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# Given $\bar{x}$ is the mean and $\sigma^2$ is the variance of n observations $x_1,x_2,x_3......x_n$ prove that the mean and variance of the observations $ax_1,ax_2,ax_3.......ax_n$ are $a \bar{x}$ and $a^2\sigma^2$, respectively , $(a\neq 0)$

Step 1:
Mean of $ax_1,ax_2,......ax_n =\large\frac{ax_1+ax_2+.........+ax_n}{n}$
=> $a \bigg(\large\frac{x_1+x_2+........+x_n}{n} \bigg)$
We know that $\large\frac{x_1+x_2+........+x_n}{n}$$=\bar {x} \therefore a \bar {x} Step 2: Variance of ax_1,ax_2..........ax_n \qquad = \large\frac{\sum (ax_i - a \bar {x} )^2}{n} \qquad= \large\frac{(ax_1-a \bar {x})^2 +(ax_2 -a \bar{x})^2+.....+(ax_n-a \bar {x} )^2}{n} \qquad=\large\frac{a^2 [(x_1-\bar{x})^2+(x_2-\bar{x})^2+..........(x_n-\bar {x})^2]}{n} We know that \large\frac{\sum (x_i-\bar {x})^2}{n}$$=\sigma^2$
$\therefore \large\frac{a^2 \in (x_i-\bar{x})^2}{n}$
$\qquad= a^2 \sigma^2$
Hence the mean and variance is proved.