Introduction
Mean, mode, and medium form the center points of a data set. mean deviation or mean absolute deviation is used to compute how far the values are in a particular data set from the center point. In simpler terms, mean absolute deviation is used to calculate the average absolute deviations from relevant data from the central point. You can calculate the mean absolute deviation for both ungrouped and grouped data.
Mean absolute deviation is a much simpler variability measurement than the standard deviation. While it can be obtained from any of the measures of central tendency, mean absolute deviation from the median and mean are the ones most commonly used for statistical considerations.
Mean Deviation about Median
To calculate the mean deviation about median, you first need to arrange data values in ascending order and find the middle value. This will help you to find the median value. As you determine the medium, you will have to subtract the median from each data value and ultimately take the average.
Mean deviation about median value ‘M’ would be the mean of absolute values of deviations of observations from ‘M.’ M.D. (M) refers to the mean deviation from ‘M.’
The mean deviation about a median value ‘M’ is the mean of the absolute values of the deviations of the observations from ‘M.’ The mean deviation from ‘M’ is denoted as M.D. (M).
M.D. (M) = Sum of absolute values of deviations from ‘M’Number of observations
Calculating mean deviation from median
There are three key steps involved in determining the mean deviation about median either for grouped or ungrouped data. These steps include:
- Finding median values for data values provided
- Subtracting median values from each data value provided, as well as taking into account absolute deviations present
- The calculating mean of deviations obtained in the former step.
Mean deviation about median for ungrouped data
For better understanding, this concept can be explained with the help of an example. You can consider x1, x2, x3…..in as n observations here. The formula to find a mean deviation from the median for this data would be:
M.D.(M)= |xi−M|
In this situation, the median of data provided is ‘M.’
E.g. 1: Find mean deviation about median for the data: 6, 7, 10, 12, 13, 4, 8, 16
To find the median in math, you firstly need to arrange this data set into ascending order.
Ascending order of the data set: 4, 6, 7, 8, 10, 12, 12, 13, 16
Number of data values : 9
Median = (n + 1)/2 th observation
= (9 + 1)/2
= 5th observation
As a result, median = 10
Absolute values of the respective deviations from the median, i.e., |xi − M|would be
= |4 – 10|, |6 – 10|, |7 – 10|, |8 – 10|, |10 – 10|, |12 – 10|, |12 – 10|, |13 – 10|, |16 – 10|
= 6, 4, 3, 2, 0, 2, 2, 3, 6
As mentioned above, M.D.(M)=1n∑ni=1|xi−M|M.D.(M)=1n∑i=1n|xi−M|
Hence, (6 + 4 + 3 + 2 + 0 + 2 + 2 + 3 + 6)/9
= 28/9
= 3.11
As a result, the mean deviation from the median for the data provided is 3.11. Knowledge of the median calculator formula is important for understanding this process.
mean deviation about median for grouped data
There are two ways to find a mean deviation from median for grouped data. These ways are:
- Mean absolute deviation about median for discrete frequency distribution
To find the mean absolute deviation about median, you need first to calculate the median of the discrete frequency distribution provided. Mean absolute deviation can be found out with the usage of the following formula
M.D.(M)= |xi−M|
In this situation:
M= Median
xi = Data values
fi = frequency
N= sum of the frequencies
To understand the concept more clearly, you can follow the following example.
e.g., Find mean deviation about median for the following distribution
xi | 2 | 5 | 6 | 8 | 10 | 12 |
fi | 2 | 9 | 12 | 4 | 8 | 5 |
Your first step for solving the problem would be to arrange the data in ascending order. However, as the data values are already in ascending order, with the usage of the median calculator you will get:
xi | 2 | 5 | 6 | 8 | 10 | 12 |
fi | 2 | 9 | 12 | 4 | 8 | 5 |
cumulative frequency (cf) | 2 | 11 | 23 | 27 | 35 | 40 |
The number of frequencies or N here is 40, which is an even number
Therefore N/2 = 40/2 = 20
Here the median shall be the average of 20th and 21st observations.
But those observations lie in cf 23, and the corresponding observation for it is 6
As a result, Median = (6 + 6)/2 = 6
Now to find absolute deviations, you shall need to prepare another table
|xi – M| | 4 | 1 | 0 | 2 | 4 | 6 |
fi | 2 | 9 | 12 | 4 | 8 | 5 |
fi |xi – M| | 8 | 9 | 0 | 8 | 32 | 30 |
Considering the formula M.D.(M)= |xi−M| , your answer would be
= (8 + 9 + 0 + 8 + 32 + 30)/40
= 87/40
= 2.175
Hence, it can be understood that the mean deviation about the median for the distribution given is 2.175.
- Mean deviation about median for continuous frequency distribution
Life before, you need to use the following formula
M.D.(M)= 1Ni=1nfi|xi−M|
M= Median
xi = Data values
fi = frequency
N= sum of the frequencies
For continuous frequency distribution, you can use the following median calculator formula:
In this situation:
Median = l +N2-Cffh
l = Lower limit of median class
N = Number of observations
Cf = Cumulative frequency of class preceding the median class
f = Frequency of median class
h = Class size
You can use the following solved example to properly understand how to calculate mean deviation about median for a continuous distribution of data.
Class interval | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 – 80 |
Frequency | 2 | 3 | 8 | 14 | 8 | 3 | 2 |
Before anything else, you need to calculate the mean deviation and mean for the data provided.
Class interval | Frequency (fi) | Cumulative frequency | Mid-points (xi) | |xi– M| | fi |xi – M| | Class interval | Frequency (fi) |
10 – 20 | 2 | 2 | 15 | 30 | 60 | 10 – 20 | 2 |
20 – 30 | 3 | 5 | 25 | 20 | 60 | 20 – 30 | 3 |
30 – 40 | 8 | 13 | 35 | 10 | 80 | 30 – 40 | 8 |
40 – 50 | 14 | 27 | 45 | 0 | 0 | 40 – 50 | 14 |
50 – 60 | 8 | 35 | 55 | 10 | 80 | 50 – 60 | 8 |
60 – 70 | 3 | 38 | 65 | 20 | 60 | 60 – 70 | 3 |
70 – 80 | 2 | 40 | 75 | 30 | 60 | 70 – 80 | 2 |
Total | ∑fi = 40 | ∑fi|xi–M| = 400 | Total | ∑fi = 40 |
In this situation,
N = ∑fi = 40
N/2 = 40/2 = 20
cf greater than or equal to N/2 is 27, and corresponding class interval is 40 – 50 for it
Median class = 40 – 50
Hence, l = 40
N = 40
cf = 13
f = 14
h = 10
Median = l +N2-Cffh
= 40 + [(20 – 13)/14] × 10
= 40 + (7/14) × 10
= 40 + 5
= 45
In this situation, median (M) = 45
As per the table provided above,
∑fi |xi – M| = 400
N = ∑fi = 40
Therefore, the needed mean deviation = 400/40 = 10
Conclusion
Mean absolute deviation basically is the mean of the absolute deviations of the observations or values from a suitable average. You need to follow either the discrete frequency distribution approach or continuous frequency distribution to find the mean deviation about median.