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

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Question

(A) 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$ $\Rightarrow a+b = 7$ $\Rightarrow a = 7-b$

Vedclass · Accepted Answer

$a=3, b=4$

Vedclass · Answer

(A) Given the mean $\bar{x} = 6$ for the numbers $a, b, 8, 5, 10$:$\frac{a+b+8+5+10}{5} = 6$$a+b+23 = 30$ $\Rightarrow a+b = 7$ $\Rightarrow a = 7-b$ $(i)$Given the variance $\sigma^2 = 6.8$:$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$6.8 = \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 + 36 = \frac{a^2+b^2+189}{5}$$42.8 	imes 5 = a^2+b^2+189$$214 = a^2+b^2+189 \Rightarrow a^2+b^2 = 25$ (ii)Substitute $(i)$ into (ii):$(7-b)^2 + b^2 = 25$$49 - 14b + b^2 + b^2 = 25$$2b^2 - 14b + 24 = 0$$b^2 - 7b + 12 = 0$$(b-3)(b-4) = 0$So,$b=3$ or $b=4$.If $b=3$,$a=4$. If $b=4$,$a=3$. Thus,the pair $(a, b)$ is $(3, 4)$ or $(4, 3)$.

List-$I$	List-$II$
$(P)$ The mean of the above data is	$(1) 2.5$
$(Q)$ The median of the above data is	$(2) 5$
$(R)$ The mean deviation about the mean of the above data is	$(3) 6$
$(S)$ The mean deviation about the median of the above data is	$(4) 2.7$
	$(5) 2.4$

The mean of the numbers $a, b, 8, 5, 10$ is $6$ and the variance is $6.8$. Then which of the following gives possible values of $a$ and $b$?

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