Question

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

• A. $3$ and $5$
• B. $2$ and $6$
• C. $4$ and $4$
• D. $8$ and $10$
• E. $1$ and $7$

Answer 1

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

Let the observations be $2,4,6,x,y$.
Now, $\overline{x}=\frac{\sum x_i}{N}$
$\Rightarrow 4=\frac{2+4+6+x+y}{5}$
$\Rightarrow x+y=8$ ...(i)
Also, ${\sigma }^2=\frac{\sum x_i^2}{N}-{\left(\overline{x}\right)}^2$
$\Rightarrow {\left(5.2\right)}^2=\frac{4+16+36+x^2+y^2}{5}-{\left(4\right)}^2$
$\Rightarrow x^2+y^2=50$ ....(ii)
From equations (i) and (ii), we get
$x^2+{(8-x)}^2=50$
$\Rightarrow 2x^2-16x+64=50$
$\Rightarrow 2x^2-16x+14=0$
$\Rightarrow x^2-8x+7=0$
$\Rightarrow (x-1)(x-7)=0$
$\Rightarrow x=1,7$
$\Rightarrow y=7,1$
Hence, remaining two observations are $1$ and $7$.