Lesson 3 of 11 · 11 min
Mean deviation for ungrouped data
NCERT §13.4, §13.4.1
Kabir's supporters say he hits big. The coach counts his sixes in the 7 innings where he faced at least one over: 2, 0, 5, 1, 9, 3, 1 (made up). He wants one number for how far a typical innings is from the centre.
The lesson in notes
In short
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.