What is the difference between mean deviation and standard deviation?

Mean deviation uses absolute values, while standard deviation squares the differences.

Which is more accurate — mean deviation or standard deviation?

Standard deviation gives more weight to large deviations, so it's more sensitive and accurate for statistical inference.

Is mean deviation always positive?

Yes. Since it involves absolute values, MD is always non-negative.

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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: $MD(from mean) = \frac{1}{N} \sum ∣ x_{i} - \overset{x}{ˉ} ∣$
From Median: $MD (from median) = \frac{1}{N} \sum ∣ x_{i} - M e d ian ∣$

For a grouped frequency distribution:

From Mean: $M D = \frac{\sum f ∣ x - x ˉ ∣}{\sum f}$

Where:

x = mid-point of the class
f = frequency
$\overset{x}{ˉ}$ = 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:

Mean = $\overset{x}{ˉ} = \frac{2 + 4 + 6 + 8 + 10}{5} = 16$
Deviations from mean: $∣2 - 6∣ = 4, ∣4 - 6∣ = 2, ∣6 - 6∣ = 0, ∣8 - 6∣ = 2, ∣10 - 6∣ = 4$
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:

x	3	4	5	6	7
f	2	3	4	3	2

Solution:

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

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

Solution:

Calculate the Mean: $\overset{x}{ˉ} = \frac{48 + 52 + 49 + 51}{5} = \frac{250}{5} = 50$
Find Deviations:

x	x – 50
48	2
52	2
50	0
49	1
51	1

Sum of absolute deviations: $= 2 + 2 + 0 + 1 + 1 = 6$
Mean Deviation: $M D = \frac{6}{5} = 1.2$

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

Class Interval	Frequency
0 – 10	5
10 – 20	8
20 – 30	15
30 – 40	16
40 – 50	6

Solution:

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.

Apply median formula: $M e d ian = l + (\frac{\frac{N}{2} - F}{f}) \times h$

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

$\Rightarrow M e d ian = 20 + (\frac{25 - 13}{15}) .10$

$\Rightarrow M e d ian = 20 + (\frac{12}{15}) .10$

$\Rightarrow M e d ian = 20 + 8 = 28$

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

Class	f	x (midpoint)	x – 28	f·\|x – 28\|
0–10	5	5	23	115
10–20	8	15	15	104
20–30	15	25	3	45
30–40	16	35	7	112
40–50	6	45	17	102

Sum of f·|x – median| = 115 + 104 + 45 + 112 + 102 = 478
Mean Deviation: $M D = \frac{\sum f ∣ x - m e d ian ∣}{\sum f} = \frac{478}{50} = 9.56$

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:

$M D = \frac{\sum ∣ x _{i} - x ˉ ∣}{n}$

$\Rightarrow 2 = \frac{\sum ∣ x _{i} - 18∣}{5}$

$\Rightarrow \sum ∣ x_{i} - 18∣ = 2 \times 5 = 10$

Correct Option: A

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

Solution:

Mean: $\overset{x}{ˉ} = \frac{- 3 + 0 + 3 + 6 + 9}{5} = \frac{15}{5} = 3$
Deviations: $∣ - 3 - 3∣ = 6, ∣0 - 3∣ = 3, ∣3 - 3∣ = 0, ∣6 - 3∣ = 3, ∣9 - 3∣ = 6$
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:

Class Interval	Frequency
0 – 10	2
10 – 20	3
20 – 30	8
30 – 40	10
40 – 50	3
50 – 60	2

Solution:

Find midpoints x for each class.
Use: $\overset{x}{ˉ} = \frac{\sum f x}{\sum f}, M D = \frac{\sum f ∣ x - x ˉ}{\sum f}$

Class	f	x	f·x	$∣ x - \overset{x}{ˉ}$	$f .∣ x - \overset{x}{ˉ} ∣$
0–10	2	5	10	25	50
10–20	3	15	45	15	40
20–30	8	25	200	5	40
30–40	10	35	350	5	50
40–50	3	45	135	15	45
50–60	2	55	110	25	50

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

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

Now compute using approximate values:

Class	x	\|x - 30.36\|
5	25.36	2	50.72
15	15.36	3	46.08
25	5.36	3	42.88
35	4.64	10	46.4
45	14.64	3	43.92
55	24.64	2	49.28

$\sum f ∣ x - \overset{x}{ˉ} ∣ \approx 279.28$

$\Rightarrow M D \approx \frac{279.28}{28} \approx 9.97$

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: $x - 2, x, x + 2, x + 4, x + 6$

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

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

$∣ x + 4 - (x + 2) ∣ = 2, ∣ x + 6 - (x + 2) ∣ = 4$

Mean deviation: $M D = \frac{4 + 2 + 0 + 2 + 4}{5} = \frac{12}{5} = 2.4$

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:

Class Interval	Frequency
0–20	4
20–40	6
40–60	10
60–80	5
80–100	3

Find the mean deviation from the median.

