Mean Deviation (original) (raw)

Last Updated : 4 Nov, 2025

The mean deviation (also known as Mean Absolute Deviation, or MAD) of the data set is the value that tells us how far each data point is from the center point of the data set. The center point of the data set can be the Mean, Median, or Mode. Thus, the mean of the deviation is the average of the absolute deviations of all data points from the chosen central value.

Example of Mean Deviation

Suppose we have to find the mean deviation of the given data set: {4, 5, 6, 7, 8}.

**Calculate Value (Mean):

Mean = (4 + 5 + 6 + 7 + 8)/5 = 6

| Values | Absolute Deviation from Mean|value - mean| | | ------ | ------------------------------------------- | | 4 | 2 | | 5 | 1 | | 6 | 0 | | 7 | 1 | | 8 | 2 |

The above table can be depicted on a number line:

Mean-Deviation

Mean Deviation

This gives the mean deviation of the given data set.

**Key Concepts of Mean Deviation

Following are some important terms related to mean deviation:

**1) Central Tendency

**2) Absolute Deviation:

**3) Mean of Deviations:

Mean Deviation Formulas

There are various types of mean deviation formulas used depending upon the types of data given and the central point chosen for the given data set.

Mean Deviation Formula

Where,

Mean Deviation Formula for Ungrouped Data

For ungrouped data or data that is not properly arranged, that is, the given data is in raw form, the mean deviation is calculated using the formula,

**Mean Deviation = ∑ i = 1 n |x i - x̄| / n

Where,

**Mean Deviation for Discrete Frequency Distribution

In the discrete series, the data of each value is represented, and its frequency is represented in the next column of the table. The table added below shows the data in a discrete frequency distribution table.

Wages Number of Workers (Frequency)
2500 7
3000 9
4000 5
4500 6
5000 3

The formula used to calculate the mean deviation in this type of data set is,

**Mean Deviation = ∑ i=1 n f i | x i - x̄ | / ∑ i=1 n f i

Where,

**Mean Deviation for Continuous Frequency Distribution

In a continuous series, data is arranged in certain definite classes. The data items contained in classes lose their identity, and the individual items are merged into one or the other class groups. The classes have continuity, that is, the end of the first class is marked by the beginning of the next class. Thus, the name is a continuous series. For instance, the continuous series is depicted using the following data,

Age Group Frequency
10-20 15
20-30 10
30-40 13
40-50 12

The formula used to calculate the mean deviation in this type of data set is,

**Mean Deviation = ∑ i=1 n | x i - x̄ | / ∑ i=1 n f i

Where,

Mean Deviation about Mean

We can easily calculate the mean deviation from the mean of the given dataset. The mean of the data set is simply the arith mean also called the average of the data set. It is calculated by taking the sum of all observations divided by the number of observations.

The formula used to calculate the mean deviation about the mean of the data set is,

**For Ungrouped Data,

**Mean Deviation = ∑ i=1 n | x i - x̄ | / n

Where **x̄ = (x 1 + x 2 + x 3 **+ ... + x n )/n

**For Continuous and Discrete Frequency Distribution,

**Mean Deviation = ∑ i=1 n | x i - x̄ | / ∑ i=1 n f i

Where **x̄ = ∑ i=1 n x i f i / ∑ i=1 n **f i

The middle point of the data set, when arranged in ascending or descending order, is called the median of the data set. It is the middle value of the data set that divides the data set into two equal halves. The formula to calculate the mean deviation of the data set about the mean is,

**For Ungrouped Data,

**Mean Deviation = ∑ i=1 n | x i - M | / n

Where **M represents the middle point or median of the data set and is calculated as,

**For Discrete Frequency Distribution,

**Mean Deviation = ∑ i=1 n f i | x i - M | / ∑ i=1 n **f i

Where **M represents the middle point or median of the data set, and is calculated in the same way as above.

**For Continuous Frequency Distribution,

**Mean Deviation = ∑ i=1 n f i | x i - M | / ∑ i=1 n **f i

Where **M represents the middle point or median of the data set and is calculated as,

**M = l + {[∑ i=1 n **f i /2 - cf] / f} × h

Where,

Mean Deviation about Mode

The term with the highest frequency in the dataset is the mode of the dataset. The formula to calculate the mean deviation of the data set about the mode is,

**For Ungrouped Data,

**Mean Deviation = ∑ i=1 n | x i - M | / n

Where **M represents the mode of the data set.

**For Discrete Frequency Distribution,

**Mean Deviation = ∑ i=1 n f ​i ​| x i - M | / ∑ i=1 n **f i

Where **M represents the mode of the data set.

**For Continuous Frequency Distribution,

**Mean Deviation = ∑ i=1 n f i | x i - M | / ∑ i=1 n **f i

where, M represents the mode of the data set and is calculated as,

**Mode = l + [(f - f 1 ) / (2f - f 1 - f 2 )] × h

Where,

**How to Calculate Mean Deviation?

**Step 1: Calculate the value of the mean, median, or mode of the series.

**Step 2: Calculate the absolute deviations about the mean, median, or mode.

**Step 3: Multiply the deviations of the given data set with their respective frequency. It is required if the frequency of the data is given.

