Question

The mean of $6$ numbers is $6.5$ and its variance is $10.25$. If $4$ numbers are $2,4,5$ and $7$, then find the other two.

Answer 1

Step 1: Find the relation between the remaining two observations.

Since, it is given that the mean of $6$ observations is $6.5$ and four observations are $2,4,5$ and $7$.

Assume that, the remaining two observations are $x_1$ and $x_2$.

Therefore,

$$\begin{gathered} \frac{{\displaystyle \underset{i=1}{\overset{6}{\sum \left(x\right)}}}}{6}=6.5 \\ \Rightarrow \underset{i=1}{\overset{6}{\sum \left(x\right)}}=39 \\ \Rightarrow 2+4+5+7+x_1+x_2=39 \\ \Rightarrow x_1+x_2=21 \\ \Rightarrow x_1=21-x_2...1 \end{gathered}$$

Also, it is given that the variance of $6$ observations is $10.25$.

$$\begin{gathered} \frac{{\displaystyle \underset{i=1}{\overset{6}{\sum \left(x^2\right)}}}}{6}-{(6.5)}^2=10.25 \\ \Rightarrow \frac{\underset{i=1}{\overset{6}{\sum \left(x^2\right)}}}{6}=52.5 \\ \Rightarrow \underset{i=1}{\overset{6}{\sum \left(x^2\right)}}=315 \\ \Rightarrow {\left(2\right)}^2+{\left(4\right)}^2+{\left(5\right)}^2+{\left(7\right)}^2+{\left(x_1\right)}^2+{\left(x_2\right)}^2=315 \\ \Rightarrow {\left(x_1\right)}^2+{\left(x_2\right)}^2=221...2 \end{gathered}$$

Step 2: Find the remaining two observations.

From equation $1$ and equation $2$.

$$\begin{gathered} {\left(21-x_2\right)}^2+x_2^2=221 \\ \Rightarrow {\left(x_2\right)}^2+441-42x_2+x_2^2=221 \\ \Rightarrow 2x_2^2-42x_2+220=0 \\ \Rightarrow x_2=\frac{42\pm \sqrt{{42}^2-4\left(2\right)\left(220\right)}}{2\left(2\right)} \\ \Rightarrow x_2=\frac{42\pm \sqrt{1764-1760}}{4} \\ \Rightarrow x_2=\frac{42\pm \sqrt{4}}{4} \\ \Rightarrow x_2=\frac{42\pm 2}{4} \\ \Rightarrow x_2=10,11 \end{gathered}$$

For $x_2=10$,

$$\begin{gathered} x_1=21-10 \\ \Rightarrow x_1=11 \end{gathered}$$

For $x_2=11$,

$$\begin{gathered} x_1=21-11 \\ \Rightarrow x_1=10 \end{gathered}$$

Therefore, the remaining two observations are $10$ and $11$.