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 Mean of $x_1,\frac{x_2}{a},x_3+a,x_4-a$ is

• A. $M_2+\frac{(a-1)(x_1-2M_1)}{4a}$
• B. $M_2+\frac{2M_1-x_1-a{x}_2}{4a}$
• C. $\frac{M_2}{4_a}+(a-1)x_1+2M_1$
• D. $\frac{M_1}{4a}+(a-1)x_2+2M_1$

Answer 1

The correct option is B $M_2+\frac{2M_1-x_1-a{x}_2}{4a}$

Given the mean of $x_1$ and $x_2$ is $M_1$ And mean of $x_1$,$x_2$,$x_2$ and $x_4$ is $M_2$

Then mean=$\frac{x_1+x_2}{2}=M_1$

$x_1+x_2=2M_1$

And mean=$\frac{x_1+x_2+x_3+x_4}{4}=M_2$.

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

$\frac{x_1+\frac{x_2}{a}+x_3+a+x_4-a}{4}=\frac{a{x}_1+x_2+a{x}_3+a{x}_4}{4}=\frac{a{x}_1+a{x}_2+a{x}_3+a{x}_4-a{x}_2+x_2}{4a}$

$\frac{a4M_2+x_1+x_2-x_1-a{x}_2}{4a}=M_2+\frac{2M_1-x_1-a{x}_2}{4a}$