Question

The means of five observations is $4$ and their variance is $5.2$. If three observations are $1,2$ and $6$, then the other two observations are

• A. $2$ and $9$
• B. $3$ and $8$
• C. $4$ and $7$
• D. $5$ and $6$

Answer 1

Answer: (C)

$4$ and $7$

Let the two missing observations be x and y

$\overline{x}=\frac{1}{n}{\sum }_{i=1}^n{x}_i$

$\Rightarrow 4=\frac{1}{5}(1+4+6+x+y)$

$\Rightarrow x+y=11$

$\Rightarrow y=11-x......\left(i\right)$

${\sigma }^2=\frac{1}{n}{\sum }_{i=1}^n{(x_i-\overline{x})}^2$

$5.2=\frac{1}{5}[{(-3)}^2+{(-2)}^2+{\left(2\right)}^2+{(x-4)}^2+{(y-4)}^2$

$\Rightarrow 26=[\left(9\right)+\left(4\right)+\left(4\right)+{(x-4)}^2+{(7-x)}^2$ [by using $\left(i\right)$]

$\Rightarrow 26=17+(-22x)+{2x}^2+65$

$\Rightarrow {2x}^2-22x+56=0$

$\Rightarrow x^2-11x+28=0$

$\Rightarrow x^2-4x-7x+28=0$

$\Rightarrow x=7,x=4$

$\Rightarrow (x-7)(x-4)=0$

$\therefore y=7,y=4$

Hence the missing observations are 4 and 7