Question

The mean and variance of $5$ observations are $4$ and $5.2$ respectively, out of them if three observation are $1,2,6$, then remaining two observations will be

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

Answer 1

The correct option is C $4,7$

$\overline{x}=4,N=5$ and $\frac{\sum {(x-\overline{x})}^2}{N}=5.2$

$\Rightarrow \sum {(x-\overline{x})}^2=\left(5.2\right)5=26\Rightarrow {(1-4)}^2+{(2-4)}^2+{(6-4)}^2+{(\alpha -4)}^2+{(\beta -4)}^2=26$
where $\alpha ,\beta$ are the other two observation

$\Rightarrow 9+4+4+{(\alpha -4)}^2+{(\beta -4)}^2=26\Rightarrow {(\alpha -4)}^2+{(\beta -4)}^2=9\Rightarrow \alpha +\beta =20-9=11$

Clearly $4,7$ only satisfy the above equation in $\alpha ,\beta$

Hence required number are $4,7$