Mean Deviation About Mean and Median
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Direct answer
Mean deviation is the average of absolute deviations from a chosen central value: MD about a equals (1/n) Σ |xi - a| for raw data, or Σ fi |xi - a| / Σ fi for a frequency distribution. Its defining property is that the median minimises it — MD about the median is the smallest such average over all choices of a, a fact that distinguishes it from the standard deviation, which the mean minimises in the squared sense. Two results JEE expects on tap: the MD of the first n natural numbers about their mean is (n^2 - 1)/4n, and for any data MD ≤ SD, with the normal distribution displaying the tidy ratios QD = 2σ/3, MD = 4σ/5, SD = σ. The coefficient of mean deviation, MD divided by the same central value, renders the measure unit-free for comparisons.
What you must remember
- Definition: MD about a = Σ|xi - a|/n (or with frequencies, Σ fi |xi - a|/Σfi); absolute values are non-negotiable — signs must not cancel.
- Median minimality: the MD about the median is the least possible; about the mean it is larger (or equal for symmetric data), a one-mark fact asked repeatedly.
- First n naturals: MD about the mean = (n^2 - 1)/4n; for n = 5 this is 24/20 = 1.2, verifiable by hand on {1, 2, 3, 4, 5}.
- Relation to SD: for any data, MD ≤ SD; equality-type questions exploit data concentrated at two values where the gap narrows.
- Normal-curve ratios: for a normal distribution, QD : MD : SD = 10 : 12 : 15, i.e., QD = 2σ/3 and MD = 4σ/5 ≈ 0.7979σ.
- Coefficient of MD: MD about mean ÷ mean, or MD about median ÷ median — a unit-free relative measure for comparing scatter across data sets.
- Grouped data shortcut: for a modest frequency table, take the median class, estimate the median, then compute Σ fi |xi - median| directly rather than building a full deviation column first.
Watching the median win
Take the data 3, 5, 7, 15. The median of the four values is (5 + 7)/2 = 6, and the mean is 30/4 = 7.5. MD about the median: |3 - 6| + |5 - 6| + |7 - 6| + |15 - 6| = 3 + 1 + 1 + 9 = 14, so MD = 14/4 = 3.5. MD about the mean: |3 - 7.5| + |5 - 7.5| + |7 - 7.5| + |15 - 7.5| = 4.5 + 2.5 + 0.5 + 7.5 = 15, so MD = 15/4 = 3.75. The median's total beats the mean's by exactly one unit here — not an accident but the visible edge of the minimisation theorem: each step of a away from the median flips the sign-contribution of the observations on one side. The outlier 15 drags the mean upward, and the mean then pays for its own displacement; the median, resistant to the outlier, stays cheaper. This is the standard exam demonstration — a small data set, both central values computed, and the comparison asked as "verify that MD about the median is least".
How the exam frames it
JEE Main asks for the MD of a small data set about the mean or median, the (n^2 - 1)/4n result for naturals, or the coefficient of MD — three to four marks of careful arithmetic, usually as a numerical value. Frequency-table versions with ten to twelve classes appear too, and the combined-formula trap lands there: students average deviations of class marks without weighting by frequencies. Advanced seldom features MD alone; it embeds the concept in comparisons (which measure of dispersion is least affected by extreme values — answer MD; which is least — range; which enters the normal machinery — SD) and in the normal ratios. The predictable losses: dropping absolute values midway and summing to zero; computing MD about the mean when the question asks about the median on symmetric-looking data; and misstating the minimiser (the mean minimises the sum of squared deviations; the median minimises the sum of absolute ones — swapping the two is the classic single-correct trap). Statistics sits explicitly in the Main syllabus; MD itself is the least examined of the dispersion measures, which is exactly why it is safe marks when it appears.
Frequently asked questions
About which central value is the mean deviation minimum?
The median — Σ|xi - a| is smallest when a is the median, making MD about the median the least of all such averages.
What is the mean deviation of the first n natural numbers about their mean?
(n^2 - 1)/4n; for the data 1 to 5 it equals 24/20 = 1.2.
How is the coefficient of mean deviation defined?
As MD divided by the central value used — MD about mean over the mean, or MD about median over the median — producing a unit-free measure of relative scatter.
How does mean deviation compare with standard deviation?
MD is always less than or equal to SD for the same data; squaring punishes large deviations more, so SD exceeds MD whenever spread is uneven.
Which measure of dispersion is least affected by extreme values?
The mean deviation (and the quartile deviation), because absolute — not squared — deviations, and medians rather than means, keep outliers from dominating.