The S.D.is not less than the mean deviation. If this is true enter 1, else enter 0.

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Solution

We take $\frac{1}{N} \sum i f_{i} {(x_{i} - ¯ x)}_{i}^{2} \geq {\frac{1}{N} \sum f_{i} | x_{i} - ¯ x |}_{i}^{2}$
Put $| x_{i} - ¯ x | = z_{i}$
$∴ \frac{1}{N} \sum i f_{i} z_{i}^{2} -^{2} \geq {\frac{1}{N} \sum i f_{i} z_{i}}_{i}^{2}$ to $\frac{1}{N} \sum f_{i} z_{i}^{2} \geq {\frac{1}{N} \sum i z_{i}}_{i}^{2}$
$\Rightarrow σ_{z}^{2} \geq 0$ which is always true.

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