The mean of five observations is 4 and their | vedclass

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(C) Let the five observations be $1, 2, 6, a,$ and $b$.Given that the mean $\bar{x} = 4$ and $n = 5$.$\bar{x} = \frac{1 + 2 + 6 + a + b}{5} = 4$$9 + a

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$4$ and $7$

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(C) Let the five observations be $1, 2, 6, a,$ and $b$.Given that the mean $\bar{x} = 4$ and $n = 5$.$\bar{x} = \frac{1 + 2 + 6 + a + b}{5} = 4$$9 + a + b = 20 \implies a + b = 11 \implies b = 11 - a$ ... $(1)$Given that the variance $\sigma^2 = 5.2$.$\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = 5.2$$\frac{(1-4)^2 + (2-4)^2 + (6-4)^2 + (a-4)^2 + (b-4)^2}{5} = 5.2$$(-3)^2 + (-2)^2 + (2)^2 + (a-4)^2 + (b-4)^2 = 26$$9 + 4 + 4 + (a-4)^2 + (b-4)^2 = 26$$17 + (a-4)^2 + (b-4)^2 = 26$$(a-4)^2 + (b-4)^2 = 9$ ... $(2)$Substitute $b = 11 - a$ into equation $(2)$:$(a-4)^2 + (11 - a - 4)^2 = 9$$(a-4)^2 + (7 - a)^2 = 9$$(a^2 - 8a + 16) + (49 - 14a + a^2) = 9$$2a^2 - 22a + 65 = 9$$2a^2 - 22a + 56 = 0$$a^2 - 11a + 28 = 0$$(a-4)(a-7) = 0$So,$a = 4$ or $a = 7$.If $a = 4$,then $b = 11 - 4 = 7$.If $a = 7$,then $b = 11 - 7 = 4$.Thus,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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