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# Mean Deviation

Mean deviation can be defined as the mean of the absolute deviations and it is denoted by Dx. The formula of the mean deviation can be given as Dx = 1/N sigma 1 to n |xi – x bar| where x bar is the mean of the given deviations and N is the sum of the frequency or number of deviations. The value of N depends upon the type of the statics data given and xi is the given deviations.

Let us discuss the process of finding the mean deviation of the given data. First we need to find out the mean of the given data then we need to find out the deviations about the mean and find out the sum of deviations now find the mean of the sum of deviations. Then we will get the required mean deviation of the given data.

Let us discuss mean deviation with some of the examples.

Consider the first example to be: A student marks in five subjects are given as follows find out the mean deviation of the marks of the student. The marks are 92, 75, 95, 90 and 98.

The given marks of the student are 92, 75, 95, 90 and 98. Now let us find out the mean of the given marks then we will get it as mean = (92 + 75 + 95 + 90 + 98) / 5 which is equal to 90. Now we got the mean of the marks as 90. Now let us find out the mean deviations, we will get the mean deviations if we subtract given observations from the mean. The mean deviations are (92 – 90) = 2, (75 – 90) = -15, (95 – 90) = 5, (90 – 90) = 0 and (98 – 90) = 8. Now if we find the modulus of these deviations then we will get the positive observations that is the observations which we will get after finding the modulus are 2, 15, 5, 0 and 8. Now let us find out the mean of these deviations, first we need to find out the sum of the deviations then we will get it as (2 + 15 + 0 + 5 + 8) which is equal to 30. If we find out the mean then we will get it as 30/5 which is equal to 6. Now the mean deviation is 6.

Consider the second example be: find out the mean deviation of observations 9, 3, 6.

The given observations are 9, 3, 6. First let us find out the mean of the given observations then we will get it as (9 + 3 + 6) / 3 which is equal to 18/3 = 6. Now let us find out the mean deviations then we will get it as (9 – 6) = 3, (3 – 6) = -3, (6 – 6) = 0. Now let us find out the modulus of the observations then we will get it as 3, 3, 0. Now let us find out the mean of these deviations for that we need to find out the sum of deviations which is equal to (3 + 3 + 0) = 6. The mean now is 6/3 = 2. Now the required mean deviation is 2.

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