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Mean deviation questions look routine until the data come as a frequency table or a set of classes, or until a question asks you to compare the deviation about the mean with the one about the median. This post works all three data types with fresh numbers, explains why the median gives the smallest possible mean deviation, and lists the slips that cost marks.
Is mean deviation still in the syllabus?
Yes. The JEE (Main) 2026 syllabus lists "mean deviation for grouped and ungrouped data" under statistics, and the JEE (Advanced) 2026 brochure lists "mean deviation, standard deviation and variance of grouped and ungrouped data". The CBSE 2026-27 mathematics curriculum keeps mean deviation in Class 11, and section 13.4 of the NCERT Class 11 chapter Statistics (2026-27 reprint) teaches it about both the mean and the median. Check the current documents for your exam.
What is mean deviation about the median?
NCERT explains why we need absolute values. Deviations from the mean add to zero, so their plain average is useless as a measure of spread. Taking the absolute value gives each observation's distance from a central value a, and the mean of those distances is the mean deviation, M.D.(a). About the median M:
Ungrouped data: M.D.(M) = Σ|xᵢ − M| / n
Frequency data: M.D.(M) = Σfᵢ|xᵢ − M| / N, where N = Σfᵢ
For classes, xᵢ is the mid-point of each class.
Worked example 1: raw data with an outlier
Sample marks of seven students: 12, 15, 9, 18, 14, 11, 47.
Sort: 9, 11, 12, 14, 15, 18, 47. With n = 7, the median is the 4th value, M = 14.
Distances from 14: 5, 3, 2, 0, 1, 4, 33. Sum = 48.
M.D.(M) = 48/7 ≈ 6.86.
For comparison, the mean is 126/7 = 18, the distances from 18 add to 58, and M.D. about the mean is 58/7 ≈ 8.29. The deviation about the median is smaller. That is not a coincidence.
Why is the median the point that gives the smallest mean deviation?
Picture the seven marks on a number line and a point a that you can slide. Move a one unit to the right. Every observation to the right of a gets one unit closer, and every observation to the left gets one unit further away. If more observations are on the right, the total distance falls, so keep moving. The total stops falling when as many observations are on each side, and that is the median.
Two consequences turn up in exam questions:
Σ|xᵢ − a| is smallest when a is the median, so M.D. about the median is never larger than M.D. about the mean, or about any other point.
With an even number of observations, every point between the two middle values gives the same total. For the sample data 3, 7, 8, 10, 13, 15, the sum of distances is 20 for a = 8, 8.5, 9 or 10, but 22 for a = 7 or 11.
Worked example 2: a discrete frequency table
Sample data:
x: 2, 4, 6, 8, 10
f: 3, 5, 9, 6, 2
N = 25, so the median is the 13th observation.
Cumulative frequencies: 3, 8, 17, 23, 25. The 13th observation falls in the group with cumulative frequency 17, so M = 6.
|x − 6|: 4, 2, 0, 2, 4.
f|x − 6|: 12, 10, 0, 12, 8. Sum = 42.
M.D.(M) = 42/25 = 1.68.
The mean here is 5.92, and the mean deviation about it works out to about 1.71, again a little larger.
Worked example 3: grouped data in classes
Sample data:
Class: 0–10, 10–20, 20–30, 30–40, 40–50
f: 5, 8, 12, 10, 5
N = 40, N/2 = 20. Cumulative frequencies: 5, 13, 25, 35, 40. The median class is 20–30.
NCERT's formula: Median = l + [(N/2 − C)/f] × h, with l = 20, C = 13, f = 12, h = 10. Median = 20 + (7/12) × 10 ≈ 25.83.
Mid-points: 5, 15, 25, 35, 45. Distances from 25.83: 20.83, 10.83, 0.83, 9.17, 19.17.
f × distance: 104.17, 86.67, 10.00, 91.67, 95.83. Sum ≈ 388.33.
M.D.(M) = 388.33/40 ≈ 9.71.
Keep the median as 25.833 or as the fraction 310/12 while you work, and round only at the end.
Which mistakes cost marks?
Not sorting the data before picking the median.
Using the class limits instead of mid-points for xᵢ.
Dividing by the number of classes instead of N = Σf.
Forgetting the absolute value, which drags the sum towards zero.
Rounding the median early in grouped data, which shifts every distance.
A practice routine
Work NCERT Exercise 13.1, questions 3, 4, 7 and 8, which ask for mean deviation about the median.
For one data set, find M.D. about both the mean and the median, and check the median gives the smaller value.
Slide-test an even-sized set: compute Σ|x − a| for three values of a between the middle pair and two outside it.
Then move on to variance and standard deviation in our post on statistics problems for JEE.
If the reason the median minimises distance still feels like a trick, the Fermi Concept Library covers every concept in mathematics with its own lesson, so you can see the idea built up step by step.
Tonight, redo example 1 with the 47 replaced by 19 and see how much each mean deviation changes.
