
Content Writer
Mean deviation is a measure of central tendency. In the field of statistics, the deviation is used to calculate the difference between the observed value and the expected value.
- In the same way, the mean deviation indicates how far the values deviate from the middle of the data set.
- As a statistical measure, it is calculated by calculating the average deviation from the mean value of a set of data.
- Central tendency can be calculated using the mean, median, or mode of the data.
- It is also known as mean absolute deviation and abbreviated as MAD.
- Mean deviation is calculated for grouped data and ungrouped data.
- You can use the concept in the field of manufacturing and production to determine the consistency of product quality.
Key Terms: Mean Deviation, Mean Deviation Formula, Statistics, Central Tendency, Median, Arithmetic Mean, Observations, Mode, Continuous Data, Discrete Data
Mean Deviation
[Click Here for Sample Questions]
Mean deviation can be defined as a statistical measure that helps in computing the average deviation from the average value of a given data collection.
- It can be calculated using the arithmetic mean, median, or mode.
- We can calculate the mean deviation using various data series, such as continuous, discrete, and individual data series.
- The deviation indicates how far the observations are situated from the average of the observed data.
- The method will provide more accurate results of the dispersion.
- Economists use mean deviation to determine variability in income distribution, price fluctuations, and consumer behavior.
Mean Deviation ExampleExample: Suppose you have a set of numbers which is given by {4, 5, 5, 6}. First, to calculate the mean deviation, we will determine the mean of the given data set.
|
Measures of Central Tendency
Read More:
Mean Deviation Formula
[Click Here for Previous Year's Questions]
The simplest and the basic formula for mean deviation for a given data set is:
Mean Deviation = \(\frac{\sum |X - \bar{X}|}{N}\)
Where,
- X refers to each value in the data set.
- denotes the mean value of the data set.
- N refers to the total number of data values.
- |\(\bar{x}\)| represents the absolute value (values without the sign).
Mean Deviation From Mean
Mean can be calculated by dividing the sum of all observations by the total number of observations. The formulas for calculating the mean deviation from the mean are given below:
- Individual Series: The formula to find the Mean Deviation for an individual series is given as follows:
Mean Derivation = \(\frac{ \sum _1^n |x_i - \mu|}{n}\)
where the mean is μ = \(\frac{x_1 + x_2 + ... + x_n}{n}\)
- Continuous/Discrete Series: The formula of Mean Deviation from Mean for a discrete series is:
Mean Derivation = \(\frac{ \sum _1^n f_i |x_i - \mu|}{\sum _1^n f_i }\)
where the mean of grouped data is μ = \(\frac{ \sum _1^n f_i x_i}{\sum _1^n f_i }\)
Mean Deviation From Median
Median refers to the value that distinguishes between the lower and upper halves of the data. The median is the middle value, in an ascending or descending list of numbers. The formulas for mean deviation about the median are listed down below:
Individual Series
The formula to find the Mean Deviation for an individual series is given below:
Mean Deviation = \(\frac{ \sum_1^n |x_i - M|}{n}\)
Where, median (M) is- If n is odd- M= (n+1)/2th observation
- If n is odd- M= ((n/2)th observation + (n/2)+1th observation)/2
Discrete Series
The formula to find the Mean Deviation for a discrete series is given below:
Mean Deviation = \(\frac{ \sum _1^n f_i |x_i - M|}{\sum _1^n f_i }\)
Continuous Series
The formula to find the Mean Deviation for a continuous series is as follows:
Mean Deviation = \(\frac{ \sum _1^n f_i |x_i - M|}{\sum _1^n f_i }\)
where the median of continuous data (M) = I + \(\frac{ \frac{\sum_1^n f_i}{2} - c.f}{f}\) x h
- c.f = It denotes the cumulative frequency preceding the median class
- l = It denotes the lower value of the median class
- f = frequency of the median class
- h = length of the median class
Mean Deviation From Mode
Mode refers to the value that occurs the most frequently in a given data collection. The formulas to calculate mean deviation about the mode are listed below:
Individual Series
The formula to find the Mean Deviation from Mode for an individual series is given as follows:
Mean Deviation = \(\frac{ \sum_1^n |x_i - mode|}{n}\)
Discrete Series
The formula to find the Mean Deviation from Mode for a discrete series is given as follows:
Mean Deviation = \(\frac{ \sum _1^n f_i |x_i - mode|}{\sum _1^n f_i }\)
Continuous Series
The formula to find the Mean Deviation from Mode for a continuous series is given as follows:
Mean Deviation = \(\frac{ \sum _1^n f_i |x_i - mode|}{\sum _1^n f_i }\)
where the mode of continuous data = I + \((\frac{f - f_1}{2f - f_1 - f_2})\) x h
Where,
- l = It denotes the lower value of the modal class
- h = size of the modal class
- f = frequency of the modal class
- f1 = It denotes the frequency of the class preceding the modal class
- f2 = It denotes the frequency of the class succeeding the modal class
Example of Mean Deviation FormulaExample: Determine the mean deviation for the data values 5, 6,7, 8, 4, 10. Ans: Given data values are 5, 6, 7, 8, 4, 10. First, find the mean for the given data: Mean, µ = ( 7+6+7+8+4+10)/6 µ = 42/6 µ = 7 Therefore, the mean value is 7. Now, subtract each mean from the given data value, and ignore the minus sign 7 – 7 = 0 7 – 6 = 1 7 – 7 = 0 8 – 7 = 1 4 – 7 = 3 10 – 7 = 3 = Now, the obtained data set is 0, 1, 0, 1, 3, 3. = Determine the mean value for the obtained data set Therefore, the mean deviation for required set is = (0 + 1 + 0 + 1 + 3 + 3) /6 = 8/6 = 1.33 |
Mean Deviation Formula
How to calculate Mean Deviation?
