Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Mean Deviation Formula

  • Last Updated : 20 Jan, 2022

The process of collecting and analyzing the data is known as statistics. The deviation in statistics is known as the variation between the other values of the variable and the observed value. Let’s study more about the deviation

Mean deviation

The mean deviation of the standard distribution is the measurement of its central tendency. It can be calculated by using three methods. They are Arithmetic mean, Median, Mode.

Mean deviation is used to show how far the observations are situated from the average of the observed data. Each one of these is to be considered of an absolute value. In this, the negative signs are totally ignored. According to this, it is said that the deviation of both sides is of equivalent nature. The suitable average of the mean deviation could be of mean, median, or mode of data. For an instance the formula for the mean deviation of individual, discrete and continuous series.

Types of mean deviation

There are three types of mean deviation. They are individual series, discrete series, and continuous.

  • Individual series

When the data is provided individually in the form of a series then it is known as the individual series. It is basically a form of raw data in the form of a series that forms an arrangement individually. In individual series, items are represented in a single form.

For example, let us assume the following scores by the players in a cricket match:

56, 97, 46, 88, 67, 59, 62, 78, 90, 58, 77

In the above data, it is not given that how many players score 56 runs and more than 78 in a single sight.

  • Discrete series

The discrete series is the series that is used to reflect each specific value of the observed variable. One of the variables corresponds to an integer value. In discrete series exact measurement of the items in the data is visible. For example, the wages of 20 workers are below in the table as,

Wages 

(number of workers)

Frequency
20006
25004
30002
35003
40003
45002

Conclusively 6 workers are getting RS 2000 wages paid, and 4 workers are getting RS 2500 paid, and so on.

  • Continuous series

A continuous series is a series that maintains the items in certain definite classes. The items in the class interval lose their individual identity, and those individual items are merged in one or another class intervals. Each class has continuity in which means that the end of one class should be the beginning for the other one. That’s why it is named a continuous series. 

For example, the continuous series is depicted as follows

AgeFrequency
10-154
15-2012
20-2516
25-3022
30-3510
35-408
40-456
45-504

Mean deviation from the arithmetic mean (formula)

  • Individual series

Mean Deviation (M.D) = ∑∣X – X̄∣ / N

Where,

∑ – Summation

x – Observation

X̄  – Mean

N – Number of observation

  • Discrete series

Mean Deviation (M.D) = ∑f∣X – X̄∣ / ∑f

Where,

∑ – Summation

x – Observation

X̄ – Mean

f – frequency of observation

  • Continuous series

Mean Deviation (M.D) = ∑f∣X – X̄∣ / ∑f

Where,

∑ – Summation

x – Mid-value of the class

X̄ – Mean

f – frequency of observation

Mean deviation from the median (formula)

  • Individual series

Mean Deviation (M.D) =  ∑|X – M| / N

Where,

∑ – Summation

x – Observation

M – Median

N – Number of observation

  • Discrete series

Mean Deviation (M.D) = ∑ f|X – M| / ∑ f

Where,

∑ – Summation

x – Observation

M – Median

N – Frequency of observations

  • Continuous series

Mean Deviation (M.D) = ∑ f∣X – X̄∣ / ∑f

Where,

∑ – Summation

x – Observation

M – Median

N – Frequency of observations

Mean deviation by mode (formula)

  • Individual series

Mean Deviation (M.D) = ∑|X – Mode| / N

Where,

∑ – Summation

x – Observation

M – Mode

N – Number of observations

  • Discrete series

Mean Deviation (M.D) = ∑ f|X – Mode| / ∑ f

Where,

∑ – Summation

x – Observation

M – Mode

N – Frequency of observations

  • Continuous series

Mean Deviation (M.D) = ∑ f |X – Mode| / ∑ f

Where,

∑ – Summation

x – Observation

M – Mode

N – Frequency of observations

Steps to calculate mean deviation

  1. Firstly we have to calculate the arithmetic mean, median, or by mode of the given data
  2. Now we have to calculate the deviation from mean, median, or mode and have to ignore the negative items
  3. Now we have to multiply the deviations with the frequency of the data. This step can only be done while solving either the discrete or the continuous series this step doesn’t work in individual series.
  4. Now sum up all the deviations
  5. Apply the formula and solve the question.

Similar Problems

Question 1: Calculate the mean deviation from the median and the co-efficient of mean deviation from the following data:

Marks of the students: 88, 14, 78, 69, 44, 54, 18, 79, 40.

