Sigma(x_{i} - bar{x}) = ldots ldots ldots | vedclass

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(D) The mean of a set of observations is defined as $\bar{x} = \frac{\Sigma x_{i}}{n}$,where $n$ is the number of observations.Therefore,$\Sigma x_{i}

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$0$

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(D) The mean of a set of observations is defined as $\bar{x} = \frac{\Sigma x_{i}}{n}$,where $n$ is the number of observations.Therefore,$\Sigma x_{i} = n\bar{x}$.Now,consider the expression $\Sigma(x_{i} - \bar{x})$.This can be expanded as $\Sigma x_{i} - \Sigma \bar{x}$.Since $\bar{x}$ is a constant,$\Sigma \bar{x} = n\bar{x}$.Substituting the values,we get $\Sigma x_{i} - n\bar{x} = n\bar{x} - n\bar{x} = 0$.Thus,the sum of deviations of observations from their mean is always $0$.

$\Sigma(x_{i} - \bar{x}) = \ldots \ldots \ldots$

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