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(D) The variance of a set of observations $x_i$ such that $a \le x_i \le b$ is given by $	ext{Var}(x) = \frac{1}{n} \sum (x_i - \bar{x})^2$.It is a k

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$(b - a)^2 \ge 	ext{Var}(x)$

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(D) The variance of a set of observations $x_i$ such that $a \le x_i \le b$ is given by $	ext{Var}(x) = \frac{1}{n} \sum (x_i - \bar{x})^2$.It is a known property that for a set of observations in the interval $[a, b]$,the standard deviation $\sigma$ satisfies $\sigma \le \frac{b - a}{2}$.Therefore,the variance $	ext{Var}(x) = \sigma^2 \le \left( \frac{b - a}{2} ight)^2 = \frac{(b - a)^2}{4}$.Since $\frac{(b - a)^2}{4} \le (b - a)^2$,the inequality $(b - a)^2 \ge 	ext{Var}(x)$ holds true.

${x_i}$	$3$	$8$	$13$	$18$	$23$
${f_i}$	$7$	$10$	$15$	$10$	$6$

Suppose values taken by a variable $x$ are such that $a \le x_i \le b$,where $x_i$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, ..., n$. Then:

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