Solve this


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 :

  1. 7

  2. 3

  3. 5

  4. 9

Correct Option: 1



$\Rightarrow \bar{x}=\frac{63+a+b}{8}=10 \Rightarrow a+b=17$ .........(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) ; $\mathrm{a}=12 \& \mathrm{~b}=5$


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