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

Answer 1

The correct option is C

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

Explanation for the correct option:

Find the mean for given values:

We know, $Mean=\frac{sumofobservations}{totalcountofobservations}$

Given that, $\frac{x_1+x_2}{2}=M_{1}and\frac{x_1+x_2+x_3+x_4}{4}=M_2$

$$\begin{gathered} x_1+x_2=2M_{1}and{x}_1+x_2+x_3+x_4=4M_2 \\ \Rightarrow x_3+x_4=4M_2-2M_1 \end{gathered}$$

Mean of given data is:

$$\begin{gathered} mean=\frac{a{x}_1+a{x}_2+\frac{x_3}{a}+\frac{x_4}{a}}{4} \\ \Rightarrow mean=\frac{\left(a\left(x_1+x_2\right)+{\displaystyle \frac{1}{a}}\left(x_3+x_4\right)\right)}{4} \\ \Rightarrow mean=\frac{a\left(2M_1\right)+{\displaystyle \frac{1}{a}}\left(4M_2-2M_1\right)}{4} \\ \Rightarrow mean=\left(\frac{2\left(M_{1}a+{\displaystyle \frac{1}{a}}(2M_2-M_1)\right)}{4}\right) \\ \Rightarrow mean=\frac{1}{2a}[\left(a^2-1\right)M_1+2M_2] \end{gathered}$$

Hence, option (C) is correct.