Calculate the mean deviation about the mean for the following frequency distribution:
| Class interval | 0 - 4 | 4 - 8 | 8 - 12 | 12 - 16 | 16 - 20 |
|---|---|---|---|---|---|
| Frequency | 4 | 6 | 8 | 5 | 2 |
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🔒 Start your 14-day free trial to unlock the full solution →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) …
Concept: Mean Deviation About Mean measures the average absolute deviation of observations from their arithmetic mean.
Solution:
First, find the class marks xi and compute the mean xˉ:
| Class | xi | fi | fixi |
|---|---|---|---|
| 0–4 | 2 | 4 | 8 |
| 4–8 | 6 | 6 | 36 |
| 8–12 | 10 | 8 | 80 |
| 12–16 | 14 | 5 | 70 |
| 16–20 | 18 | 2 | 36 |
| Total | 25 | 230 |
Mean: xˉ=∑fi∑fixi=25230=9.2
Next, calculate absolute deviations ∣xi−xˉ∣ and their weighted sum:
| xi | fi | ∣xi−9.2∣ | fi∣xi−9.2∣ |
|-------|-------|---------------|---------------------|
| 2 | 4 | 7.2 | 28.8 |
| 6 | 6 | 3.2 | 19.2 | …
Using class midpoints, the mean is xˉ=9.2 and the mean deviation about the mean is 3.84.
Step-by-step solution
1. Midpoints and frequencies
| Class interval | Midpoint xi | Frequency fi |
|---|---|---|
| 0 - 4 | 2 | 4 |
| 4 - 8 | 6 | 6 |
| 8 - 12 | 10 | 8 |
| 12 - 16 | 14 | 5 |
| 16 - 20 | 18 | 2 |
Total frequency N=∑fi=25.
2. Mean
xˉ=N∑fixi=252(4)+6(6)+10(8)+14(5)+18(2)=258+36+80+70+36=25230=9.2.
3. Absolute deviations and weighted sum
| xi | fi | ∣xi−xˉ∣ | fi∣xi−xˉ∣ |
|---|---|---|---|
| 2 | 4 | 7.2 | 28.8 |
| 6 | 6 | 3.2 | 19.2 |
| 10 | 8 | 0.8 | 6.4 |
| 14 | 5 | 4.8 | 24.0 |
- KEAM 2026Set eng-2026-04174 marksMCQQ.The mean of the data : 4, 7, x, 13, 16 is 10. Then the mean deviation of the data is (A) 3 (B) 3.3 (C) 3.6 (D) 3.8 (E) 4
›Reveal solutionSolution
Find x from the mean, then compute n1∑∣xi−xˉ∣.
Mean =10: 54+7+x+13+16=10⇒40+x=50⇒x=10. …
- KEAM 2026Set eng-2026-04184 marksMCQQ.The following set of data is given. 72, 72, 74, 80, 82, 82, 86, 88, 92, 92. The mean deviation from the median is (A) 5 (B) 6 (C) 7 (D) 8 (E) 9
›Reveal solutionSolution
Median is 82; the average absolute deviation from it is 60/10=6.
The ordered data 72,72,74,80,82,82,86,88,92,92 has 10 values, so the median is the mean of the 5th and 6th: 282+82=82. …
- KEAM 2026Set eng-2026-04194 marksMCQQ.The mean deviation about the mean for the data: 5, 6, 14, 15 is (A) 3.5 (B) 4.5 (C) 4.2 (D) 3.8 (E) 4.0
›Reveal solutionSolution
Compute the mean, then average the absolute deviations.
Mean =45+6+14+15=440=10. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.Let x1,x2,…,xn be the data with respective frequencies f1,f2,…,fn. If ∑i=1nfi(∣xi−xˉ∣)=400 and the mean deviation from the mean xˉ is 10, then ∑i=1nfi is equal to (A) 55 (B) 50 (C) 36 (D) 40 (E) 25
›Reveal solutionSolution
Mean deviation =∑fi∑fi∣xi−xˉ∣; solve for ∑fi. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.The mean deviation about the mean from the data 400,410,420,430,440 is (A) 14 (B) 10 (C) 20 (D) 12 (E) 16
›Reveal solutionSolution
Mean of the data is 420; average of absolute deviations {20,10,0,10,20} is 60/5=12.
The values 400,410,420,430,440 have mean 5400+410+420+430+440=52100=420. …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The mean deviation about the mean for the following data
x 2 4 6 10 f 7 4 5 4 (A) 2 (B) 2.5 (C) 4 (D) 6 (E) 5 ›Reveal solutionSolution
The mean is 5; summing f∣x−5∣=50 over N=20 gives mean deviation 2.5.
Data: x=2,4,6,10 with f=7,4,5,4, so N=20.
Mean: ∑fx=2(7)+4(4)+6(5)+10(4)=14+16+30+40=100, xˉ=20100=5. …
- KEAM 2025Set eng-2025-04284 marksMCQQ.The mean deviation from mean of the five numbers 2,4,6,8,10 is (A) 2.4 (B) 3.6 (C) 4.8 (D) 6 (E) 0
›Reveal solutionSolution
With mean 6, the average absolute deviation of 2,4,6,8,10 is 512=2.4.
The mean of 2,4,6,8,10 is
xˉ=52+4+6+8+10=6. …
- KEAM 2024Set eng-2024-06094 marksMCQQ.The mean deviation of the numbers 3, 10, 10, 4, 7, 10 and 5 from the mean is (A) 2 (B) 2.5 (C) 2.57 (D) 3 (E) 3.75
›Reveal solutionSolution
Compute the mean, then average the absolute deviations.
Data: 3,10,10,4,7,10,5. Sum =49, so mean =749=7. …
- KEAM 2023Set eng-2023-P2-B24 marksMCQQ.The mean deviation about the median for the data 3, 5, 9, 3, 8, 10, 7 is (A) 723 (B) 74 (C) −74 (D) 716 (E) 717
›Reveal solutionSolution
For the data with median 7, the mean deviation about the median is 16/7.
Concept and Intuition
Sort the values, find the median (the middle of 7 values), take absolute deviations from it, and average them.
Step-by-Step Solution
- Sorted: 3,3,5,7,8,9,10; median = 4th value = 7.
- Absolute deviations: 4,4,2,0,1,2,3; sum = 16.
- Mean deviation = 16/7. …
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