Question

The mean and variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6,7,10,12,12 and 13, find the remaining two observations.

Answer 1

Let the remaining two observations are a and b

Given, $\overline{x}=9$ and ${\sigma }^2=9.25$

$\because \overline{x}=9$

$\Rightarrow \frac{Sumofallobservations}{Numberofobservations}=9$

$\Rightarrow Sumofallobservations=9\times 8$

$\Rightarrow 6+7+10+12+12+13+a+b=72$

$\Rightarrow 60+a+b=72$

$\Rightarrow a+b=72-60$

$\Rightarrow a+b=12$ ...(i)

Again ${\sigma }^2=\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)or\frac{\sum x_i^2}{n}-(\overline{x}{)}^2$

$\Rightarrow 9.25=\frac{36+49+100+144+144+169+a^2+b^2}{8}-(9{)}^2$

$\Rightarrow 9.25=\frac{642+a^2+b^2}{8}-81$ $\Rightarrow 9.25+81=\frac{642+a^2+b^2}{8}$

$\Rightarrow 90.25\times 8=642+a^2+b^2$ $\Rightarrow a^2+b^2=722-642$

$\Rightarrow a^2+b^2=80$ ....(ii)

Now, from Eq. (i) put b = 12 -a in Eq. (ii),

$a^2+(12-a{)}^2=80$

$\Rightarrow a^2+144+a^2-24a=80$ $\Rightarrow 2a^2-24a+144-80=0$

$\Rightarrow 2a^2-24a+64=0$ $a^2-24a+32=0$

$\Rightarrow a^2-8a-4a+32=0$ $\Rightarrow a(a-8)-4(a-8)=0$

$\Rightarrow (a-4)(a-8)=0$ $\Rightarrow a=4ora=8$

From Eq. (i) b = 8 or b = 4

Hence, observations are 4 and 8.