Question

The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12, 14,find the remaining two observations.

Answer 1

Let the remaining two observations be x and y. Then,

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

$\Rightarrow$ 42 + x + y = 56

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

Also, variance =16

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

$[\because {\sigma }^2=\frac{\sum x_i^2}{n}-(\overline{x}{)}^2]$

$\Rightarrow \frac{1}{7}(460+x^2+y^2)=80$

$\Rightarrow 460+x^2+y^2=560$

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

Now, $(x+y{)}^2+(x-y{)}^2=2(x^2+y^2)$

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

$\Rightarrow (x-y{)}^2=(2\times 100)-(14{)}^2=(200-196)=4$

$\Rightarrow x-y=\pm 2$

Now, x + y = 14, x - y = 2 $\Rightarrow$ x = 8, y = 6;

x + y = 14, x - y =-2 $\Rightarrow$ x = 6, y = 8.

Hence, the remaining two observations are 6 and 8.