Mean Deviation Questions 

Mean Deviation, also known as Average Deviation, is a statistical measure that calculates the average distance of data points from the mean or median. It helps assess the variability or dispersion in a data set. Mean deviation questions test your ability to apply formulas, interpret results, and analyze data consistency. Commonly featured in exams like JEE and board-level mathematics, these questions strengthen your understanding of data behavior beyond just central tendency.

1.0What is Mean Deviation?

The average of the absolute deviations from a central point (mean or median) is called Mean Deviation. It gives an idea of how spread out the data is.

2.0Formula for Mean Deviation

For a discrete series:

• From Mean: $\text{MD(from mean)}=\frac{1}{N}\sum |x_i-\bar{x}|$
• From Median: $\text{MD (from median)}=\frac{1}{N}\sum |x_i-Median|$

For a grouped frequency distribution:

• From Mean: $MD=\frac{\sum f|x-\bar{x}|}{\sum f}$

Where:

• x = mid-point of the class
• f = frequency
• $\bar{x}$ = mean of the data

Also Learn: Median for Grouped Data

3.0Solved Questions and Answers on Mean Deviation

Example 1: Find the mean deviation about the mean for the data: 2, 4, 6, 8, 10

Solution:

1. Mean = $\bar{x}=\frac{2+4+6+8+10}{5}=16$
2. Deviations from mean: $|2-6|=4,|4-6|=2,|6-6|=0,|8-6|=2,|10-6|=4$
3. MD = $\frac{4+2+0+2+4}{5}=\frac{12}{5}=2.4$

Example 2: Find the mean deviation about the mean for the following data:

Solution:

1. Calculate ($\bar{x}$): $\bar{x}=\frac{2\times 3+3\times 4+4\times 5+3\times 6+2\times 7}{2+3+4+3+2}=\frac{66}{14}=4.714$
2. Find $f|x-\bar{x}|$ for each row and sum it.
3. MD = $\frac{\sum f|x-\bar{x}|}{\sum f}\approx \frac{11.57}{14}\approx 0.83$

Example 3: The weights (in kg) of 5 students are: 48, 52, 50, 49, 51. Find the mean deviation from the mean.

Solution:

1. Calculate the Mean: $\bar{x}=\frac{48+52+49+51}{5}=\frac{250}{5}=50$
2. Find Deviations:

3. Sum of absolute deviations:$=2+2+0+1+1=6$
4. Mean Deviation: $MD=\frac{6}{5}=1.2$

Example 4: Find the mean deviation about the median for the following frequency distribution:

Solution:

1. Calculate cumulative frequency and find median class:

• Total frequency N = 5 + 8 + 15 + 16 + 6 = 50
• $\frac{N}{2}=25$, so the median class is 20–30.

2. Apply median formula: $Median=l+(\frac{\frac{N}{2}-F}{f})\times h$

Where:

• l = 20, lower boundary of median class
• F = 5 + 8 = 13, cumulative frequency before median class
• f = 15, frequency of median class
• h = 10, class width

$\Rightarrow Median=20+(\frac{25-13}{15}).10$

$\Rightarrow Median=20+(\frac{12}{15}).10$

$\Rightarrow Median=20+8=28$

3. Find mid-points and deviations from median (28):

4. Sum of f·|x – median| = 115 + 104 + 45 + 112 + 102 = 478
5. Mean Deviation: $MD=\frac{\sum f|x-median|}{\sum f}=\frac{478}{50}=9.56$

Example 5: If the mean of five observations is 18 and the mean deviation from mean is 2, then what is the sum of absolute deviations?

Options:
A) 10
B) 12
C) 8
D) 6

Solution: 

$MD=\frac{\sum |x_i-\bar{x}|}{n}$

$\Rightarrow 2=\frac{\sum |x_i-18|}{5}$

$\Rightarrow \sum |x_i-18|=2\times 5=10$

Correct Option: A

Example 6: Find the mean deviation about mean for: -3, 0, 3, 6, 9 

Solution:

1. Mean: $\bar{x}=\frac{-3+0+3+6+9}{5}=\frac{15}{5}=3$
2. Deviations:$|-3-3|=6,|0-3|=3,|3-3|=0,|6-3|=3,|9-3|=6$
3. MD = $\frac{6+3+0+3+6}{5}=\frac{18}{5}=3.6$

Example 7: Find the mean deviation about the mean for the following frequency distribution:

Solution:

1. Find midpoints x for each class.
2. Use: $\bar{x}=\frac{\sum fx}{\sum f},MD=\frac{\sum f|x-\bar{x}}{\sum f}$

$\sum f=28,\sum fx=850$

$\Rightarrow \bar{x}=\frac{850}{28}\approx 30.36$

$\sum f|x-\bar{x}|\approx 279.28$

$\Rightarrow MD\approx \frac{279.28}{28}\approx 9.97$

Example 8: If the mean deviation from the mean for the set of numbers: x - 2, x, x + 2, x + 4, x + 6 is 2.4, find the value of x.

Solution:

The numbers are: $x-2,x,x+2,x+4,x+6$

1. Mean: $\bar{x}=\frac{(x-2)+x+(x+2)+(x+4)+(x+6)}{5}=\frac{5x+10}{5}=x+2$
2. Deviations from x + 2:

$|x-2-(x+2)|=4,|x-(x+2)|=2,|x+2-(x+2)|=0$

$|x+4-(x+2)|=2,|x+6-(x+2)|=4$ 

3. Mean deviation: $MD=\frac{4+2+0+2+4}{5}=\frac{12}{5}=2.4$

Verified! This means MD is independent of xx in symmetric series! Nice JEE trick!

Example 9: If a dataset is symmetrical around the mean, which of the following is true?

A) Mean deviation from the mean = 0
B) Mean deviation from the mean = Standard deviation
C) Mean deviation is less than standard deviation
D) Standard deviation is always less than mean deviation

Answer: C
(For any data, mean deviation is always less than or equal to standard deviation.)

Example 10: 

Find the mean deviation from the median.

Solution: 

Step 1: Calculate Cumulative Frequency

Total Frequency:

$N=\sum f=28$

$\frac{N}{2}=14$

So, the median class is 40–60 (since 14 lies between 10 and 20 in the cumulative frequency).

Step 2: Use Median Formula

$Median=l+(\frac{\frac{N}{2}-F}{f})\times h$

Where:

• l = 40 = lower boundary of median class
• F = 10 = cumulative frequency before median class
• f = 10 = frequency of median class
• h = 20 = width of class

$\Rightarrow Median=40+(\frac{14-10}{10})\times 20$

$\Rightarrow Median=40+(\frac{4}{10})\times 20$

$\Rightarrow Median=40+8=48$

Step 3: Find Midpoints (x) and |x − Median|

Step 4: Apply Mean Deviation Formula

$MD=\frac{\sum f|x-Median|}{\sum f}$

$\sum f|x-48|=152+108+20+110+126=516$

$MD=\frac{516}{28}\approx 18.43$

Final Answer:

Mean Deviation from Median=18.43  

Also Solve: Mean Median Mode Questions

4.0Practice Mean Deviation Questions

1. Find the mean deviation about the mean for: 5, 10, 15, 20, 25
2. The marks obtained by 5 students are: 12, 18, 14, 10, 16. Find MD from mean.
3. Find the MD from median for the dataset: 8, 7, 10, 5, 9, 6, 11
4. Calculate MD from mean for the following frequency distribution: