Q.Find the mean deviation about the mean for the following data: 6, 7, 10, 12, 13, 4, 8, 12
Concept understanding — Mean Deviation About Mean
Mean Deviation About Mean – The Intuition First
Imagine you have a small set of numbers: the marks of five students in a test: 4, 6, 8, 10, 12. The average (mean) is 8. Now, each student is some distance away from this average. The student who scored 4 is 4 marks below the mean; the one who scored 12 is 4 marks above. The student who scored 8 is exactly at the mean.
If you simply add these distances, the positives and negatives cancel out — you get zero. That's not useful. So instead, we ask: on average, how far is each data point from the mean? That's the mean deviation about mean.
Mean deviation is a measure of spread or dispersion. It tells you how scattered the data is around the central value. A small mean deviation means most data points are close to the mean; a large one means they are spread out.
The Precise Definition
For a set of n observations x1,x2,…,xn with mean xˉ, the mean deviation about mean (often written as MD or M.D.) is:
MD(xˉ)=n1∑i=1n∣xi−xˉ∣
That vertical bars mean absolute value — we take the distance without caring about direction. So every deviation is positive.
Mean Deviation about Mean=n∑∣xi−xˉ∣
Step-by-Step Calculation
Let's use the marks example: 4, 6, 8, 10, 12.
Step 1: Find the mean.
xˉ=54+6+8+10+12=540=8
Step 2: Find each absolute deviation ∣xi−xˉ∣.
| xi | xi−xˉ | ∣xi−xˉ∣ |
|------|----------------|-------------------|
| 4 | -4 | 4 |
| 6 | -2 | 2 |
| 8 | 0 | 0 |
| 10 | 2 | 2 |
| 12 | 4 | 4 |
Step 3: Sum the absolute deviations.
4+2+0+2+4=12
Step 4: Divide by n=5.
MD=512=2.4
So, on average, each student's mark is 2.4 marks away from the mean of 8.
Notice that the mean deviation is always less than or equal to the standard deviation (another measure of spread). For this data, standard deviation is about 2.83, which is larger than 2.4. This is because standard deviation squares deviations, giving more weight to extreme values.
Why Use Absolute Values?
You might wonder: why not just average the plain deviations (without absolute value)? Because the sum of (xi−xˉ) is always zero — that's a property of the mean. The absolute value is the simplest way to make all deviations positive so they don't cancel.
A common mistake: forgetting to take absolute values and getting zero. Always check: if your sum of deviations is zero, you forgot the absolute value.
When Is This Used?
Mean deviation is intuitive and easy to explain. It's used in:
- Quality control (checking how consistent a manufacturing process is)
- Economics (measuring income inequality)
- Early statistics courses (before introducing variance and standard deviation)
However, it has a limitation: absolute values are mathematically tricky to work with in advanced statistics (they aren't differentiable at zero). That's why standard deviation (which squares the deviations) is more common in higher-level work.
Quick Summary
| Concept | Meaning |
|---|---|
| Mean deviation about mean | Average absolute distance from the mean |
| Formula | $\frac{1}{n} \sum |
| Tells you | How spread out the data is, in the same units as the data |
| Key property | Always non-negative; zero only if all values are identical |
Final answer: Mean deviation about mean is the average of the absolute differences between each data point and the arithmetic mean. For the set {4,6,8,10,12}, it equals 2.4.
Mean Deviation about Mean is one of the measures of dispersion covered in the NCERT Class 11 Mathematics chapter on Statistics, matching searches like "mean deviation: formula and examples" or "statistics important questions class 11 maths". It's a regularly tested, calculation-based topic in CBSE boards, and understanding it also builds the intuition needed for standard deviation and variance questions in JEE Main and CET exams.
Concept: Mean Deviation About Mean
The mean deviation about the mean measures the average absolute distance of each observation from the arithmetic mean.
Step 1. Calculate the mean xˉ:
xˉ=86+7+10+12+13+4+8+12=872=9
Step 2. Find the absolute deviations ∣xi−xˉ∣ for each observation:
∣6−9∣=3,∣7−9∣=2,∣10−9∣=1,∣12−9∣=3,∣13−9∣=4,∣4−9∣=5,∣8−9∣=1,∣12−9∣=3
Step 3. Sum the absolute deviations and divide by n:
M.D.=83+2+1+3+4+5+1+3=822=2.75
The mean deviation about the mean is 2.75.
The mean of the data is 9, and the mean deviation about the mean is 2.75.
Understanding Mean Deviation About the Mean
Mean deviation measures how far, on average, each observation lies from the arithmetic mean. It uses absolute values (not squares), so it directly answers "how far is a typical data point from the centre?"
M.D.(xˉ)=n∑∣xi−xˉ∣
where xˉ is the mean and n is the number of observations.
