Question

The mean of five numbers is $0$ and their variance is $2.$ If three of those numbers are $-1,1$ and $2,$ then the other two numbers are :

• A. $-5$ and $3$
• B. $-4$ and $2$
• C. $-3$ and $1$
• D. $-2$ and $0$
• E. $-1$ and $-1$

Answer 1

The correct option is D $-2$ and $0$

Let the other two numbers be $x$ and $y$
Thus mean $=\frac{-1+1+2+x+y}{5}$
$\Rightarrow 0=\frac{2+x+y}{5}$
$\Rightarrow x+y=2...\left(i\right)$
and variance $=\frac{\sum (x_i-\overline{x}{)}^2}{n}$
$\Rightarrow 2=\frac{\left[\right(-1-0{)}^2+(1-0{)}^2+(2-0{)}^2+(x-0{)}^2+(y-0{)}^2]}{5}$
$\Rightarrow 10=1+1+4+x^2+y^2$
$\Rightarrow x^2+y^2=4...\left(ii\right)$
On squaring Eq. (i), we get
$x^2+y^2+2xy=4$
$\Rightarrow 4+2xy=4$ .....[From Eq. (ii)]
$\Rightarrow xy=0$
Now, $x-y=\sqrt{(x+y{)}^2-4xy}$
$=\sqrt{(-2{)}^2-4\times 0=2}$ ....[ From Eq. (i)]
$\Rightarrow x-y=2$ ...(iii)
On solving Eqs. (i) and (iii), we get
$2x=0\Rightarrow x=0$
and $y=-2$