Question

Question 16
If $\overline{x}$ is the mean of $x_1,x_2,\dots \dots x_n$, then for $a\ne 0$, the mean of $a{x}_1,a{x}_2,\dots \dots a{x}_n,\frac{x_1}{a},\frac{x_2}{a},\dots \dots \frac{x_n}{a}$ is:

A) $(a+\frac{1}{a})\overline{x}$
B) $(a+\frac{1}{a})\frac{\overline{x}}{2}$
C) $(a+\frac{1}{a})\frac{\overline{x}}{n}$
D) $\frac{(a+\frac{1}{a})\overline{x}}{2n}$

Answer 1

The answer is B.
Given, mean of $x_1,x_2\dots \dots x_{n}is\overline{x}$.
$\therefore {\sum }_{i=1}^n{x}_i=n\overline{x}\dots \dots \left(i\right)$
Now, let the mean of $(a{x}_1+a{x}_2+\dots \dots +a{x}_n)+\frac{x_1}{a},\frac{x_2}{a},\dots \dots \frac{x_n}{a}is\overline{z}$.
Then, $\overline{z}=\frac{(a{x}_1+a{x}_2+\dots \dots +a{x}_n)+(\frac{x_1}{a}+\frac{x_2}{a}+\dots \dots \frac{x_n}{a})}{n+n}$
$=\frac{a(x_1+x_2+\dots \dots +x_n)+\frac{1}{a}(x_1+x_2+\dots \dots +x_n)}{2n}$
$=\frac{(a+\frac{1}{a})(x_1+x_2+\dots \dots +x_n)}{2n}$
$=\frac{(a+\frac{1}{a}){\sum }_{i=1}^n{x}_i}{2n}=\frac{(a+\frac{1}{a})n\overline{x}}{2n}$ [from Eq. (i)]
$=\frac{(a+\frac{1}{a})\overline{x}}{2}$