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$If$ $\bar{x}$ $is \ the \ mean \ of$ $x_{1},x_{2},........,x_{n}$ , $\text{then for}$ $a\neq 0$ $\text{,the mean of }$ $ax_{1},ax_{2},........ax_{n},\frac{x_{1}}{a},\frac{x_{2}}{a},........,\frac{x_{n}}{a}$ $is$

$(A)$ $\left ( a+\frac{1}{a} \right )\bar{x}$

$(B)$ $\left ( a+\frac{1}{a} \right )\frac{\bar{x}}{2}$

$(C)$ $\left ( a+\frac{1}{a} \right )\frac{\bar{x}}{n}$

$(D)$ $\frac{\left ( a+\frac{1}{a} \right )\bar{x}}{2n}$

Answers (1)

Answer : B

$\frac{x_{1}+x_{2}...+x_{n}}{n}=\bar{x} \; \; \; \; \; \; \; \; ....(i)$

Multiply both sides by a

$\frac{ax_{1}+ax_{2}...+ax_{n}}{n}=a\bar{x} \; \; \; \; \; \; \; \; ....(i)$

$ax_{1}+ax_{2}......ax_{n}=na\bar{x}\; \; \; \; \; \; \; \; \; \; .....(ii)$

Dividing (i) by a, we get

$\frac{\frac{x_{1}}{a}+\frac{x_{2}}{a}.......+\frac{x_{n}}{a}}{n}=\frac{\bar{x}}{a}$

$\frac{x_{1}}{a}+\frac{x_{2}}{a}.......+\frac{x_{n}}{a}=\frac{n\bar{x}}{a}\; \; \; \; \; \; \; .....(iii)$

The mean of

$ax_{1}+ax_{2}.........+ax_{n}+$ $\frac{x_{1}}{a},\frac{x_{2}}{a}.......\frac{x_{n}}{a}$ $\text{is :}$

$\frac{ax_{1}+ax_{2}+ax_{n}+\frac{x_{1}}{a}+\frac{x_{2}}{a}+.......+\frac{x_{n}}{a}}{2n}$

from (ii) and (iii) we get

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

$=\frac{\bar{x}n}{2n}\left ( a+\frac{1}{a} \right )$

$=\frac{\bar{x}}{2}\left ( a+\frac{1}{a} \right )$

Therefore option (B) is correct.

Posted by

infoexpert23

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