Let the mean of 6 observations 1, 2, 4, 5, $$x$$ and $$y$$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to

We are given 6 observations: 1, 2, 4, 5, $$x$$, $$y$$ with mean = 5 and variance = 10.

$$ \frac{1 + 2 + 4 + 5 + x + y}{6} = 5 $$

$$ 12 + x + y = 30 \Rightarrow x + y = 18 $$

$$ \text{Variance} = \frac{\sum x_i^2}{n} - \bar{x}^2 = 10 $$

$$ \frac{1 + 4 + 16 + 25 + x^2 + y^2}{6} = 10 + 25 = 35 $$

$$ 46 + x^2 + y^2 = 210 \Rightarrow x^2 + y^2 = 164 $$

From $$x + y = 18$$ and $$x^2 + y^2 = 164$$:

$$ (x + y)^2 = x^2 + 2xy + y^2 = 324 $$

$$ 2xy = 324 - 164 = 160 \Rightarrow xy = 80 $$

So $$x, y$$ are roots of $$t^2 - 18t + 80 = 0$$:

$$ t = \frac{18 \pm \sqrt{324 - 320}}{2} = \frac{18 \pm 2}{2} $$

$$ x = 10, \; y = 8 \text{ (or vice versa)} $$

$$ \text{M.D.} = \frac{1}{6}\sum|x_i - \bar{x}| $$

$$ = \frac{|1-5| + |2-5| + |4-5| + |5-5| + |10-5| + |8-5|}{6} $$

$$ = \frac{4 + 3 + 1 + 0 + 5 + 3}{6} = \frac{16}{6} = \frac{8}{3} $$

The mean deviation about the mean is Option C: $$\frac{8}{3}$$.