Step-by-Step Solution
1. Find the mean
∑xi=6+7+10+12+13+4+8+12=72,n=8
xˉ=872=9
2. Absolute deviations from the mean
| xi | ∣xi−9∣ |
|-------|-------------|
| 6 | 3 |
| 7 | 2 |
| 10 | 1 |
| 12 | 3 |
| 13 | 4 |
| 4 | 5 |
| 8 | 1 |
| 12 | 3 |
3. Sum the absolute deviations
∑∣xi−xˉ∣=3+2+1+3+4+5+1+3=22
4. Divide by n
M.D.(xˉ)=822=2.75
The mean deviation about the mean is 2.75.
- CBSE 2025Set ANNUAL1 markMCQQ.Mean Deviation about mean of 5, 8, 9, 10, 11, 13, 14, 18 is:(a) 2.5(b) 3(c) 3.5(d) 4
›Reveal solutionSolution
Find the mean, take absolute deviations of each value from it, then average those deviations.
Data: 5,8,9,10,11,13,14,18 (n=8).
Mean xˉ=85+8+9+10+11+13+14+18=888=11
Absolute deviations ∣xi−xˉ∣: 6,3,2,1,0,2,3,7
Sum of deviations =6+3+2+1+0+2+3+7=24
Mean deviation =824=3
✓Final answerMean deviation about the mean =3 — option (b).
- CBSE 2023Set ANNUAL1 markQ.Write the formula of mean deviation about mean for grouped data.
›Reveal solutionSolution
Mean deviation averages the absolute deviations of each class mark from the mean, weighted by frequency.
For grouped data with class marks xi, frequencies fi and N=∑fi, mean deviation about the mean xˉ is:
M.D.(xˉ)=N1∑ifi∣xi−xˉ∣
Each class mark's distance from the mean is taken as positive (absolute value) so that deviations above and below the mean don't cancel out, and the whole is weighted by how many observations (fi) fall in that class.
✓Final answerM.D.=N1∑fi∣xi−xˉ∣.
- CBSE 2023Set ANNUAL1 markQ.Fill in the blank: The formula for mean deviation from the mean, for ungrouped data, is ____.
›Reveal solutionSolution
For ungrouped data, mean deviation about the mean is n∑∣xi−xˉ∣.
Mean deviation measures the average absolute spread of the data from its mean.
For ungrouped data x1,x2,…,xn with mean xˉ, we take the absolute deviation ∣xi−xˉ∣ of each observation, sum them, and divide by the number of observations n.
✓Final answerMean deviation from mean =n∑∣xi−xˉ∣, where xˉ is the mean and n is the number of observations.
- CBSE 2023Set ANNUAL1 markQ.What is the mean deviation from the mean of the data 3,4,5,6,7?
›Reveal solutionSolution
The mean deviation of 3,4,5,6,7 about their mean is 1.2.
Mean xˉ=53+4+5+6+7=525=5.
Absolute deviations: ∣3−5∣=2, ∣4−5∣=1, ∣5−5∣=0, ∣6−5∣=1, ∣7−5∣=2.
Sum of absolute deviations =2+1+0+1+2=6.
Mean deviation =56=1.2.
✓Final answerMean deviation =1.2.
- CBSE 2022Set ANNUAL1 markQ.The mean deviation of 5, 6, 9, 11, 12, 3, 7, 11 about Mean is ............ .
›Reveal solutionSolution
The mean is 8; averaging ∣xi−8∣ over all 8 values gives a mean deviation of 2.75.
Data: 5,6,9,11,12,3,7,11 (n = 8)
Mean:
xˉ=85+6+9+11+12+3+7+11=864=8
Absolute deviations ∣xi−xˉ∣:
∣5−8∣=3, ∣6−8∣=2, ∣9−8∣=1, ∣11−8∣=3, ∣12−8∣=4, ∣3−8∣=5, ∣7−8∣=1, ∣11−8∣=3
Sum of deviations: 3+2+1+3+4+5+1+3=22
Mean deviation:
MD=822=2.75
✓Final answerMean deviation about the mean =2.75.
- CBSE 2019Set ANNUAL1 markMCQQ.The mean deviation about mean of the following data 14, 17, 18, 19, 20, 22, 23, 27 is:(a) 20(b) 0(c) 3(d) √14
›Reveal solutionSolution
Compute the mean, then find n1∑∣xi−xˉ∣.
Data: 14,17,18,19,20,22,23,27 (n=8).
Mean xˉ=814+17+18+19+20+22+23+27=8160=20
Absolute deviations from mean: 6,3,2,1,0,2,3,7
Sum of deviations =6+3+2+1+0+2+3+7=24
Mean deviation =824=3
✓Final answerMean deviation about the mean =3, option (c).
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