Question

The mean of 5 obsservations 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 and y be the other 2 observation

Mean is 4.4

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

9+x+y = 22

$\therefore x+y=13$ ....(i)

Let var (X) be the variance of those observations, which is given to be 8.24

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

8.24 = $\frac{1}{5}(1^2+2^2+6^2+x^2+y^2)-(\overline{x}{)}^2$

= $\frac{1}{5}(1+4+36+x^2+y^2)-(4.4{)}^2$

= $\frac{1}{5}(41+x^2+y^2)-(4.4{)}^2$

= $\frac{1}{5}(41+x^2+y^2)-19.36$

$x^2+y^2=97$ .... (2)

$(x+y{)}^2+(x-y{)}^2=2(x^2+y^2)$

$\Rightarrow {13}^2+(x-y{)}^2=2\times 97$ [using (1) and (2)]

$(x-y{)}^2=194-169=25$

$\Rightarrow x-y\pm 5$ ...(3)

Solving eq. (1) and (30) for x - y = -5 and x = y = 13

2x = 18

$\Rightarrow x=9$

y = 4

Thus the other 2 observations are 9 and 4.