The mean of 6 distinct observations is 6 . 5 and their variance is...

$\dfrac{a + b + 2 + 4 + 5 + 7}{6} = 6 . 5$

$⇒ a + b = 21$

$⇒ b = 21 - a$

Then, other two numbers are a and \left(21 - a\right)$</p><p>Now, variance is$10 . 25$, therefore</p><p>$10 . 25 = \dfrac{\left(4 + 16 + 25 + 49 + {a}^{2} + {\left(21 - a\right)}^{2}\right)}{6} - {\left(6 . 5\right)}^{2}$</p><p>$⇒ 6\left(10 . 25\right) + 6{\left(6 . 5\right)}^{2} = 94 + {a}^{2} + {\left(21 - a\right)}^{2}$</p><p>$⇒ {a}^{2} + {\left(21 - a\right)}^{2} = 221$</p><p>∴ a = 10 and \left(21 - a\right) = 21 - 10 = 11$

So, remaining two observations are $10, 11$.