If the arithmetic mean of the numbers x_1, x_ | vedclass

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(D) Given that the arithmetic mean of $x_1, x_2, ..., x_n$ is $\bar{x}$,we have $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$,which implies $\sum_{i=1}^{n}

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$a \bar{x} + b$

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(D) Given that the arithmetic mean of $x_1, x_2, ..., x_n$ is $\bar{x}$,we have $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$,which implies $\sum_{i=1}^{n} x_i = n \bar{x}$.The required mean of the numbers $ax_1 + b, ax_2 + b, ..., ax_n + b$ is given by:$	ext{Mean} = \frac{(ax_1 + b) + (ax_2 + b) + ... + (ax_n + b)}{n}$$= \frac{a(x_1 + x_2 + ... + x_n) + nb}{n}$$= a \left( \frac{\sum_{i=1}^{n} x_i}{n} ight) + \frac{nb}{n}$$= a \bar{x} + b$

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

If the arithmetic mean of the numbers $x_1, x_2, x_3, ..., x_n$ is $\bar{x}$,then the arithmetic mean of the numbers $ax_1 + b, ax_2 + b, ax_3 + b, ..., ax_n + b$,where $a$ and $b$ are two constants,would be:

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