Let the mean of 6 observations 1, 2, 4, 5, x, | vedclass

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(D) Given observations are $1, 2, 4, 5, x, y$. The mean is $\overline{x} = 5$.$\frac{1+2+4+5+x+y}{6} = 5 \implies 12+x+y = 30 \implies x+y = 18$ $(i)$

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$\frac{8}{3}$

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(D) Given observations are $1, 2, 4, 5, x, y$. The mean is $\overline{x} = 5$.$\frac{1+2+4+5+x+y}{6} = 5 \implies 12+x+y = 30 \implies x+y = 18$ $(i)$.Variance $\sigma^2 = 10 = \frac{\sum x_i^2}{n} - (\overline{x})^2$.$10 = \frac{1^2+2^2+4^2+5^2+x^2+y^2}{6} - 25$.$35 = \frac{1+4+16+25+x^2+y^2}{6} \implies 210 = 46 + x^2+y^2 \implies x^2+y^2 = 164$ (ii).From $(x+y)^2 = x^2+y^2+2xy$,we have $18^2 = 164 + 2xy \implies 324 - 164 = 2xy \implies 2xy = 160 \implies xy = 80$.Solving $x+y=18$ and $xy=80$,we get $x=8, y=10$ (or vice versa).The observations are $1, 2, 4, 5, 8, 10$.Mean deviation about mean $	ext{M.D.}(\overline{x}) = \frac{\sum |x_i - 5|}{6}$.$	ext{M.D.} = \frac{|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|}{6}$.$	ext{M.D.} = \frac{4+3+1+0+3+5}{6} = \frac{16}{6} = \frac{8}{3}$.

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 $........$.

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