The mean of 5 observations is 4.4 and their variance is 8.24. If three of those observations are 1,2 and 6 ,…
The mean of 5 observations is 4.4 and their variance is 8.24. If three of those observations are 1,2 and 6 , then the other two observations are
9,4
9,5
9,2
9,13
Solution
Let other two observation be $x$ and $y$
$4.4=\frac{1+2+6+x+y}{5} \Rightarrow x+y=13...(i)$
and $8.24=\frac{1}{5}\left(1^2+2^2+6^2+x^2+y^2\right)-(4.4)^2$
$\Rightarrow x^2+y^2=97$
Now, $x^2+y^2=\frac{(x+y)^2+(x-y)^2}{2}$
$\Rightarrow 2 \times 97-13^2=(x-y)^2 \Rightarrow(x-y)=5...(ii)$
On solving (i) \& (ii) we get, $x=9$ and $y=4$