Question

The mean and the variance of five observations are $4$ and $5.20$, respectively. If three of observations are $3,4$ and $4;$ then the absolute value of the difference of the other two observations, is

• A. $7$
• B. $5$
• C. $1$
• D. $3$

Answer 1

The correct option is A $7$

Let observations are $3,4,4,a,b.$

$\text{Mean}=\frac{11+a+b}{5}=4$
$\Rightarrow a+b=9\cdots \left(1\right)$

$\text{Variance}=\frac{\sum (x_i{)}^2}{n}-(\overline{x}{)}^2$
$\Rightarrow 5.2=\frac{9+16+16+a^2+b^2}{5}-(4{)}^2$
$\Rightarrow 21.2\times 5=41+a^2+b^2$
$\Rightarrow a^2+b^2=65$

From equation $\left(1\right),$ we have
$\Rightarrow a^2+b^2+2ab=81$
$\Rightarrow 2ab=16$
$\Rightarrow ab=8$
$\Rightarrow a=\frac{8}{b}\cdots \left(3\right)$

Substitute value of $a$ from equation $\left(3\right)$ to equation $\left(1\right)$

$\Rightarrow \frac{8}{b}+b=9$
$\Rightarrow b^2-9b+8=0$
$\therefore b=8,1$
$\therefore a=1,\frac{1}{8}$

Absolute value of difference is
$\Rightarrow 8-1=7$