Box Plots from Cumulative Frequency and Outliers
A box plot built from cumulative frequency data is a diagram showing the five-number summary, minimum, lower quartile, median, upper quartile and maximum, estimated from a cumulative frequency graph rather than raw data. An outlier is a value more than 1.5 times the interquartile range beyond a quartile, marked as a separate point rather than at the end of a whisker. This is assessed on GCSE Statistics Higher tier.
Before you start
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Method
- From a cumulative frequency graph or table, estimate the median (position n/2), lower quartile (position n/4) and upper quartile (position 3n/4), interpolating within grouped data if needed.
- Record the five-number summary: minimum, lower quartile, median, upper quartile, maximum.
- Calculate the interquartile range: IQR = Q3 - Q1.
- Calculate the outlier boundaries: lower boundary = Q1 - 1.5 x IQR, upper boundary = Q3 + 1.5 x IQR.
- Test the minimum, maximum and any other suspicious values against these boundaries: any value beyond a boundary is an outlier.
- Draw the box plot with the box from Q1 to Q3 (median marked inside), whiskers extending to the most extreme values that are NOT outliers, and outliers marked individually beyond the whiskers.
Worked example
The cumulative frequency table below shows delivery times, in minutes, for 80 parcels: 0 < t <= 20, frequency 10; 20 < t <= 40, frequency 20; 40 < t <= 60, frequency 40; 60 < t <= 100, frequency 10. The minimum delivery time was 2 minutes and the maximum was 95 minutes. Estimate the quartiles, then determine whether the minimum and maximum should be plotted as outliers.
- With n = 80, the lower quartile is at position n/4 = 20, the median at position n/2 = 40, and the upper quartile at position 3n/4 = 60.
- The cumulative frequencies are 10, 30, 70, 80 at the class boundaries 20, 40, 60, 100. The lower quartile falls in the class 20 < t <= 40 (cumulative frequency 10 to 30): Q1 = 20 + (20 - 10)/20 x 20 = 30 minutes.
- The median falls in the class 40 < t <= 60 (cumulative frequency 30 to 70): median = 40 + (40 - 30)/40 x 20 = 45 minutes.
- The upper quartile also falls in the class 40 < t <= 60: Q3 = 40 + (60 - 30)/40 x 20 = 55 minutes.
- IQR = 55 - 30 = 25. Lower boundary = 30 - 1.5 x 25 = -7.5. Upper boundary = 55 + 1.5 x 25 = 92.5. The minimum (2) is above -7.5, so it is not an outlier; the maximum (95) is above 92.5, so it is an outlier.
- Final answer: Q1 = 30, median = 45, Q3 = 55 minutes; the maximum value of 95 minutes is an outlier and should be plotted as a separate point, with the whisker drawn only to the highest non-outlier value.
Practice questions
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Exam-style questions
Written in the style of a GCSE Statistics exam paper, with a full mark scheme.
A data set has Q1 = 24 and Q3 = 40. Calculate the IQR and both outlier boundaries.
The masses, in kg, of 50 parcels have quartiles Q1 = 3.2 kg and Q3 = 7.6 kg. The heaviest parcel has a mass of 15 kg. (a) Calculate the IQR and the upper outlier boundary. (b) Determine whether the heaviest parcel is an outlier.
A cumulative frequency table for the times, in minutes, of 120 runners gives Q1 = 45, median = 55, Q3 = 65. The slowest runner finished in 140 minutes. (a) Calculate the IQR. (b) Calculate the upper outlier boundary. (c) Determine whether the slowest runner should be shown as an outlier, and explain how their result should be plotted.
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