Question

The mean and variance of $7$ observations are $8\text{and}16$, respectively. If five of the observations are $2,4,10,12,14$. Find the remaining two observations.

Answer 1

Let the other two observations are $x\text{and}y$

$\therefore$ Our observations are $2,4,10,12,14,x,y$

Finding a relation between $x\text{and}y$ using mean

Given Mean $=8$

i.e. $\frac{\text{Sum of observations}}{\text{Number of observations}}=8$

$\Rightarrow \frac{2+4+10+12+14+x+y}{7}=8$

$\Rightarrow 42+x+y=7\times 8$

$\Rightarrow x+y=56-42$

$\Rightarrow x+y=14$ ....(1)

Finding another relation between $x\text{and}y$ using variance

$\therefore$ Variance $=16$

$\Rightarrow \frac{1}{N}\sum (x_i-\overline{x}{)}^2=16$

$\Rightarrow \frac{1}{7}\left[\right(2-8{)}^2+(4-8{)}^2+(10-8{)}^2+(12-8{)}^2+(14-8{)}^2+(x-8{)}^2+(y-8{)}^2]=16$

$\Rightarrow \frac{1}{7}\left[\right(-6{)}^2+(-4{)}^2+(2{)}^2+(4{)}^2+(6{)}^2+(x-8{)}^2+(y-8{)}^2]=16$

$\Rightarrow \frac{1}{7}[36+16+4+16+36+x^2+(8{)}^2-2\left(8\right)x+y^2+(8{)}^2-2(8\left)y\right]=16$

$\Rightarrow [108+x^2+64-16x+y^2+64-16y]=16\times 7$

$\Rightarrow [236+x^2+y^2-16y-16x]=112$

$\Rightarrow [236+x^2+y^2-16(x+y\left)\right]=112$

$\Rightarrow [236+x^2+y^2-16(14\left)\right]=112$ (From (1))

$\Rightarrow 236+x^2+y^2-224=112$

$\Rightarrow x^2+y^2=112-236+224$

$\Rightarrow x^2+y^2=100$ ...(2)

From (1)

$\Rightarrow x+y=14$

Squaring both sides

$\Rightarrow (x+y{)}^2={14}^2$

$\Rightarrow x^2+y^2+2xy=196$

$\Rightarrow 100+2xy=196$ (from (2))

$\Rightarrow 2xy=196-100$

$\Rightarrow 2xy=96$

$\Rightarrow xy=48$

$\Rightarrow x=\frac{48}{y}$ ...(3)

Finding values of $x\text{and}y$

Substituing (3) in (1), we get :

$\Rightarrow \frac{48}{y}+y=14$

$\Rightarrow 48+y^2=14y$

$\Rightarrow y^2-14y+48=0$

$\Rightarrow y^2-6y-8y+48=0$

$\Rightarrow y(y-6)-8(y-6)=0$

$\Rightarrow (y-6)(y-8)=0$

So, $y=6\text{or}y=8$

Now, substituting $y=6$ in equation (1), we get
$x=8$

Now, substituting $y=8$ in equation (1), we get
$x=6$

So, $x=6\text{or}x=8$