The mean of five numbers is 0 and their variance is 2 . If three of...

Let the other two numbers be $x and y$.
According to the question,
Mean$= \dfrac{- 1 + 1 + 2 + x + y}{5} = 0 ⇒ x + y = - 2 \ldots \left(i\right)<br />Also,${\sigma}^{2} = 2
⇒ \dfrac{{\left(- 1 - 0\right)}^{2} + {\left(1 - 0\right)}^{2} + {\left(2 - 0\right)}^{2} + {\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2}}{5} =2$<br />$⇒ 1 + 1 + 4 + x^{2} + y^{2} = 10 ⇒ x^{2} + y^{2} = 4 \ldots \left(ii\right)$<br />$⇒ { \left(x + y\right)}^{2} - 2xy = 4$<br />$⇒ 4 - 2xy = 4 ⇒ xy = 0 \ldots \left(iii\right)
Now, ${ \left(x - y\right)}^{2} = x^{2} + y^{2} - 2xy = 4 - 0 = 4${using \left(ii\right)$and \left(iii\right)$ } \ldots \left(iv\right)$<br />$⇒ x - y = \pm2$<br />Solving \left(i\right)$ and \left(iv\right)$, we get,<br />If$x - y = 2, x = 0, y = - 2$<br />If$x - y = - 2, x = - 2, y = 0$<br />So, the other two numbers are$- 2, 0