Question

The mean of five observations is $4$ and their variance is $\mathrm{5.2.}$ If three of these observations are $1,2$ and $6,$ then the other two 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 the other two numbers be $a$ and $b.$
Given mean, $\overline{x}=4$
$\Rightarrow \frac{1+2+6+a+b}{5}=4$
$\Rightarrow a+b=11\dots \left(i\right)$

Variance, ${\sigma }^2=5.2$
$\Rightarrow \frac{\sum _{i=1}^N(x_i-\overline{x}{)}^2}{N}=5.2$
$\Rightarrow \frac{(1-4{)}^2+(2-4{)}^2+(6-4{)}^2+(a-4{)}^2+(b-4{)}^2}{5}=5.2$
$\Rightarrow (a-4{)}^2+(b-4{)}^2=9\Rightarrow a^2+b^2+16+16-8(a+b)=9$
$\Rightarrow a^2+b^2=65\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we get
$a=4,b=7$