Solution:

Step 1: Calculate Cumulative Frequency

Class Interval	f	Cumulative Frequency (CF)
0–20	4	4
20–40	6	10
40–60	10	20
60–80	5	25
80–100	3	28

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

$M e d ian = l + (\frac{\frac{N}{2} - F}{f}) \times h$

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

$\Rightarrow M e d ian = 40 + (\frac{14 - 10}{10}) \times 20$

$\Rightarrow M e d ian = 40 + (\frac{4}{10}) \times 20$

$\Rightarrow M e d ian = 40 + 8 = 48$

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

Class Interval	f	Midpoint (x)	\|x - 48\|	$f .∣ x - 48∣$
0–20	4	10	38	4 × 38 = 152
20–40	6	30	18	6 × 18 = 108
40–60	10	50	2	10 × 2 = 20
60–80	5	70	22	5 × 22 = 110
80–100	3	90	42	3 × 42 = 126

Step 4: Apply Mean Deviation Formula

$M D = \frac{\sum f ∣ x - M e d ian ∣}{\sum f}$

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

$M D = \frac{516}{28} \approx 18.43$

Final Answer:

Mean Deviation from Median=18.43

Also Solve: Mean Median Mode Questions

4.0Practice Mean Deviation Questions

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

x	1	2	3	4	5
f	4	6	10	6	4

Frequently Asked Questions

Join ALLEN!

(Session 2026 - 27)

Name

Mobile Number

Class

Choose class

Goal

Choose your goal

Preferred Programs

Preferred Mode

State

Choose State

I agree to

I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

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: $MD(from mean) = \frac{1}{N} \sum ∣ x_{i} - \overset{x}{ˉ} ∣$
From Median: $MD (from median) = \frac{1}{N} \sum ∣ x_{i} - M e d ian ∣$

For a grouped frequency distribution:

From Mean: $M D = \frac{\sum f ∣ x - x ˉ ∣}{\sum f}$

Where:

x = mid-point of the class
f = frequency
$\overset{x}{ˉ}$ = 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:

Mean = $\overset{x}{ˉ} = \frac{2 + 4 + 6 + 8 + 10}{5} = 16$
Deviations from mean: $∣2 - 6∣ = 4, ∣4 - 6∣ = 2, ∣6 - 6∣ = 0, ∣8 - 6∣ = 2, ∣10 - 6∣ = 4$
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:

x	3	4	5	6	7
f	2	3	4	3	2

Solution:

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

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

Solution:

Calculate the Mean: $\overset{x}{ˉ} = \frac{48 + 52 + 49 + 51}{5} = \frac{250}{5} = 50$
Find Deviations:

x	x – 50
48	2
52	2
50	0
49	1
51	1

Sum of absolute deviations: $= 2 + 2 + 0 + 1 + 1 = 6$
Mean Deviation: $M D = \frac{6}{5} = 1.2$

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

Class Interval	Frequency
0 – 10	5
10 – 20	8
20 – 30	15
30 – 40	16
40 – 50	6

Solution:

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.

Apply median formula: $M e d ian = l + (\frac{\frac{N}{2} - F}{f}) \times h$

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

$\Rightarrow M e d ian = 20 + (\frac{25 - 13}{15}) .10$

$\Rightarrow M e d ian = 20 + (\frac{12}{15}) .10$

$\Rightarrow M e d ian = 20 + 8 = 28$

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

Class	f	x (midpoint)	x – 28	f·\|x – 28\|
0–10	5	5	23	115
10–20	8	15	15	104
20–30	15	25	3	45
30–40	16	35	7	112
40–50	6	45	17	102

Sum of f·|x – median| = 115 + 104 + 45 + 112 + 102 = 478
Mean Deviation: $M D = \frac{\sum f ∣ x - m e d ian ∣}{\sum f} = \frac{478}{50} = 9.56$

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:

