Question

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

• A. $2$, $9$
• B. $4$, $7$
• C. $5$, $6$
• D. $2$, $10$

Answer 1

Answer: (B)

Let the other two observations be $a$, $b$. Then, the five observations are, $1$, $2$, $6$, $a$, $b$.
We have, mean $\bar{x}=4$
$\Rightarrow \frac{a+b+1+2+6}{5}=4$
$\Rightarrow a+b=11\dots \left(i\right)$
Also, ${\sigma }^2=\frac{\sum {\left(x_i-\bar{x}\right)}^2}{n}=5.2$
$\Rightarrow \frac{{\left(1-4\right)}^2+{\left(2-4\right)}^2+{\left(6-4\right)}^2+{\left(a-4\right)}^2+{\left(b-4\right)}^2}{5}=5.2$
$\Rightarrow {\left(a-4\right)}^2+{\left(b-4\right)}^2=26-9-4-4=9$
$\Rightarrow a^2+16-8a+b^2+16-8b=9$
$\Rightarrow a^2+b^2-8\left(a+b\right)=-23$
$\Rightarrow a^2+b^2-88=-23\left(using\left(i\right)\right)$
$\Rightarrow a^2+b^2=65$
$\Rightarrow {\left(a+b\right)}^2-2ab=65$
$\Rightarrow 2ab=121-65$
$\Rightarrow ab=28\dots \left(ii\right)$
Now, ${\left(a-b\right)}^2={\left(a+b\right)}^2-4ab={\left(11\right)}^2-4\times 28=9$
$\Rightarrow a-b=\pm 3\dots \left(iii\right)$
Solving $\left(i\right)$ and $\left(iii\right)$, we get
$a=7$, $b=4$, if $a-b=3$
$\Rightarrow a=4$, $b=7$, if $a-b=-3$