Let the mean of 6 observations 1, 2, 4, 5, x and y be 5 and their...

Consider the data $1, 2, 4, 5, x, y$.

So,

Mean = \dfrac{1 + 2 + 4 + 5 + x + y}{6}$</p><p>$⇒ 5 = \dfrac{x + y + 12}{6}$</p><p>$⇒ x + y = 18$</p><p>And,</p><p>Variance = \left(\dfrac{\mathop{∑}_{}{x}_{i}^{2}}{n}\right) - {\left(\dfrac{\mathop{∑}_{}{x}_{i}}{n}\right)}^{2}$

$⇒ 10 = \left(\dfrac{{1}^{2} + {2}^{2} + {4}^{2} + {5}^{2} + {x}^{2} + {y}^{2}}{6}\right) - {\left(5\right)}^{2}$

$⇒ {x}^{2} + {y}^{2} = 164$

We have $x + y = 18$ and ${x}^{2} + {y}^{2} = 164$.

Now solving above equations we get, $x = 10, y = 8$,

Now mean deviation about mean will be $= \dfrac{\mathop{∑}_{i = 1}^{6}\left|{x}_{i} - x\right|}{6}$

$= \dfrac{\left|1 - 5\right| + \left|2 - 5\right| + \left|4 - 5\right| + \left|5 - 5\right| + \left|10 - 5\right| + \left|8 - 5\right|}{6}$

$= \dfrac{16}{6}$

$= \dfrac{8}{3}$

Therefore, the required answer is $\dfrac{8}{3}$