If mu is the mean of a distribution (y_i, f_i | vedclass

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(C) We know that the mean $\mu$ is defined as $\mu = \frac{\sum f_i y_i}{\sum f_i}$,which implies $\sum f_i y_i = \mu \sum f_i$.Now,consider the expre

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(C) We know that the mean $\mu$ is defined as $\mu = \frac{\sum f_i y_i}{\sum f_i}$,which implies $\sum f_i y_i = \mu \sum f_i$.Now,consider the expression $\sum f_i(y_i - \mu)$.Expanding the summation,we get $\sum f_i y_i - \sum f_i \mu$.Since $\mu$ is a constant,this becomes $\sum f_i y_i - \mu \sum f_i$.Substituting $\sum f_i y_i = \mu \sum f_i$,we get $\mu \sum f_i - \mu \sum f_i = 0$.

$x_i$	$5$	$8$	$11$	$14$	$17$
$f_i$	$4$	$5$	$6$	$10$	$20$

If $\mu$ is the mean of a distribution $(y_i, f_i)$,then $\sum f_i(y_i - \mu) = $

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