Question

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

• A. $1$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$

$\mu =4$
${\sigma }^2=5.2$
Let the unknown observations be $x$ and $y$.
$\therefore \frac{3+4+4+x+y}{5}=4\Rightarrow x+y=9$
and $\frac{3^2+4^2+4^2+x^2+y^2}{5}-(4{)}^2=5.2$
$\Rightarrow 41+(x+y{)}^2-2xy-80=26$
$\Rightarrow 41+81-2xy-80=26$
$\Rightarrow xy=8$

$|x-y|=\sqrt{(x+y{)}^2-4xy}=\sqrt{81-32}=7$