**Step 4: Find the sum of all the deviations and divide it by the number of observations to get the mean deviation.

**Example: Find the mean deviation of the given data about their mean, {4, 6, 7, 3, 5, 5}.

**Solution:

**Step 1: Find the mean of the given data.

Mean = (4 + 6 + 7 + 3 + 5 + 5) / 6 = 30 / 6 = 5

**Step 2: Find the absolute deviation

Given Data Set Absolute Deviation of Mean
4 4-5 = |-1
6 6-5 = |1
7 7-5 = |2
3 3-5 = |-2
5 5-5 = |0
5 5-5 = |0

**Step 3: Find the mean deviation of the absolute value so obtained.

**Mean Deviation = (1 + 1 + 2 + 2 + 0 + 0)/6 = 6/6 = 1

Thus, the mean deviation from the mean of the given data set is 1.

Mean Deviation vs Standard Deviation

The basic difference between the mean deviation and the standard deviation is discussed in the table below.

Mean Deviation Standard Deviation
All the central points (mean, median, and mode) are used to find the mean deviation. The mean is only used to find the standard deviation.
The absolute value of the deviation is used to find the mean deviation. The square of the deviation is used to find the standard deviation.
Mean deviation is a less frequently used data measure. Standard Deviation is a highly used data measure that is used to find various central measures.

Merits and Demerits of Mean Deviation

Mean deviation is a highly used data measure, and the various merits and demerits of the mean deviation are,

Merits of Mean Deviation

Demerits of Mean Deviation

**Also Check:

**Solved Examples of Mean Deviation

**Example 1: Find the mean deviation of the following data: 7, 5, 1, 3, 6, 4, 10.

**Solution:

For Finding the mean deviation,

First, calculate the mean value of the given data

Mean = Sum of all the observations / Number of observations

= \frac{7 \ + \ 5 \ + \ 1 \ + \ 3 \ + \ 6 \ + \ 4 \ + \ 9}{7}

Mean = 5.14 ≅ 5

Now finding the absolute deviation.

Given Value Absolute Deviation of Mean
7 |7 - 5 =
5 |5 - 5 =
1 |1 - 5 =
3 |3 - 5 =
6 |6 - 5 =
4 |4 - 5 =
10 |10 - 5 =

Further the mean of Absolute Values

= 2, 0, 4, 2, 1, 1, 5

Mean = \frac{2 \ + \ 0 \ + \ 4 \ + \ 2 \ + \ 1 \ + \ 1 \ + \ 5}{7}

Mean Deviation = 2.14

Therefore,
Mean deviation for 7, 5, 1, 3, 6, 4, 10 is 2.14

**Example 2: Find the mean deviation of the following data: 11, 9, 7, 3, 2, 8, 10, 12, 15, 13.

**Solution:

For Finding the mean deviation,

First, calculate the mean value of the given data

Mean = = \frac{11 \ + \ 9\ + \ 7 \ + \ 3 \ + \ 2 \ + \ 8 \ + \ 10 \ + \ 12 \ + \ 15 \ + \ 13}{10}

Mean = 9

Now finding the absolute deviation.

Given Value Absolute Deviation of Mean
11 |11 - 9 =
9 |9 - 9 =
7 |7 - 9 =
3 |3 - 9 =
2 |2 - 9 =
8 |8 - 9 =
10 |10 - 9 =
12 |12 - 9 =
15 |15 - 9 =
13 |13 - 9 =

Further find the mean of these values obtained are 2, 0, 2, 6, 7, 1, 1, 3, 6, 4.

Mean Deviation ⇒\frac{2 \ + \ 0 \ + \ 2 \ + \ 6 \ + \ 7 \ + \ 1 \ + \ 1 \ + \ 3 \ + \ 6 \ + \ 4 }{10}\\=\frac{32}{10}

⇒ Mean = 3.2

Therefore, Mean deviation for 11, 9, 7, 3, 2, 8, 10, 12, 15, 13 is 3.2.

**Example 3: Find the mean deviation of the following data table:

Class Interval Frequency
5-15 8
15-25 12
25-35 6
35-45 4

**Solution:

The mean of the following data is,

Class Interval Frequency(fi) Mid Point(xi) fixi
5-15 8 10 80
15-25 12 20 240
25-35 6 30 180
35-45 4 40 160
∑fi = 30 ∑fixi = 660

x̄ = ∑fixi / ∑fi = 660/30 = 22

| Class Interval | Frequency(fi) | Mid Point(xi) | |xi - x̄| | | -------------- | ----------------- | ------------- | ----------- | | 5-15 | 8 | 10 | 12 | | 15-25 | 12 | 20 | 2 | | 25-35 | 6 | 30 | 8 | | 35-45 | 4 | 40 | 18 | | ∑fi = 30 | ∑|xi - x̄| = 40 | | |

Mean Deviation = 40/30 = 1.33