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Mean Deviation: In statistics, deviation means the difference between the observed and expected values of a variable. In simple words, the deviation is the distance from the centre point. The centre point can be median, mean, or mode. Similarly, the mean deviation definition in statistics or the mean absolute deviation is used to compute how far the values fall from the middle of the data set. The standard deviation meaning states the dispersion of the dataset as compared to its mean.
The continuous data series, discrete and individual data series are used to calculate the mean deviation. The mean deviation formula is an integral part of mathematics; therefore, students need to understand this concept properly. Also, the mean and standard deviation calculator is essential for board exams. In this article, let us understand mean deviation, its definition, formula, and examples in detail.
Mean deviation can be described as a statistical measure that is used to compute the average deviation from the average value of the given data set. The mean deviation of the data values can be easily calculated using the method given below:
– 1st Step: Find the mean or the average value for the given data values.
– 2nd Step: Thereafter, find the difference of the mean value from each of the data values given, ignoring the signs.
– 3rd Step: Now, find the mean or average of those values obtained in step. The result obtained in step \(3\) is the mean deviation.
The formula to calculate mean deviation for the given data set is provided below:
Mean Deviation \( = \frac{{\sum \left| {x – \mu } \right|}}{N}\)
Here,
\(\sum \) denotes the sum of the values
\(x\) denotes each value in the data set
\(\mu \) denotes the mean value of the data set
\(N\) denotes the number of data values
\(\left| {} \right|\) represents the absolute value
To represent the data in a more condensed form, we group it and mention the frequency distribution of each group. These groups are known as class intervals.
Grouping of data is possible in two ways:
The number of times an observation occurs within a class interval is called frequency distribution. If the data collection is huge, for example, if we need to analyze students’ marks, then the representation can be obtained easily by using the concept of ‘Grouping of Data’ according to the class intervals.
Now, we will talk about mean deviation for group data, continuous frequency data, mean deviation formula for group data, mean deviation formula for the ungrouped data, mean deviation formula for continuous series, etc.
In frequency distribution of continuous type, the class intervals or groups are organized so that there are no spaces between the classes, and each class in the table has its distinct frequency. The class intervals are selected to be mutually exclusive and exhaustive.
Mean Deviation for Continuous Frequency Distribution
To understand the concept of continuous frequency distribution, we will look at the example given below.
The following table represents the group of students studying in a school:
| Marks group | Number of Students |
| \(30 – 40\) | \(130\) |
| \(40 – 50\) | \(100\) |
| \(50 – 60\) | \(50\) |
| \(60 – 70\) | \(20\) |
This representation is continuous. The frequency is mentioned according to the class interval:
| Marks group | \({x_i}\) | Number of Students \(\left( {{f_i}} \right)\) |
| \(30 – 40\) | \(35\) | \(130\) |
| \(40 – 50\) | \(45\) | \(100\) |
| \(50 – 60\) | \(55\) | \(50\) |
| \(60 – 70\) | \(65\) | \(20\) |
(Mean) \(\;\bar x = \frac{1}{N}\mathop \sum \limits_{i = 1}^n {x_i}{f_i}\)
The mean deviation \(({x_d}) = \frac{1}{N}\mathop \sum \limits_{i = 1}^n f_i\left|x_i-\overline x\right|\)
In a discrete frequency distribution, the values of the variable are arranged individually. The frequencies of the various values are the number of times each value occurs.
For example,
The following is the number of male employees in different branches of public sector banks. Make a frequency distribution.
\(2,4,6,1,3,5,3,6,4,4,4,2,1,3,6,4,2,5\)
| Number of employees \(\left( {{x_i}} \right)\) | Branches\(\left( {{f_i}} \right)\) |
| \(1\) | \(2\) |
| \(2\) | \(3\) |
| \(3\) | \(3\) |
| \(4\) | \(5\) |
| \(5\) | \(2\) |
| \(6\) | \(3\) |
Mean Deviation for Discrete Frequency Distribution
Discrete means are non-continuous. In such a distribution, the frequency given in the set of data is discrete.
To calculate the mean deviation for grouped data and particularly for discrete distribution data, the following steps are followed:
1st Step: The measure of central tendency about which mean deviation is to be calculated. Let this measure be \(a\).
