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 $\bar{x} = 4$
$\Rightarrow \frac{a+b+1+2+6}{5} = 4$
$\Rightarrow a + b = 11 \quad \ldots\left(i\right)$
Also, $\sigma^{2} = \frac{\sum\left(x_{i}-\bar{x}\right)^{2}}{n} = 5.2$
$\Rightarrow \frac{\left(1-4\right)^{2} + \left(2-4\right)^{2} + \left(6-4\right)^{2} + \left(a -4 \right)^{2} + \left(b-4\right)^{2}}{5} = 5.2$
$\Rightarrow \left( a -4 \right)^{2} + \left(b -4 \right)^{2} = 2 6 - 9 - 4 - 4 = 9$
$\Rightarrow a^{2} + 16 - 8a + b^{2} + 16 - 8b = 9$
$\Rightarrow a^{2} + b^{2} - 8\left(a + b\right) = - 23$
$\Rightarrow a^{2} + b^{2 }- 88 = - 23 \left(using \left(i\right)\right)$
$\Rightarrow a^{2} + b^{2} = 65$
$\Rightarrow \left(a + b\right)^{2 }- 2ab = 65$
$\Rightarrow 2ab = 121 - 65$
$\Rightarrow ab = 28 \quad\ldots\left(ii\right)$
Now, $\left(a - b\right)^{2}= \left(a + b\right)^{2} - 4ab = \left(11\right)^{2} - 4 \times 28 = 9$
$\Rightarrow a - b = \pm 3\quad\ldots\left(iii\right)$
Solving $\left(i\right)$ and $\left(iii\right)$, we get
$a = 7$, $b= 4$, if $a-b = 3$
$\Rightarrow a = 4$, $b = 7$, if $a - b = - 3$