The mean of the numbers a, b, 8, 5, 10 is 6 a | vedclass

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

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$a=3, b=4$

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(D) Given the mean $\bar{x} = 6$ for the numbers $a, b, 8, 5, 10$:$\frac{a+b+8+5+10}{5} = 6$$a+b+23 = 30 \Rightarrow a+b = 7$.The variance is given by $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} = 6.80$.$\frac{(a-6)^2 + (b-6)^2 + (8-6)^2 + (5-6)^2 + (10-6)^2}{5} = 6.80$$(a-6)^2 + (b-6)^2 + 4 + 1 + 16 = 34$$(a-6)^2 + (b-6)^2 = 13$.Since $a+b=7$,let $b = 7-a$. Substituting this into the variance equation:$(a-6)^2 + (7-a-6)^2 = 13$$(a-6)^2 + (1-a)^2 = 13$$a^2 - 12a + 36 + 1 - 2a + a^2 = 13$$2a^2 - 14a + 24 = 0$$a^2 - 7a + 12 = 0$$(a-3)(a-4) = 0$.So,$a=3$ or $a=4$. If $a=3$,then $b=4$. If $a=4$,then $b=3$. Thus,the pair $(3, 4)$ is a possible solution.

The mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.80$. Then which one of the following gives possible values of $a$ and $b$?

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