The means of 5 observations is 4.4 and variance is 8.24 . If three of the five observations are 1,2 and 6 ,…
The means of 5 observations is 4.4 and variance is 8.24 . If three of the five observations are 1,2 and 6 , then the values of other two observations are
5,7
4,9
3,9
4,8
Solution
Let the two observation be $\mathrm{x}$ and $\mathrm{y}$
$\begin{aligned}
& 1+2+6+x+y=5 \times 4.4=22 \\
& \Rightarrow x+y=13
\end{aligned}$
Also $\frac{1^2+2^2+6^2+x^2+y^2}{5}-(4.4)^2=8.24$
$\begin{aligned}
& \Rightarrow \frac{41+x^2+y^2}{5}=8.24+19.36=27.60 \\
& \Rightarrow x^2+y^2=138-41 \\
& \Rightarrow x^2+(13-x)^2=97 \\
& \Rightarrow x^2-13 x+36=0 \\
& \Rightarrow x=4,9
\end{aligned}$
$\Rightarrow y=9,4$