Question

The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two observations.

Answer 1

Let $x\mathrm{and}y$ be the other two observations.

Mean is 4.4.

$\therefore \frac{1+2+6+x+y}{5}=4.4$

$$\begin{gathered} \Rightarrow 9+x+y=22 \\ \Rightarrow x+y=13...\left(1\right) \end{gathered}$$

Let Var (X) be the variance of these observations, which is given to be 8.24.

If $\overline{\mathit{X}}$ is the mean, then we have:

$$\begin{gathered} 8.24=\frac{1}{5}\left(1^2+2^2+6^2+x^2+y^2\right)-{\left(\overline{x}\right)}^2 \\ =\frac{1}{5}\left(1+4+36+x^2+y^2\right)-{\left(4.4\right)}^2 \\ =\frac{1}{5}\left(41+x^2+y^2\right)-19.36 \\ \Rightarrow x^2+y^2=97...\left(2\right) \\ {\left(x+y\right)}^2+{\left(x-y\right)}^2=2\left(x^2+y^2\right) \\ \Rightarrow {13}^2+{\left(x-y\right)}^2=2\times 97\left[\mathrm{using}\mathrm{eq}\left(1\right)\mathrm{and}\mathrm{eq}\hspace{0.17em}\left(2\right)\right] \\ \Rightarrow {\left(x-y\right)}^2=194-169=25 \\ \Rightarrow x-y=\pm 5....\left(3\right) \\ \mathrm{Solving}\mathrm{eq}\hspace{0.17em}\left(1\right)\mathrm{and}\mathrm{eq}\left(3\right)\mathrm{for}x-y=-5\mathrm{and}x+y=13 \\ 2x=18 \\ \Rightarrow x=9 \\ \Rightarrow y=4 \end{gathered}$$

Thus, the other two observations are 9 and 4.