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Q. If the mean of the numbers $a, b, 8,5,10$ is $6$ and their variance is $6.8$, then $ab$ is equal to

KEAMKEAM 2016Statistics

Solution:

Given, $\frac{a+b+8+5+10}{5}=6$
$\Rightarrow a+b+23=30$
$\Rightarrow a+b=7 \, ......(i)$
and $\frac{\Sigma\left(x_{i}-\bar{x}\right)^{2}}{n}=\sigma^{2}$
$\frac{(a-6)^{2}+(b-6)^{2}+4+1+16}{5}=\frac{34}{5}$
$\Rightarrow (a-6)^{2}+(b-6)^{2}+21=34$
$\Rightarrow a^{2}+b^{2}-84+93=34 [\because a+b=7]$
$\Rightarrow a^{2}+b^{2}=25$
From Eq. (i), $(a+b)^{2}=(7)^{2}\,....(ii)$
$\Rightarrow a^{2}+b^{2}+2 a b=49$
$\Rightarrow 25+2 a b=49 [$ from $E q .(ii)]$
$\Rightarrow 2 a b=49-25$
$\Rightarrow 2 a b=24 \Rightarrow a b=12$