$M D = \frac{\sum ∣ x _{i} - x ˉ ∣}{n}$

$\Rightarrow 2 = \frac{\sum ∣ x _{i} - 18∣}{5}$

$\Rightarrow \sum ∣ x_{i} - 18∣ = 2 \times 5 = 10$

Correct Option: A

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

Solution:

Mean: $\overset{x}{ˉ} = \frac{- 3 + 0 + 3 + 6 + 9}{5} = \frac{15}{5} = 3$
Deviations: $∣ - 3 - 3∣ = 6, ∣0 - 3∣ = 3, ∣3 - 3∣ = 0, ∣6 - 3∣ = 3, ∣9 - 3∣ = 6$
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:

Class Interval	Frequency
0 – 10	2
10 – 20	3
20 – 30	8
30 – 40	10
40 – 50	3
50 – 60	2

Solution:

Find midpoints x for each class.
Use: $\overset{x}{ˉ} = \frac{\sum f x}{\sum f}, M D = \frac{\sum f ∣ x - x ˉ}{\sum f}$

Class	f	x	f·x	$∣ x - \overset{x}{ˉ}$	$f .∣ x - \overset{x}{ˉ} ∣$
0–10	2	5	10	25	50
10–20	3	15	45	15	40
20–30	8	25	200	5	40
30–40	10	35	350	5	50
40–50	3	45	135	15	45
50–60	2	55	110	25	50

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

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

Now compute using approximate values:

Class	x	\|x - 30.36\|
5	25.36	2	50.72
15	15.36	3	46.08
25	5.36	3	42.88
35	4.64	10	46.4
45	14.64	3	43.92
55	24.64	2	49.28

$\sum f ∣ x - \overset{x}{ˉ} ∣ \approx 279.28$

$\Rightarrow M D \approx \frac{279.28}{28} \approx 9.97$

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: $x - 2, x, x + 2, x + 4, x + 6$

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

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

$∣ x + 4 - (x + 2) ∣ = 2, ∣ x + 6 - (x + 2) ∣ = 4$

Mean deviation: $M D = \frac{4 + 2 + 0 + 2 + 4}{5} = \frac{12}{5} = 2.4$

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:

Class Interval	Frequency
0–20	4
20–40	6
40–60	10
60–80	5
80–100	3

Find the mean deviation from the median.

Solution:

Step 1: Calculate Cumulative Frequency

Class Interval	f	Cumulative Frequency (CF)
0–20	4	4
20–40	6	10
40–60	10	20
60–80	5	25
80–100	3	28

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

$M e d ian = l + (\frac{\frac{N}{2} - F}{f}) \times h$

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

$\Rightarrow M e d ian = 40 + (\frac{14 - 10}{10}) \times 20$

$\Rightarrow M e d ian = 40 + (\frac{4}{10}) \times 20$

$\Rightarrow M e d ian = 40 + 8 = 48$

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

Class Interval	f	Midpoint (x)	\|x - 48\|	$f .∣ x - 48∣$
0–20	4	10	38	4 × 38 = 152
20–40	6	30	18	6 × 18 = 108
40–60	10	50	2	10 × 2 = 20
60–80	5	70	22	5 × 22 = 110
80–100	3	90	42	3 × 42 = 126

Step 4: Apply Mean Deviation Formula

$M D = \frac{\sum f ∣ x - M e d ian ∣}{\sum f}$

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

$M D = \frac{516}{28} \approx 18.43$

Final Answer:

Mean Deviation from Median=18.43

Also Solve: Mean Median Mode Questions

4.0Practice Mean Deviation Questions

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

x	1	2	3	4	5
f	4	6	10	6	4

Frequently Asked Questions

What is the difference between mean deviation and standard deviation?

Which is more accurate — mean deviation or standard deviation?

Is mean deviation always positive?

Join ALLEN!

Mean Deviation Questions

1.0What is Mean Deviation?

2.0Formula for Mean Deviation

3.0Solved Questions and Answers on Mean Deviation

4.0Practice Mean Deviation Questions

Table of Contents

Frequently Asked Questions

What is the difference between mean deviation and standard deviation?

Which is more accurate — mean deviation or standard deviation?

Is mean deviation always positive?

Join ALLEN!

Mean Deviation Questions

1.0What is Mean Deviation?

2.0Formula for Mean Deviation

3.0Solved Questions and Answers on Mean Deviation

4.0Practice Mean Deviation Questions

Table of Contents