If this measure is mean, then it is calculated as,
Mean \(\;\bar x = \frac{1}{N}\mathop \sum \limits_{i = 1}^n {x_i}{f_i}\)
2nd Step: Calculate the absolute deviation of each observation from the measure of central tendency calculated in step (I)
3rd Step: The mean absolute deviation around the measure of central tendency is then calculated by using the formula
\(({x_a}) = \frac{1}{N}\mathop \sum \limits_{i = 1}^n f_i\left|x_i-\overline a\right|\)
If the central tendency is mean then,
The mean deviation \(({x_d}) = \frac{1}{N}\mathop \sum \limits_{i = 1}^n f_i\left|x_i-\overline x\right|\), when \(\overline x\) is the mean.
If the measure is median, then the given set of data is ordered in ascending order, the cumulative frequency is found, and the observations whose cumulative frequency is equal to or just greater than \(N/2\) are chosen as the median for the given discrete frequency distribution. This value is observed to be in the centre of the frequency distribution. In the case of median,
\(({x_M}) = \frac{1}{N}\mathop \sum \limits_{i = 1}^n f_i\left|x_i-M\right|\), when \(M\) is the median.
| Mean deviation | Standard deviation |
| We use mean, median to find the mean deviation | We only use mean to find the standard deviation |
| To calculate the dispersion of the data, the mean deviation uses the mod of deviations. | To calculate the dispersion of the data, the standard deviation uses the square of deviations. |
Q.1. Determine the mean deviation for the data values \(4,3,\,5,\,8,\,1,\,9\)
Ans: Given data values are \(4,3,\,5,\,8,\,1,\,9\)
First, we will find the mean for the given data:
Mean, \({\rm{\mu \;}} = \frac{{4 + 3 + 5 + 8 + 1 + 9}}{6}\)
\( \Rightarrow {\rm{\mu }} = \frac{{30}}{6}\)
\( \Rightarrow {\rm{\mu }} = 5\)
Therefore, the mean value is \(5.\)
Now, subtract each mean from the data value, and ignore the minus symbol
\(4 – 5 = 1\)
\(3 – 5 = 2\)
\(5 – 5 = 0\)
\(8 – 5 = 3\)
\(1 – 5 = 4\)
\(9 – 5 = 4\)
Now, the obtained data set is \(1,\;2,\;0,\;3,\;4,\;4\)
Let us find the mean value for the obtained data set.
Therefore, the mean deviation is
\(\frac{{1\, + \,2\, + \,0\, + \,3\, + \,4\, + \,4}}{6} = \frac{{14}}{6} = 2.3\)
Hence, the mean deviation for \(4,3,5,8,1,9\,{\rm{is\; 2}}{\rm{.3}}\)
Q.2. Find the mean deviation for the following data set.
| Class Interval | Frequency |
| \(0 – 2\) | \(4\) |
| \(2 – 4\) | \(3\) |
| \(4 – 6\) | \(5\) |
| \(6 – 8\) | \(7\) |
Ans:
| Class Interval | Classmark/midpoint \(\left( {{x_i}} \right)\) | Frequeny \(\left( {{f_i}} \right)\) | \({x_i}{f_i}\) |
| \(0 – 2\) | \(1\) | \(4\) | \(4\) |
| \(2 – 4\) | \(3\) | \(3\) | \(9\) |
| \(4 – 6\) | \(5\) | \(5\) | \(25\) |
| \(6 – 8\) | \(7\) | \(7\) | \(49\) |
| \(\sum {f_i}\, = 19\) | \(\sum {x_i}{f_{i\,}} = \,87\) |
(Mean) \(\;\bar x = \frac{1}{N}\mathop \sum \limits_{i = 1}^n {x_i}{f_i} = 4.57\) when \(N = \mathop \sum \limits_{i = 1}^n {f_i}\)
The mean deviation \(({x_d}) = \frac{1}{N}\mathop \sum \limits_{i = 1}^n f_i\left|x_i-\overline x\right| = \,\frac{{38.15}}{{19}}\, = 2.007\)
Q.3. For the given data set, if the mean of the dataset is \(31.\) Find the mean absolute deviation of the data set.
\(30,14,35,55,45,21,x\)
Ans:
To find: Mean absolute deviation of the data set
Given:
Data set \(\left\{ {30,14,35,55,45,21,x} \right\}\)
Mean of the data set \( = 31\)
Mean of the data \(\left( \mu \right)\, = \,\frac{{30 + 14 + 35 + 45 + 21 + x}}{7}\, = 31\)
\(\Rightarrow \,200\, + x\, = \,217\)
\(x\, = \,17\)
Using Mean Deviation Formula,
Mean Deviation \( = \left[ {\sum \frac{{\left| {x – {\rm{\mu }}} \right|}}{N}} \right] = \frac{{\left| {30 – 31} \right| + \left| {14 – 31} \right| + \left| {35 – 31} \right| + \left| {55 – 31} \right| + \left| {45 – 31} \right| + \left| {21 – 31} \right| + \left| {17 – 31} \right|}}{7} = 12\)
Hence, the mean absolute deviation of the data set is 12.
