Show that the sum of the deviations of all the given observations from the mean is zero.

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Solution

Let A.M. of $x_{1}, x_{2}, . . . . x_{n} = \frac{x_{1} + x_{2} + . . . . + x_{n}}{n} = K$
Mean deviation $= \frac{\sum_{1}^{n} (x - K)}{n} = \frac{x_{1} - K + x_{2} - K + . . . . + x_{n} - K}{n} = \frac{x_{1} + x_{2} + . . . . + x_{n} - n K}{n} = \frac{x_{1} + x_{2} + . . . . + x_{n}}{n} - k = K - K = 0$
Hence, the algebraic sum of the deviations of a set of n values from their mean is $0$

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