[Click Here for Sample Questions]
Steps to calculate mean deviation are as follows:
- First, we need to determine the mean of the given data set.
- After determining the mean, you need to subtract each value of the data set from the required mean, ignoring the negative sign.
- If you have discrete or continuous series, then multiply the mean deviation of data with the frequency of the data.
- Lastly, determine the mean of all values of the deviation to get the required result.
Also Read:
Things to Remember
- In statistics, the mean is one of the measures of central tendency, along with the mode and median.
- Mean derivation is used to check the spread of data with respect to the central value.
- It is used in the fields of education, finance and health care.
- Individual, discrete and continuous series are three types of mean deviation.
- The method is used in case the data has a greater number of outliers.
Previous Years Questions
- The mean deviation from the mean of the data is :...[JEE Main 2016]
- A ratio of other two observations is :….[JEE Main 2019]
- The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, find the value of P(X=1)….[JKCET 2014]
- The correct mean of the 100 items is...[JKCET 2016]
- If M. D. is 12, the value of S.D. will be
- The average speed of a car running at the rate of 15 km/hour during the first 30 km and at 20 km/hour during the second 30 km and at 25 km/hour during the third 30 km, will be
- The correct average is...
- Find correct standard deviation.
- The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below. Which of these three subjects shows the highest variability in marks and which shows the lowest?
- The mean deviation from the median of the following set of observations 5, 3, 9, 12, 3, 10, 12, 21, 18, 12, 21 is
- The mean marks of 120 students is 20. It was later discovered that two marks were wrongly taken as 50 and 80 instead of 15 and 18. The correct mean of marks is
Sample Questions
Ques. Find the mean deviation from mean for the data values 5, 3,7, 8, 4, 9? (5 marks)
Ans. Given values are 5, 3, 7, 8, 4, 9
First, calculate the mean for the given data
Mean, µ = ( 5+3+7+8+4+9)/6
µ = 36/6
µ = 6
Therefore, the mean value is 6.
Subtract each mean from the data value
(Ignore”-”)
5 – 6 = 1
3 – 6 = 3
7 – 6 = 1
8 – 6 = 2
4 – 6 = 2
9 – 6 = 3
The obtained data set is 1, 3, 1, 2, 2, 3
Lastly, let us find the mean value for the obtained data set
Therefore, the mean deviation is
= (1+3 + 1+ 2+ 2+3) /6
= 12/6
= 2
Mean deviation for 5, 3,7, 8, 4, 9 is 2.
Ques. Calculate the Mean Deviation and the coefficient of Mean Deviation using the Data given below: Test Marks of 9 students are as follows: 86, 25, 87, 65, 58, 45, 12, 71, 35 respectively? (5 marks)
Ans. Arrange them into ascending order, i.e., 12, 25, 35, 45, 58, 65, 71, 86, 87.
Then find out the median so, median = Value of the (N+1)th/2 term
value of the (9+1)th/2 = 58
Now we have to calculate the Mean Deviation
| X | |X-M| |
|---|---|
| 12 | 46 |
| 25 | 33 |
| 35 | 23 |
| 45 | 13 |
| 58 | 0 |
| 65 | 7 |
| 71 | 13 |
| 86 | 28 |
| 87 | 29 |
| N=9 | ∑∣X−M∣=460 |
MD= ∑∣X−M∣/N
MD=460/9
51.11
Coefficient of the Mean Deviation from median =M.D/M
=51.11/58
=0.881
Ques. Find the mean data deviation values for 5, 3, 7, 8, 4, 9? (3 marks)
Ans. To begin, calculate the Mean of the Data
5+3+7+8+4+9/6 is the average.
