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Q. The mean of five observations is $4$ and their variance is $5.2.$ If three of these observations are $2, 4$ and $6$, then the other two observations are

KEAMKEAM 2015Statistics

Solution:

Let the observations be $2,4,6, x, y$.
Now, $\bar{X}=\frac{\Sigma x_{i}}{N}$
$\Rightarrow 4=\frac{2+4+6+x+y}{5}$
$\Rightarrow x+y=8 \,...(i)$
Also, $\sigma^{2}=\frac{\Sigma x_{i}^{2}}{N}-(\bar{X})^{2}$
$\Rightarrow (5.2)^{2}=\frac{4+16+36+x^{2}+y^{2}}{5}-(4)^{2}$
$\Rightarrow x^{2}+y^{2}=50\, ....(ii)$
From Eqs. (i) and (ii), we get
$x^{2}+(8-x)^{2}=50$
$\Rightarrow 2 x^{2}-16 x+64=50$
$\Rightarrow 2 x^{2}-16 x+14=0$
$\Rightarrow x^{2}-8 x+7=0$
$\Rightarrow (x-1)(x-7)=0$
$\Rightarrow x=1,7$
$\Rightarrow y=7,1$
Hence, remaining two observations are 1 and 7 .