Question

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

Answer 1

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

Given that,
$\frac{x_1+x_2}{2}=M_1\Rightarrow x_1+x_2=2M_1$

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

Mean of $a{x}_1,a{x}_2,\frac{x_3}{a}and\frac{x_4}{a}is$

$=\frac{a{x}_1+a{x}_2+\frac{x_3}{a}+\frac{x_4}{a}}{4}=\frac{a}{4}(x_1+x_2)+\frac{1}{4a}(x_3+x_4)=\frac{a}{4}.2M_1+\frac{1}{4a}(4M_2-2M_1)=\frac{a{M}_1}{2}+\frac{M_2}{a}-\frac{M_1}{2a}=\frac{1}{2a}\left[\right(a^2-1)M_1+2M_2]$.