Question

The mean of five observations is $4$ and their variance is $\mathrm{5.2.}$ If three of these observations are $2,4$ and $6$, then the other two observations are

• A. 3 and 5
• B. 2 and 6
• C. 4 and 4
• D. 8 and 10
• E. 1 and 7

Answer 1

Answer: (E)

Let the observations be $2,4,6,x,y$.
Now, $\bar{X}=\frac{\Sigma x_i}{N}$
$\Rightarrow 4=\frac{2+4+6+x+y}{5}$
$\Rightarrow x+y=8...(i)$
Also, ${\sigma }^2=\frac{\Sigma x_i^2}{N}-(\bar{X}{)}^2$
$\Rightarrow (5.2{)}^2=\frac{4+16+36+x^2+y^2}{5}-(4{)}^2$
$\Rightarrow x^2+y^2=50....(ii)$
From Eqs. (i) and (ii), we get
$x^2+(8-x{)}^2=50$
$\Rightarrow 2x^2-16x+64=50$
$\Rightarrow 2x^2-16x+14=0$
$\Rightarrow x^2-8x+7=0$
$\Rightarrow (x-1)(x-7)=0$
$\Rightarrow x=1,7$
$\Rightarrow y=7,1$
Hence, remaining two observations are 1 and 7 .