36/6 = 6
The average value is 6.
Now, Subtract each Mean from the Data value, ignoring any minus symbols that may appear
6 + 5 = 1
3 – 6 =3
1 = 7 – 6
2 = 8 – 6
2 = 4 – 6
3 = 9 – 6
The resulting data set is now 1, 3, 1, 2, 2, 3.
Finally, calculate the Mean value for the Data set you've gathered.As a result, the standard deviation is
(1+3 + 1+ 2+ 2+ 2+3)/6
12/6 = 2
Mean Deviation for the numbers 5, 3,7, 8, 4, 9 is calculated as 2.
Ques. Find mean deviation from median from given data? (5 marks)

Ans. We begin by calculating cumulative frequency
| Height (cms) (xi) | No. of students (fi) | Cumulative frequency |
|---|---|---|
| 147 | 8 | 8 |
| 148 | 12 | 20 |
| 150 | 15 | 35 |
| 152 | 10 | 45 |
| 155 | 5 | 50 |
| N=50 |
As N=50 which is an even number
Therefore,
Median of the given data (M) = Average of the 25th and 26th observations
Median of the given data (M) = Average of the 25th and 26th observations
= 25th + 26th observation/2
=150+150/2
=300/2
=150 cm
Now, find the absolute values of the deviations of the heights from the median 150 and the products of the deviations and their frequencies.
| Height (cms) (xi) | No. of students (fi) | Cumulative frequency | |xi-x| = |xi-150| | fi|xi-M| |
|---|---|---|---|---|
| 147 | 8 | 8 | 3 | 24 |
| 148 | 12 | 20 | 2 | 24 |
| 150 | 15 | 35 | 0 | 0 |
| 152 | 10 | 45 | 2 | 20 |
| 155 | 5 | 50 | 5 | 25 |
| N=50 | i=1∑5fi| x- M|= 93 |
Sum of the products of the deviations and their frequencies = 93
Mean deviation about the median = \(\frac{1}{n}\)i=1∑nfi| xi - M | = \(\frac{93}{50}\)
= 1.86 cm.
Ques. Calculate the mean deviation from the mean for the following data? (5 marks)

Ans. The process is as follows:
| x | f | x.f | |x – µ| | f. |x – µ| |
|---|---|---|---|---|
| 12 | 7 | 84 | 2.619 | 18.33 |
| 9 | 3 | 27 | 0.381 | 1.143 |
| 6 | 8 | 48 | 3.381 | 27.048 |
| 18 | 1 | 18 | 8.619 | 8.619 |
| 10 | 2 | 20 | 0.619 | 1.238 |
| Total | 21 | 197 | 56.378 |
Mean (µ)= \(\frac{ \sum_1^5fixi}{\sum_1^5fi}\) = \(\frac{197}{21}\)= 9.381
Then, we substitute values in the mean deviation about mean formula,
Mean Deviation= \(\frac{ \sum_1^5fi|xi - \mu|}{\sum_1^5fi}\) = \(\frac{56.378}{21}\)= 2.684
Hence, the mean deviation about the mean is 2.684
Ques. Calculate the mean deviation for the following data? (5 marks)

Ans. The process is as follows:
| Class Interval | Mid-point (x) | Frequency (f) | f.x | |x – µ| = |x – 4| | f. |x – µ| |
|---|---|---|---|---|---|
| 0 – 2 | 1 | 4 | 4 | 3 | 12 |
| 2 – 4 | 3 | 2 | 6 | 1 | 2 |
| 4 – 6 | 5 | 5 | 25 | 1 | 5 |
| 6 – 8 | 7 | 3 | 21 | 3 | 9 |
| Total | 14 | 56 | 28 |
Mean (µ)= \(\frac{ \sum_1^nfixi}{\sum_1^nfi}\) = \(\frac{56}{14}\)= 4
Then we substitute the values in the formula
Mean Deviation= \(\frac{ \sum_1^nfi|xi - \mu|}{\sum_1^nfi}\) = \(\frac{28}{14}\)= 2
Ques. Find mean derivation of following data 36,72,46,42,60,45,53,46,51,49? (3 marks)
Ans. Given observations are = 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Total Number of observations = n = 10
In order to calculate median for given data we have to arrange observations in ascending order as follows:
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Median = [(n/2)the observations + ((n/2) + 1)th observation] / 2
