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

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Question

$6.80=\frac{(6-a)^{2}+(6-b)^{2}+(6-8)^{2}+(6-5)^{2}+(6-10)^{2}}{5}$

$\Rightarrow(6-a)^{2}+(6-b)^{2}+4+1+16=34$

$(6-a)^{2}+(6-b)^{2}=34-21$

$(

Vedclass · Accepted Answer

$a=3 ,b=4$

Vedclass · Answer

$6.80=\frac{(6-a)^{2}+(6-b)^{2}+(6-8)^{2}+(6-5)^{2}+(6-10)^{2}}{5}$

$\Rightarrow(6-a)^{2}+(6-b)^{2}+4+1+16=34$

$(6-a)^{2}+(6-b)^{2}=34-21$

$(6-a)^{2}+(6-b)^{2}=13$

$(6-a)^{2}+(6-b)^{2}=9+4$

$(6-a)^{2}+(6-b)^{2}=3^{2}+2^{2}$

$(6-a)^{2}=3^{2}(6-b)^{2}=2^{2}$

$6-a=3 \quad 6-b=2$

$-a=-3 \quad-b=-4$

$a=3$

$b=4$

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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