Q.4. Find the mean absolute deviation for the following data set: \(302,\,140,\,352,\,563,\,455,\,215,\,213.\)
Ans:
To find: Mean absolute deviation for the given data set.
Given:
Data set\( = \,\left\{ {302,\,140,\,352,\,563,\,455,\,215,\,213} \right\}\)
Mean of the data \(\left( \mu \right) = \frac{{302\, + \,140\, + \,352\, + \,563\, + \,455\, + \,215\, + 213}}{7} = \,320\)
Using mean deviation formula,
\( = \left[ {\sum \frac{{\left| {x – \mu } \right|}}{N}} \right] = \frac{{\left| {302 – 320} \right| + \left| {140 – 320} \right| + \left| {352 – 320} \right| + \left| {563 – 320} \right| + \left| {455 – 320} \right| + \left| {215 – 320} \right| + \left| {213 – 320} \right|}}{7}\)
\( = \,117.14\)
Hence, the mean deviation is \( = \,117.14\)
Q.5: In a class, there are subjects, and the frequencies of students learning the subjects and the frequency of lectures per week are given as:
| Subjects | Math | Science | Hindi | English |
| No.of students\(\left( {{x_i}} \right)\) | \(6\) | \(5\) | \(9\) | \(12\) |
| Frequency of subjects \(\left( {{f_i}} \right)\) | \(5\) | \(7\) | \(4\) | \(9\) |
Ans: Calculate the mean deviation about the mean for the given data.
The following table gives us a tabular representation of data and the calculations
| \(\left( {{x_i}} \right)\) | \(\left( {{f_i}} \right)\) | \({x_i}{f_i}\) | \(\left| {{x_i} – \overline x } \right|\) | \({f_i}\left| {{x_i} – \overline x } \right|\) |
| \(6\) | \(5\) | \(30\) | \(2.26\) | \(11.8\) |
| \(5\) | \(7\) | \(35\) | \(3.36\) | \(23.52\) |
| \(9\) | \(4\) | \(36\) | \(0.64\) | \(2.56\) |
| \(12\) | \(9\) | \(108\) | \(3.64\) | \(32.76\) |
| \(\sum {f_i}\, = \,25\) | \(\overline x \, = \,\frac{1}{N}\sum\limits_{i = 1}^n {{x_i}{f_i}} \) \( = 8.36\) | \(\overline x \, = \,\frac{1}{N}\sum\limits_{i = 1}^n {{f_i}} \left| {{x_i}\, – \,\overline x } \right|\) \( = 2.82\) |
We have learned about mean deviation in statistics and mean deviation formula in this article. We have also discussed how to find the mean deviation about mean and the mean deviation about median, and the difference between mean deviation and standard deviation.
Here are some of the most common FAQs on mean deviation formula and definition:
Q1: How do we calculate the mean deviation?
Ans: Follow the steps mentioned below to calculate the mean deviation:
Step \(1\): Find the mean or the average value for the given data values.
Step \(2\): Then, find the difference of the mean value from each of the data values given ignoring the signs.
Step \(3\): Now, find the mean or average of those values obtained in step
The result obtained in step \(3\) is the mean deviation.
Q2: What is mean deviation in statistics also called?
Ans: Mean deviation in statistics is also called the mean absolute deviation.
Q3: Is mean and mean deviation definition the same?
Ans: No, mean and mean deviation is not the same.
Q4: Which part of Mathematics does mean deviation come under?
Ans: The mean deviation concept comes under Statistics.
Q5: How do we calculate the mean deviation from the median?
Ans: If the measure is median, then the given set of data is ordered in ascending order, the cumulative frequency is found, and the observations whose cumulative frequency is equal to or just greater than \(N/2\) are chosen as the median for the given discrete frequency distribution. This value is observed to be in the centre of the frequency distribution. In the case of median,
\(({x_M}) = \frac{1}{N}\mathop \sum \limits_{i = 1}^n f_i\left|x_i-M\right|\), when \(M\) is the median.
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