The question asks about the property of the sum of squares of deviations when calculated from the mean of a dataset.
For any set of observations $x_1, x_2, \dots, x_n$, let $\bar{x}$ be their arithmetic mean.
The sum of the squares of deviations from the mean is given by:
$ S_{\bar{x}} = \sum_{i=1}^{n} (x_i - \bar{x})^2 $Now, consider the sum of the squares of deviations from any arbitrary value '$a$':
$ S_a = \sum_{i=1}^{n} (x_i - a)^2 $It can be shown mathematically that:
$ S_a = \sum_{i=1}^{n} (x_i - \bar{x})^2 + n(\bar{x} - a)^2 $Since $n$ (the number of observations) is positive and $(\bar{x} - a)^2$ is always non-negative (zero only if $a = \bar{x}$), the term $n(\bar{x} - a)^2$ is always greater than or equal to zero.
Therefore, $S_a \ge S_{\bar{x}}$. Equality holds only when $a = \bar{x}$.
This proves that the sum of the squares of deviations is minimized when the deviations are taken from the arithmetic mean.
Hence, the sum of squares of deviation taken from the Mean is always the Least.
If for a moderately symmetrical distribution mean deviation is 12, then the value of standard deviation is
Variance is independent of change of :