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 observations are

Let the five observations be $x_1, x_2, x_3, x_4, x_5$. Given $x_1 = 1, x_2 = 2, x_3 = 6$. Let the other two observations be $a$ and $b$.

The mean $\bar{x} = \dfrac{1+2+6+a+b}{5} = 4$.

$9 + a + b = 20 \Rightarrow a + b = 11$.

The variance $\sigma^2 = \dfrac{\sum x_i^2}{5} - (\bar{x})^2 = 5.2$.

$\dfrac{1^2 + 2^2 + 6^2 + a^2 + b^2}{5} - 4^2 = 5.2$.

$\dfrac{1 + 4 + 36 + a^2 + b^2}{5} - 16 = 5.2$.

$\dfrac{41 + a^2 + b^2}{5} = 21.2$.

$41 + a^2 + b^2 = 106 \Rightarrow a^2 + b^2 = 65$.

Using $(a+b)^2 = a^2 + b^2 + 2ab$, we have $11^2 = 65 + 2ab$.

$121 = 65 + 2ab \Rightarrow 2ab = 56 \Rightarrow ab = 28$.

The quadratic equation with roots $a$ and $b$ is $t^2 - (a+b)t + ab = 0$, which is $t^2 - 11t + 28 = 0$.

$(t-4)(t-7) = 0$.

The roots are $t = 4$ and $t = 7$. Thus, the other two observations are 4 and 7.

Answer: 4, 7