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Statistics

Measures of dispersion

Compute the range and the mean deviation of a list of numbers.

The mean, median and mode each give one number that stands for the whole data. That is useful, but it hides something important: how spread out the values are.

Two classes each have mean mark 50:

  • Class A: 40, 50, 60. Everyone is close to 50.
  • Class B: 10, 50, 90. The marks are far apart.
  • Same centre, very different pictures. A number that measures the spread is called a measure of dispersion.

Range = largest value – smallest value. Class A: 60 – 40 = 20; class B: 90 – 10 = 80. It is quick, but it uses only the two end values and ignores everything in between.

Why not just average the distances from the mean? Because the plus and minus distances cancel: the deviations from the mean always add up to 0. For 3, 5, 7, 9, 11, 13 (mean 8) the deviations are –5, –3, –1, 1, 3, 5, and their sum is 0.

So we drop the signs and use the absolute values (sizes) of the deviations. Their average is the mean deviation:

  • M.D.(a) = Σ|xᵢ – a| / n, where a is the centre used, usually the mean or the median.
  • Steps: find the centre, subtract it from every value, drop the signs, add, divide by n.

About the mean: 3, 5, 7, 9, 11, 13. Mean = 48/6 = 8.

  • |deviations|: 5, 3, 1, 1, 3, 5. Sum 18.
  • M.D.(mean) = 18/6 = 3: on average a value is 3 away from the mean.

About the median: 2, 4, 5, 7, 9, 12, 16 (already in order, 7 values).

  • The median is the 4th value, 7. (With an even number of values, take the average of the middle two.)
  • |deviations|: 5, 3, 2, 0, 2, 5, 9. Sum 26.
  • M.D.(median) = 26/7 ≈ 3.71.

Always put the data in order before finding the median, and check your sums: adding a long column is where most slips happen.

Average distance from the centre.

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