Solution:

 Arrange the data in ascending order: 14, 18, 40, 44, 54, 69, 78, 79, 88.

Median = Value of the (N + 1)TH / 2 term

= Value of the (9 + 1)TH / 2 term = 54

Calculation of mean deviation:                                                

X |X – M|
1440
1836
4014
4410
540
6915
7824
7925
8834
N = 9  ∑|X–M|=198

M.D. = ∑|X – M| / N

= 198/9

= 22

Co-efficient of Mean Deviation from Median = M.D./M

= 22/54

= 0.4074

Question 2: Calculate the mean deviation about the mean using the following data

5, 8, 14, 16, 20, 6, 8, 19.

Solution:

First, we have to find the mean of the data that we are provided with

Mean of the given data =  Sum of all the terms total number of terms

X̄ = 5 + 8 + 14 + 16 + 20 + 6 + 8 + 19

= 96/8

= 12

Next, find the mean deviation

Xi Xi – x̄ |Xi – x̄|
55 – 12 = -7|-7| = 7
88 – 12 = -4|-4| = 4
1414 – 12 = 2|2| = 2
1616 – 12 = 4|4| = 4
2020 – 12 = 8|8| = 8
66 – 12 = -6|-6| = 6
88 – 12 = -4|-4| = 4
1919 – 12 = 7|7| = 7
  ∑|Xi − x̄| = 42

Mean deviation about mean = ∑|Xi − X̄| / 8

= 42/8

= 5.25

Question 3: Find the mean deviation about the median for the following data.

ClassFrequency (f)
5-1516
15-255
25-358
35-456
45-553

Solution:

Classfcfxi |x – x̄|f. |x – x̄|
5-1516161011176
15-255212015
25-3582930972
35-456354019114
45-55338502987
TotalN = 38    

To find the median class,

N/2 = 38/2 = 19

thus cf is nearest to 20

Thus, median class is 15 – 25.

l = 15, i = 10, f = 5, cf = 16,  ∑51 fi/2= 19

Substituting these values in the formula,

M =  l+(∑51 fi/2 − cf)/f × h

= 21

Mean deviation about median =  ∑51 fi|xi − M| / ∑51fi

= 295.6 / 38 = 7.778

Answer: Mean deviation about median = 7.778

Question 4: Find the mean deviation about the mean for {17, 24, 37, 18, 4}

Solution: 

The data is ungrouped, thus mean = (17 + 24 + 37 + 18 + 4) / 5 = 20

x|x – x̄|
173
244
3717
182
416
Total42

Using the formula,  

51|xi − μ|/5

= 42 / 5 = 8.4

Mean deviation about mean = 8.4

Question 5: Determine the mean deviation for the data values 4, 2, 9, 7, 3, 5.

Solution:

Given data values are 4, 2, 9, 7, 3, 5.

We know that the procedure to calculate the mean deviation.

First, find the mean for the given data:

Mean, µ = (4 + 2 + 9 + 7 + 3 + 5)/6

µ = 30/6

µ = 5

Therefore, the mean value is 5.

Now, subtract each mean from the data value, and ignore the minus symbol if any

(Ignore”-”)

4 – 5 = 1

2 – 5 = 3

9 – 5 = 4

7 – 5 = 2

3 – 5 = 2

5 – 5 = 0

Now, the obtained data set is 1, 3, 4, 2, 2, 0.

Finally, find the mean value for the obtained data set

Therefore, the mean deviation is  

= (1 + 3 + 4 + 2 + 2 + 0) /6

= 12/6

= 2

Hence, the mean deviation for 4, 2, 9, 7, 3, 5 is 2.

Question 6: Find the mean deviation from the mean of given series also calculate its coefficient,

C.I.F
0-23
2-45
4-66
6-84
8-102

Solution:

C-Ifxfxx – x̄|x – x̄|f|x – x̄|
0-23131-4.7=-3.73.711.1
2-453153-4.7=-1.71.78.5
4-665305-4.7=0.30.31.8
6-847287-4.7=2.32.39.2
8-1029189-4.7=4.34.38.6
∑f = 20∑fx = 94    ∑f|x – x̄| = 39.2

 x̄= ∑fx/∑f = 94/20 = 4.7

M,D = ∑f|x – x̄|/∑f = 39.2/20 = 1.96

Coefficient of M,D, = M.D./mean = 1.96/4.7 = 0.417


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!