Let x_1, x_2, ldots, x_{11} be 11 distinct po | vedclass

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(C) Let the $11$ distinct positive integers be arranged in increasing order: $x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9 < x_{10} < x_{11}$.T

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the median decreases

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(C) Let the $11$ distinct positive integers be arranged in increasing order: $x_1 The median of these $11$ integers is $x_6$.The other $10$ integers are $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}$.The median of these $10$ integers is $m = \frac{x_5 + x_6}{2}$.Since $x_5 Replacing the largest integer $x_{11}$ with $m$,the new set of $11$ integers in increasing order is $x_1, x_2, x_3, x_4, x_5, m, x_6, x_7, x_8, x_9, x_{10}$.The new median is the $6^{th}$ term,which is $m$.Since $m < x_6$,the median decreases.

$x_i$	$3$	$6$	$10$	$12$	$7$	$15$
$f_i$	$3$	$4$	$2$	$8$	$13$	$10$

Let $x_1, x_2, \ldots, x_{11}$ be $11$ distinct positive integers. If we replace the largest of these integers by the median of the other $10$ integers,then:

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