Simulation · Maths · Class 11

Where is the total distance least?

From the lesson Mean deviation for ungrouped data in Statistics. Change the values and watch what happens.

The idea behind it

NCERT §13.4, §13.4.1

The deviation of x from a fixed value a is x − a. Some deviations are negative and some positive, and the deviations from the mean always add up to 0, so their plain average is useless as a measure of spread.
Taking absolute values turns each deviation into a distance on the number line. Mean deviation about a: M.D.(a) = (1/n) Σ |xᵢ − a|.
It can be taken about any central value, but mean deviation about the mean, M.D.(x̄) = (1/n) Σ |xᵢ − x̄|, and about the median, M.D.(M) = (1/n) Σ |xᵢ − M|, are the ones in common use. M stands for the median.
Steps: find the central value a; write each deviation xᵢ − a; drop the signs; take the mean of these absolute values.
Arjun (made up), about x̄ = 50: absolute deviations 8, 2, 0, 2, 5, 5, 0, 8 add to 30, so M.D.(x̄) = 30/8 = 3.75 runs. Kabir's absolute deviations add to 270, giving 33.75 runs: about nine times the spread for the same mean.
About the median: Kabir's sixes in 7 innings (made up) are 2, 0, 5, 1, 9, 3, 1. In order 0, 1, 1, 2, 3, 5, 9, so M = 2 (the 4th value). The distances from 2 add to 15, so M.D.(M) = 15/7 ≈ 2.14.
The same data about the mean x̄ = 3 give distances adding to 16, so M.D.(x̄) = 16/7 ≈ 2.29. The sum of absolute deviations is least when taken about the median.
NCERT's worked value for comparison: the data 6, 7, 10, 12, 13, 4, 8, 12 have mean 9 and M.D.(x̄) = 22/8 = 2.75.

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