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Q. If $A.M.$ of $x_1$, $x_2$, $....$, $x_n$ is $\bar{x}$, then $A.M.$ of $ax_1 + b$, $ax_2 + b$, $ax_3 + b$, $...$, $ax_n + b$ is

Statistics

Solution:

$A.M. = \frac{\left(ax_{1}+b\right)+\left(ax_{2}+b\right)+....+\left(ax_{n}+b\right)}{n}$
$= \frac{a\left(x_{1}+x_{2}+.... + x_{n}\right)+nb}{n}$
$= \frac{a\left(x_{1}+x_{2}+ .... +x_{n}\right)}{n}+b$
$= a\bar{x}+b $