Question

The mean of $x_1$ and $x_2$ is $M_1$ and that of $x_1,x_2,x_3.....x_4$ is $M_2$,

then the mean of $a{x}_1,a{x}_2,\frac{x_3}{a},\frac{x_4}{a}$ is?

• A. $\frac{a{M}_1+M_2}{2}$
• B. $\frac{a{M}_1+\left(\frac{M_2}{a}\right)}{2}$
• C. $\frac{1}{2a}\left[\right(a^2-1)M_1+M_2]$
• D. $\frac{1}{2a}\left[\right(a^2-1)M_1+2M_2]$

Answer 1

The correct option is D $\frac{1}{2a}\left[\right(a^2-1)M_1+2M_2]$

mean of $x_1$ and $x_2$ is $M_1$, or $x_1+x_2$ is $2M_1$

mean of $x_1,x_2,x_3.....x_4$ is $M_2$, or

$x_1+x_2+x_3+x_4$ = $4M_2$

$2M_1+x_3+x_4$ = $4M_2$

$x_3+x_4$ = $4M_2-2M_1$

Mean of $a{x}_1,a{x}_2,\frac{x_3}{a},\frac{x_4}{a}$ = $\frac{a(x_1+x_2)+\frac{1}{a}(x_3+x_4)}{4}$

= $\frac{a\left(2M_1\right)+\frac{1}{a}(4M_2-2M_1)}{4}$

= $\frac{a{M}_1+\frac{1}{a}(2M_2-M_1)}{2}$

= $\frac{(a^2-1)M_1+2M_2}{2a}$