Question:
If the mean and variance of eight numbers $3,7,9,12,13$, $20, x$ and $y$ be 10 and 25 respectively, then $x \cdot y$ is equal to___________.
Solution:
Mean $=\bar{x}=\frac{3+7+9+12+13+20+x+y}{8}=10$
$\Rightarrow x+y=16$...(i)
Variance $=\sigma^{2}=\frac{\Sigma\left(x_{i}\right)^{2}}{8}-(\bar{x})^{2}=25$
$\sigma^{2}=\frac{9+49+81+144+169+400+x^{2}+y^{2}}{8}-100=25$
$\Rightarrow x^{2}+y^{2}=148$
From eqn. (i), $(x+y)^{2}=(16)^{2}$
$\Rightarrow x^{2}+y^{2}+2 x y=256$
Using eqn. (ii), $148+2 x y=256$
$\Rightarrow \quad x y=52$