The mean of five observations is 4 and their | vedclass

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(C) Let the two unknown observations be $x$ and $y$.Given the mean of $5$ observations is $4$,we have:$\frac{1 + 2 + 6 + x + y}{5} = 4$$9 + x + y = 20

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$4$ and $7$

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(C) Let the two unknown observations be $x$ and $y$.Given the mean of $5$ observations is $4$,we have:$\frac{1 + 2 + 6 + x + y}{5} = 4$$9 + x + y = 20$$x + y = 11$ .....$(i)$Given the variance is $5.2$,we use the formula $	ext{Variance} = \frac{\sum x_i^2}{n} - (	ext{mean})^2$:$5.2 = \frac{1^2 + 2^2 + 6^2 + x^2 + y^2}{5} - (4)^2$$5.2 = \frac{1 + 4 + 36 + x^2 + y^2}{5} - 16$$21.2 = \frac{41 + x^2 + y^2}{5}$$106 = 41 + x^2 + y^2$$x^2 + y^2 = 65$ .....$(ii)$From $(i)$,$y = 11 - x$. Substituting into $(ii)$:$x^2 + (11 - x)^2 = 65$$x^2 + 121 - 22x + x^2 = 65$$2x^2 - 22x + 56 = 0$$x^2 - 11x + 28 = 0$$(x - 4)(x - 7) = 0$Thus,$x = 4$ or $x = 7$. If $x = 4$,then $y = 7$. If $x = 7$,then $y = 4$.Therefore,the other two observations are $4$ and $7$.

The mean of five observations is $4$ and their variance is $5.2$. If three of these observations are $1, 2$ and $6$,then the other two are

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