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 $a{x}_1,a{x}_2,,x_3/a,x_4/a$ is

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

Answer 1

Answer: (C)

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

$\Rightarrow$ $\frac{x_1+x_2}{2}=M_1$ [ Given ]

$\therefore$ $x_1+x_2=2M_1$ ----- ( 1 )

$\Rightarrow$ Now, $\frac{x_1+x_2+x_3+x_4}{4}=M_2$

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

$\Rightarrow$ $2M_1+x_3+x_4=4M_2$ [ From ( 1 ) ]

$\therefore$ $M_2=\frac{M_1}{2}+\frac{x_3+x_4}{4}$

$\Rightarrow$ Now, $a{x}_1+a{x}_2=2a\left(\frac{x_1+x_2}{2}\right)=2a{M}_1$ and

$\Rightarrow$ $\frac{x_3}{a}+\frac{x_4}{a}=\frac{4}{a}\left(\frac{x_3+x_4}{4}\right)=\frac{4}{a}(M_2-\frac{M_1}{2})$

$\therefore$ The mean of $a{x}_1,a{x}_2,\frac{x_3}{a},\frac{x_4}{a}$ is,

$\Rightarrow$ $2a{M}_1+\frac{4}{a}(M_2-\frac{M_1}{2})$

$\Rightarrow$ $2a{M}_1+\frac{4}{a}{M}_2-\frac{4}{a}\times \frac{M_1}{2}$

$\Rightarrow$ $2a{M}_1+\frac{4}{a}{M}_2-\frac{2}{a}{M}_1$

$\Rightarrow$ $2M_1(a-\frac{1}{a})+\frac{4}{a}{M}_2$

$\Rightarrow$ $\frac{1}{2a}\left[\right(a^2-1)M_1+2M_2]$