If the mean of the numbers a, b, 8, 5, 10 is | vedclass

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(D) Given the mean $\bar{x} = 6$ for the $5$ numbers $a, b, 8, 5, 10$:$\frac{a + b + 8 + 5 + 10}{5} = 6$$a + b + 23 = 30 \Rightarrow a + b = 7$  $(1)$

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$a = 3, b = 4$

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(D) Given the mean $\bar{x} = 6$ for the $5$ numbers $a, b, 8, 5, 10$:$\frac{a + b + 8 + 5 + 10}{5} = 6$$a + b + 23 = 30 \Rightarrow a + b = 7$  $(1)$Given the variance $\sigma^2 = 6.80$:$\frac{\sum x_i^2}{n} - (\bar{x})^2 = \sigma^2$$\frac{a^2 + b^2 + 8^2 + 5^2 + 10^2}{5} - 6^2 = 6.8$$\frac{a^2 + b^2 + 64 + 25 + 100}{5} - 36 = 6.8$$\frac{a^2 + b^2 + 189}{5} = 42.8$$a^2 + b^2 + 189 = 214$$a^2 + b^2 = 25$  $(2)$From $(1)$,$b = 7 - a$. Substituting into $(2)$:$a^2 + (7 - a)^2 = 25$$a^2 + 49 - 14a + a^2 = 25$$2a^2 - 14a + 24 = 0$$a^2 - 7a + 12 = 0$$(a - 3)(a - 4) = 0$So,$a = 3$ or $a = 4$. If $a = 3$,$b = 4$; if $a = 4$,$b = 3$. Thus,$(a, b) = (3, 4)$ is a possible solution.

If the mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.80$,then which of the following is a possible value for $a$ and $b$?

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