Question:
The mean and variance of 8 observations are 10 and $13.5$, respectively. If 6 of these observations are $5,7,10,12,14,15$, then the absolute difference of the remaining two observations is :
Correct Option: , 4
Solution:
Let the two remaining observations be $x$ and $y$.
$\because \bar{x}=\frac{5+7+10+12+14+15+x+y}{8}$
$\Rightarrow 10=\frac{63+x+y}{8}$
$\Rightarrow x+y=80-63$
$\Rightarrow x+y=17$ ...(i)
$\because \operatorname{var}(x)=13.5$
$=\frac{25+49+100+144+196+225+x^{2}+y^{2}}{8}-(10)^{2}$
$\Rightarrow x^{2}+y^{2}=169$.....(ii)
From (i) and (ii) we get
$(x, y)=(12,5)$ or $(5,12)$
So, $|x-y|=7$.