# Mean Deviation For Continuous Frequency Distribution

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## Introduction to Mean Deviation For Continuous Frequency Distribution

Frequency Distribution is the representation of data in a tabular form or a graphical form that indicates the frequency (the number of times any given observation occurs within a given particular interval). Assuming that the data is huge, for example, if we need to analyze the marks of 100 students, then it is not practical to represent this data in random. So based on class intervals, we use the concept of ‘Grouping of Data’.

## Mean Deviation of Grouped Data

In frequency distribution of grouped data of continuous type, the class intervals or groups are arranged in a manner that there are no gaps between the classes and each class in the table has its corresponding frequency. The class intervals are chosen in such a way that they need to be mutually exclusive and exhaustive.

In order to understand how this concept of continuous frequency works, look at the following table that is given below.

The following table represents the age group of teachers working in a certain store:

 Age Group Number of People 15-25 25 25-35 54 35-45 34 45-55 20

This above table represents the continuous frequency in nature and the frequency is mentioned according to the interval of the classes.

## How to Calculate the Mean Deviation of Continuous Frequency Distribution?

The following steps that are given below will help you calculate the mean deviation for continuous frequency distribution. These steps are:

Step 1: Consider the midpoint of each class to be its frequency. Then, the mean is calculated for these points. Using the above-mentioned table as an example again, the mid-points would be:

 Age Group x xi Number of People (fi) 15-25 20 25 25-35 30 54 35-45 40 34 45-55 50 20

You can calculate the mean using the formula:

$\overline{x} = \frac{1}{N} \sum_{i=1}^{n}$

Step 2: Find the absolute mean deviation using the formula below:

$M.A.D (\overline{x}) = \frac{1}{N} \sum_{i=1}^{n} f_{i} | x_{i} - \overline{x}|$

Tabulating the above formula, we get:

 Age Group x xi Number of People (fi) xifi $x_{i} - \overline{x}$ $f_{i} |x_{i} - \overline{x}|$ 15-25 20 25 500 13.684 324.1 25-35 30 54 1620 3.684 198.936 35-45 40 34 1360 6.316 214.744 45-55 50 20 1000 16.316 352.32 $\sum f_{i} = 133$ $\overline{x} = \frac{1}{N} \sum_{i=1}^{n}$$\overline{x}$ $\sum_{i=1}^{n} f_{i} | x_{i} - \overline{x}|$= 1090.1

Now, let us find the mean absolute deviation.

$M.A.D (\overline{x}) = \frac{1}{N} \sum_{i=1}^{n} f_{i} | x_{i} - \overline{x}| = \frac{1090.1}{133} = 8.196$

This might be a little complex method to solve but there is also another method called the step deviation method to find the mean absolute deviation. The result that you obtain using any of these methods is always the same or something very close to it. The step deviation method is less complicated than the other method. The formula used in Step Deviation method is:

$M.A.D (\overline{x}) = a + \frac{h}{N} \sum_{i=1}^{n} f_{i}d_{i}$

Here,

a = assumed mean

h = common factor

d = $\frac{x_{i} - a}{h}$

Now, to calculate the mean deviation, we need to know the median of the given set of data using a cumulative frequency that is given as:

$M = l + \frac{\frac{N}{2} - C}{f} \times h$

Where,

l = Median class’ lower limit

f = Median class’ frequency

h = Median class’ width

C = Cumulative frequency of the next or preceding class

The formula used to calculate the mean deviation is:

$M.A.D (M) = \frac{1}{N} \sum_{(i=1)}^{n} f_{i} |x_{i} - M|$

The mean and mean deviation is calculated for the above-used example as shown below:

 Class Frequency Cumulative Frequency Mid-Point |xi− M| fi |xi − M| 5-15 5 5 10 17.42 87.1 15-25 9 14 20 7.42 66.78 25-35 7 21 30 2.58 18.06 35-45 3 24 40 12.58 37.74 45-55 8 32 50 22.58 180.64 32 390.32

We know that N/2 = 16. Hence, we will pick classes 25 - 35 as the median class.

$M = l + \frac{\frac{N}{2} - C}{f} \times h$

$\Rightarrow 25 + \frac{16 - 14}{7} \times 10 = 27.42$

The mean deviation of the mean is:

$M.A.D (M) = \frac{1}{N} \sum_{(i=1)}^{n} f_{i} |x_{i} - M| = \frac{390.32}{32} = 12.19$