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(B) Let the two unknown observations be $x$ and $y$.Given that the mean of $5$ observations is $4.4$,we have:$\frac{1 + 2 + 6 + x + y}{5} = 4.4$$9 + x

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$4$ and $9$

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(B) Let the two unknown observations be $x$ and $y$.Given that the mean of $5$ observations is $4.4$,we have:$\frac{1 + 2 + 6 + x + y}{5} = 4.4$$9 + x + y = 22$$x + y = 13$ ..... $(i)$Given that the variance is $8.24$,we use the formula $	ext{Variance} = \frac{\sum x_i^2}{n} - (	ext{mean})^2$:$8.24 = \frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4.4)^2$$8.24 = \frac{1 + 4 + 36 + x^2 + y^2}{5} - 19.36$$8.24 + 19.36 = \frac{41 + x^2 + y^2}{5}$$27.6 = \frac{41 + x^2 + y^2}{5}$$138 = 41 + x^2 + y^2$$x^2 + y^2 = 97$ ..... $(ii)$From $(i)$,$y = 13 - x$. Substituting into $(ii)$:$x^2 + (13 - x)^2 = 97$$x^2 + 169 - 26x + x^2 = 97$$2x^2 - 26x + 72 = 0$$x^2 - 13x + 36 = 0$$(x - 9)(x - 4) = 0$Thus,$x = 9$ or $x = 4$. If $x = 9$,then $y = 4$. If $x = 4$,then $y = 9$. The observations are $4$ and $9$.

The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three observations are $1, 2$ and $6$,the other two observations are

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