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: 1
Solution:
$\overline{\mathrm{X}}=10$
$\Rightarrow \bar{x}=\frac{63+a+b}{8}=10 \Rightarrow a+b=17$ .........(1)
Since, variance is independent of origin.
So, we subtract 10 from each observation.
So, $\sigma^{2}=13.5=\frac{79+(\mathrm{a}-10)^{2}+(\mathrm{b}-10)^{2}}{8}-(10-10)^{2}$
$\Rightarrow a^{2}+b^{2}-20(a+b)=-171$
$\Rightarrow a^{2}+b^{2}=169 \quad \ldots(2)$
From (i) & (ii) ; $\mathrm{a}=12 \& \mathrm{~b}=5$