Question

The means of five observations is $4$ and their variance is $5.2$. If three of these observations are $1,2$ and $6$, then the other two observations are

• A. $2$ and $9$
• B. $3$ and $8$
• C. $4$ and $7$
• D. $5$ and $6$

Answer 1

The correct option is C $4$ and $7$

Let two observations be $a$ and $b.$
Mean $=\frac{1+2+6+a+b}{5}=4$
$\Rightarrow a+b=11\dots \left(i\right)$

Variance $=5.2$
$\because {\sigma }^2=\frac{\sum x_i^2}{n}-(\overline{x}{)}^2\Rightarrow {5.2}^2=\frac{1+4+36+a^2+b^2}{5}-4^2\Rightarrow 26=41+a^2+b^2-80\Rightarrow a^2+b^2=65\dots (ii)$

From $\left(i\right)$ and $\left(ii\right)$
$a^2+(11-a{)}^2=65$
$\Rightarrow a^2+121-22a+a^2=65$
$\Rightarrow 2a^2-22a+56=0$
$\Rightarrow a^2-11a+28=0$
$\Rightarrow (a-4)(a-7)=0$
$\Rightarrow a=4\text{or}7$
$\Rightarrow b=7\text{or}4$

Hence, remaining two observations are $4$ and $7.$