Descriptive statistics: averages, range, IQR and standard deviation

Averages

• Mean: total / number of values. Uses every value but is distorted by extreme values (anomalies).
• Median: the middle value when ordered (for an even number of values, the mean of the middle two). Not affected by extremes.
• Mode: the most frequent value; for grouped data, the modal class is the class with the highest frequency. The January 2024 paper asked for the modal class of mangrove tree heights ('more than 7 m') and of building floors.

The June 2025 paper asked for the range of % bare sand read from a graph: 90 - 55 = 35 (from 2025 a correct answer with no working earns both marks).

Spread: range, IQR, standard deviation

• Range = highest - lowest. Simple, but uses only two values, so one anomaly distorts it.
• Interquartile range (IQR) = upper quartile - lower quartile: the spread of the middle 50%. Order the data; the median splits it in half; the lower quartile is the median of the lower half and the upper quartile the median of the upper half (here the overall median is left out of both halves when n is odd). The IQR ignores extremes and is shown in a box and whisker plot.
• Standard deviation measures the average spread around the mean: sigma = square root of (sum of (x - mean) squared / n). A small value means data are clustered around the mean (consistent); a large value means widely spread.

Practice data set

Practice data (not real): % bare sand at 11 sites along a dune transect, from the sea inland.

Worked example

For the practice data, find the median and the interquartile range.

1. Ordered: 55, 57, 58, 60, 65, 68, 70, 72, 85, 88, 90.
2. n = 11, so the median is the 6th value = 68.
3. Lower half (5 values): 55, 57, 58, 60, 65; lower quartile = 58.
4. Upper half: 70, 72, 85, 88, 90; upper quartile = 85.
5. IQR = 85 - 58 = 27.

Answer: Median 68%, IQR 27 percentage points.

Check yourself

1. Calculate the mean % bare sand for the practice data set. Give your answer to 1 decimal place.
   Show answer

   69.8 %. Total 768 / 11 = 69.82, so 69.8%.

2. What is the range of the practice data set?
   Show answer

   35 percentage points. 90 - 55 = 35.

3. Calculate the standard deviation of the practice data set (divide by n). Give your answer to 1 decimal place.
   Show answer

   12.1. Mean = 69.818; sum of squared deviations = 1619.64; divide by 11 and square root = 12.13.

4. Tree heights were grouped: 0 to 2 m (4 trees), 2.1 to 4 m (9), 4.1 to 7 m (11), more than 7 m (16). What is the modal class?
   • 4.1 to 7 m
   • More than 7 m
   • 2.1 to 4 m
   • 0 to 2 m
   Show answer

   More than 7 m. The class with the highest frequency (16) is 'more than 7 m'.

5. Why might the median be a better average than the mean for a site where one noise reading was unusually high because a lorry passed?
   • The median uses every value
   • The median is always larger
   • The mean cannot be calculated for noise
   • The median is not distorted by one extreme value
   Show answer

   The median is not distorted by one extreme value. Extremes pull the mean but not the median.

6. Which statement about the interquartile range is correct?
   • It is the highest value minus the lowest
   • It measures the spread of the middle 50% of the data
   • It is the same as the mean
   • It is always larger than the range
   Show answer

   It measures the spread of the middle 50% of the data. UQ - LQ covers the middle half.

7. Practice data: on a graph of noise against distance, the highest reading is 82 dB and the lowest 54 dB. Calculate the range in dB.
   Show answer

   28 dB. 82 - 54 = 28 dB.

Practise with instant feedback Print this lesson

Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Geography specification, Issue 3 (September 2025)
• Pearson WGE02/01 mark scheme, June 2025
• Pearson WGE02/01 mark scheme, January 2024
• Pearson WGE02/01 mark scheme, June 2024

Last updated 8 November 2026. Spotted a mistake? Tell us.