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How to calculate the mean deviation?

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  • Last Updated : 09 Nov, 2021

The science of collecting, analyzing, presenting data is known as statistics. In statistics, the deviation is known as the variation obtained between the observed value and other values of a variable. Let’s learn more about this topic and learn what a men deviation is in statistics,

Mean Deviation

The mean deviation of a given standard distribution is a measure of the central tendency. It can be computed using the Arithmetic Mean, Median, or Mode of the data. It is used to show how far the observations are situated from the average of the data observations. Each of these deviations is considered to be an absolute value. This implies that the negative signs are ignored. In addition to this, the deviations on both sides of the mean value are equivalent in nature. Therefore, the mean deviation of the given data distribution is the mean of the absolute deviations of the observations or values from a suitable average. This suitable average can be either mean, median, or mode of the data. For instance, the mean deviation formula for an individual series or a continuous series, etc.

Mean Deviation Types

Individual Series

For a given set of raw data, when the data is presented individually in the form of a series, it is known as individual series. It is basically a sequence of raw data in such a form that forms an arrangement individually. Items are represented singly in the individual series. It is used to depict numerical values. For instance, let’s assume the following marks out of 100 to be secured by students in a class : 

56, 77, 88, 49, 60, 92, 70, 81, 69, 51,

The above data is not conclusive about how many students got 56 marks or more than 77 in a single sight.

Discrete Series

The discrete series is used to reflect data for each specific value of the observation variable. Each of the variables corresponds to an integer value. The exact measurement of the items in the data is visible. For instance, the wages earned by 30 workers are depicted below in the tables as:  

WagesNumber of workers (frequency)
25007
30009
40005
45006
50003

Conclusively, 7 workers are getting wages equivalent to Rs. 2500, and 9 workers are getting Rs. 3000 and so on.

Continuous Series

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

Age GroupFrequency
10-2015
20-3010
30-4013
40-5012
50-6018
60-704
70-808
  • Individual Series

Mean Deviation (M.D) =\frac{\sum|x-\bar x|}{N}

Where,

∑ – Summation

x – Observation

\bar x  – Mean

N – Number of observation

  • Discrete Series

Mean Deviation (M.D) =\frac{\sum f|x-\bar x|}{\sum f}

Where,

∑ – Summation

x – Observation

\bar x – Mean

f – frequency of observation

  • Continuous Series

Mean Deviation (M.D) =\frac{\sum f|x-\bar x|}{\sum f}

Where,

∑ – Summation

x – Mid-value of the class

\bar x  – Mean

f – frequency of observation

Mean deviation from Median

  • Individual Series

Mean Deviation (M.D) = \frac{\sum |x-M|}{N}

Where,

∑ – Summation

x – Observation

M – Median

N – Number of observation

  • Discrete Series

Mean Deviation (M.D) =\frac{\sum f |x-M|}{\sum f}

Where,

∑ – Summation

x – Observation

M – Median

N – Frequency of observations

  • Continuous Series

Mean Deviation (M.D) =\frac{\sum f |x-M|}{\sum f}

Where,

∑ – Summation

x – Observation

M – Median

N – Frequency of observations

Mean deviation from Mode

  • Individual Series

Mean Deviation (M.D) =\frac{\sum  |x-Mode|}{\sum N}

Where,

∑ – Summation

x – Observation

M – Mode

N – Number of observations

  • Discrete Series

Mean Deviation (M.D) =\frac{\sum f |x-Mode|}{\sum f}

Where,

∑ – Summation

x – Observation

M – Mode

N – Frequency of observations

  • Continuous Series

Mean Deviation (M.D) =\frac{\sum f |x-Mode|}{\sum f}

Where,

∑ – Summation

x – Observation

M – Mode

N – Frequency of observations

Steps to Calculate the Mean Deviation

  1. Calculate the mean, median, or mode of the series.
  2. Calculate the deviations from the Mean, median, or mode and do ignore the minus signs.
  3. Multiply the deviations with the frequency. The step is required only in the discrete and continuous series.
  4. Sum up all the deviations.
  5. Apply the formula.

Examples;

Example 1: What are the advantages of using the mean deviation?

Solution:

The advantages of using mean deviation are:

  • It is based on all the data values given, and hence it provides a better measure of dispersion.
  • It is easy to understand and calculate.

Example 2: Mention the procedure to find the mean deviation.

Solution:

The procedure to find the mean deviation are:

Step 1: Calculate the mean value for the data given.

Step 2: Then, the mean is subtracted from each data value (distance).

Step 3: Finally, the mean is found for the distance.

Example 3: Find the mean deviation of the following data?

7, 5, 1, 3, 6, 4, 10

Solution:

To find the mean deviation,

First, calculate the mean value of the given data

Mean = \frac{sum\ of\ all\ the\ observations}{Number\ of\ observations}\\ = \frac{7+ 5+ 1+ 3+ 6+ 4+ 9}{7}\\ =\frac{35}{7}

Mean = 5

Now subtract the mean from each data value {ignore negative (-)}

  • 7 – 5 = 2
  • 5 – 5 = 0
  • 1 – 5 = 4
  • 3 – 5 = 2
  • 6 – 5 = 1
  • 4 – 5 = 1
  • 10 – 5 = 5

Further find the mean of these values obtained

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

Mean =  \frac{sum\ of\ all\ the\ observations}{Number\ of\ observations}\\ = \frac{2+ 0+ 4+ 2+ 1+ 1+ 5}{7}\\ =\frac{15}{7}

Mean = 2.14

Therefore,

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

Example 4: Find the mean deviation of the following data?

11, 9, 7, 3, 2, 8, 10, 12, 15, 13

Solution:

To find the mean deviation,

First, calculate the mean value of the given data

Mean =  \frac{sum\ of\ all\ the\ observations}{Number\ of\ observations}\\ = \frac{11+ 9+ 7+ 3+ 2+ 8+ 10+ 12+ 15+ 13}{10}\\ =\frac{80}{10}

Mean = 8

Now subtract the mean from each data value {ignore negative (-)}

  • 11 – 8 = 3
  • 9 – 8 = 1
  • 7 – 8 = 1
  • 3 – 8 = 5
  • 2 – 8 = 6
  • 8 – 8 = 0
  • 10 – 8 = 2
  • 12 – 8 = 4
  • 15 – 8 = 7
  • 13 – 8 = 5

Further find the mean of these values obtained

3, 1, 1, 5, 6, 0, 2, 4, 7, 5.  

Mean =  \frac{sum\ of\ all\ the\ observations}{Number\ of\ observations}\\ = \frac{3+ 1+ 1+ 5+ 6+ 0+ 2+ 4+ 7+ 5 }{10}\\ =\frac{34}{10}

Mean = 3.4

Therefore,

Mean deviation for 11, 9, 7, 3, 2, 8, 10, 12, 15, 13 is 3.4.


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