Q60 / 100 · BITSAT 2022 · Mathematics
The mean 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 are
- A$2$ and $9$
- B$3$ and $8$
- Correct answerC$4$ and $7$
- D$5$ and $6$
Solution
Let the two unknown items be $x$ and $y$. Then,
$\operatorname{Mean}=4 \Rightarrow \frac{1+2+6+x+y}{5}=4$
$\Rightarrow \quad x+y=11\dots(i)$
and variance $=5.2$
$\Rightarrow \frac{1^2+2^2+6^2+x^2+y^2}{5}-(\text { mean })^2=5.2$
$\Rightarrow \quad 41+x^2+y^2=5(5.2+16)$
$\begin{array}{ll}\Rightarrow & 41+x^2+y^2=106 \\ \Rightarrow & x^2+y^2=65\end{array}$
On solving Eqs. (i) and (ii), we get
$x=4, y=7 \text { or } x=7, y=4$
Hence, the correct answer is option: 'C'.