For (2n+1) observations {x_1}, -{x_1}, {x_2}, | vedclass

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(B) The given observations are ${x_1}, -{x_1}, {x_2}, -{x_2}, \dots, {x_n}, -{x_n}, 0$. There are $(2n+1)$ observations.$1$. Median $(M.D.)$: Arrangin

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$S.D. > M.D.$

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(B) The given observations are ${x_1}, -{x_1}, {x_2}, -{x_2}, \dots, {x_n}, -{x_n}, 0$. There are $(2n+1)$ observations.$1$. Median $(M.D.)$: Arranging the observations in ascending order,the middle term is $0$. Thus,$M.D. = 0$.$2$. Standard Deviation $(S.D.)$: The formula for $S.D.$ is $\sqrt{\frac{1}{N} \sum (x_i - \bar{x})^2}$.Here,the mean $\bar{x} = \frac{x_1 - x_1 + x_2 - x_2 + \dots + x_n - x_n + 0}{2n+1} = 0$.So,$S.D. = \sqrt{\frac{1}{2n+1} \sum_{i=1}^n (x_i^2 + (-x_i)^2 + 0^2)} = \sqrt{\frac{2 \sum x_i^2}{2n+1}}$.Since all $x_i$ are distinct and non-zero,$\sum x_i^2 > 0$,which implies $S.D. > 0$.Comparing the two,$S.D. > 0$ and $M.D. = 0$,therefore $S.D. > M.D.$

$X$	$1$	$3$	$5$	$7$	$9$
$f$	$4$	$24$	$28$	$\alpha$	$8$

For $(2n+1)$ observations ${x_1}, -{x_1}, {x_2}, -{x_2}, ....., {x_n}, -{x_n}$ and $0$,where all $x_i$ are distinct,let $S.D.$ and $M.D.$ denote the standard deviation and median respectively. Which of the following is always true?

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