If the mean and variance of six observations 7, 10, 11, 15, $$a$$, $$b$$ are 10 and $$\frac{20}{3}$$, respectively, then the value of $$|a - b|$$ is equal to:

We are given six observations 7, 10, 11, 15, $$a$$, $$b$$ with mean 10 and variance $$\frac{20}{3}$$.

From the mean: $$\frac{7 + 10 + 11 + 15 + a + b}{6} = 10$$, so $$43 + a + b = 60$$, giving $$a + b = 17$$.

From the variance: $$\frac{\sum x_i^2}{6} - \bar{x}^2 = \frac{20}{3}$$, so $$\frac{49 + 100 + 121 + 225 + a^2 + b^2}{6} = 100 + \frac{20}{3} = \frac{320}{3}.$$ $$495 + a^2 + b^2 = 640, \quad \text{hence } a^2 + b^2 = 145.$$

Now, $$(a + b)^2 = a^2 + 2ab + b^2$$, so $$289 = 145 + 2ab$$, giving $$ab = 72$$.

Therefore, $$(a - b)^2 = (a+b)^2 - 4ab = 289 - 288 = 1$$, so $$|a - b| = 1$$.