(n/2)th observation = 10 / 2 = 5th observation= 46
((n/2) + 1)th observation = (10/2) + 1 = 6th observation = 49
Therefore, Median = (46 + 49) / 2 = 47.5
Ques. Find mean derivation of mean from following data? (5 marks)

Ans. The process is as follows:
| xi | fi | xi*fi | |xi-a| | fi * |xi – a| |
|---|---|---|---|---|
| 5 | 7 | 35 | 9 | 63 |
| 10 | 4 | 40 | 4 | 16 |
| 15 | 6 | 90 | 1 | 6 |
| 20 | 3 | 60 | 6 | 18 |
| 25 | 5 | 125 | 11 | 55 |
| Total | 25 | 350 | 158 |
Now calculate the sum of given data
N = ∑ fi = (7 + 4 + 6 + 3 + 5) = 25
∑ xi * fi = (35 + 40 + 90 + 60 + 125) = 350
Mean using following formula
Mean (a) = ∑(xi * fi)/ N = 350 / 25 = 14
Then,
∑ fi *|xi – a| = (63 + 16 + 6 + 18 + 55) = 158
Using formula, M.D (a) = ∑(fi * |xi – a|)/ N
= 158 / 25
= 6.32
So, mean deviation for given observations is 6.32
Ques. Calculate mean deviation from median using given data? (5 marks)

Ans. The process is as follows:
| xi | fi | C.F | |xi-M| | fi*|xi-M| |
|---|---|---|---|---|
| 5 | 8 | 8 | 2 | 16 |
| 7 | 6 | 14 | 0 | 0 |
| 9 | 2 | 16 | 2 | 4 |
| 10 | 2 | 18 | 3 | 6 |
| 12 | 2 | 20 | 5 | 10 |
| 15 | 6 | 26 | 8 | 48 |
| Total | 26 | 84 |
Now look out for the observation whose cumulative frequency is equal
to or just greater than N / 2 and then Find the median.
N = ∑fi = 26 is even. We divide N by 2. Thus, 26/2 = 13
The cumulative frequency for greater than 13 is 14, for which corresponding observation is 7
Median = [(N/2)th observation + ((N/2) + 1)th observation] / 2
= (13th observation + 14th observation) / 2
= (7 + 7) / 2 = 7
Now, the absolute values of the deviations from median
i.e., |xi – M| and fi * |xi – M| as shown in the above table.
∑ fi * |xi – M| = (16 + 4 + 6 + 10 + 48) = 84
Using formula, M.D (M) = ∑ (fi * |xi – M|)/ N
= 84 / 26
= 3.23
So, mean deviation about median for given observations is 3.23
Ques. The mean of 2,7,4,6,8 and p is 7. Find the mean deviation from the median of these observations? (5 marks)
Ans. Observations are 2, 7, 4, 6, 8, and p which is 6 in number
n=6
The near of these observations is 7
\(\frac{2+7+4+6+8+p}{6}\)=7
27+p=42
p=15
Arrange the observations in ascending order 2,4,6,7,8,15
Median (M) = \(\frac{\frac{n}{2}th observation + (\frac{n}{2}+1)th observation}{2}\)
= \(\frac{3rd observation + 4th observation}{2}\)
=\(\frac{6+7}{2}\)
=6.5
Calculation of mean deviation about Median.
| xi | xi-M | |xi-M| |
|---|---|---|
| 2 | -4.5 | 4.5 |
| 4 | -2.5 | 2.5 |
| 6 | -0.5 | 0.5 |
| 7 | 0.5 | 0.5 |
| 8 | 1.5 | 1.5 |
| 15 | 8.5 | 8.5 |
| Total | 18 |
Media’s deviation about median
\(\frac{18}{6}\)=3
Ques.Find the mean deviation for the following data points: 9, 3, 7, 8 and 3? (3 marks)
Ans. The values of the data are 9, 3, 7, 8, and 3.
Find the mean for the supplied data first:
Mean = ( 9+3+7+8+3)/5
Mean = 30/6
Mean = 6
Consequently, 6 is the mean value.
Next, deduct each mean from the data value, ignoring any minus signs.
9 – 6 = 3
3 – 3 = 0
7 – 3 = 4
8 – 6 = 2
3 – 6 = 3
The collected data set is currently 3,0,4,2,3.
the average deviation is
= (3+0+4+2+3) /5
= 12/5
= 2.4
Read Also:
| Maths Study Guides | ||
|---|---|---|
| Maths MCQs | NCERT Solutions for Class 11 Math | Comparison Topics in Maths |






Comments