, and a clear yes/no conclusion with the comparison.

AQA 20213 marksA cumulative frequency graph is drawn for 200200 exam scores. Describe how you would use the graph to estimate the interquartile range.
Show worked answer β†’

Read up to the lower quartile at cumulative frequency n4=2004=50\frac{n}{4} = \frac{200}{4} = 50: go across from 5050 on the vertical axis to the curve, then down to the score axis to read Q1Q_1.

Repeat at 3n4=150\frac{3n}{4} = 150 to read Q3Q_3.

The interquartile range is Q3βˆ’Q1Q_3 - Q_1.

Markers reward using n4\frac{n}{4} and 3n4\frac{3n}{4} (not n+14\frac{n+1}{4} for a graph), the read-across-and-down method, and the final subtraction.

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