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

We have eight numbers: $$3,\,7,\,9,\,12,\,13,\,20,\,x \text{ and } y.$$

The mean (average) is stated to be $$10.$$ By definition,

$$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}.$$

There are $$8$$ observations, so

$$\dfrac{3+7+9+12+13+20+x+y}{8}=10.$$

Multiplying both sides by $$8$$ we obtain

$$3+7+9+12+13+20+x+y=80.$$

Adding the known numbers first,

$$3+7+9+12+13+20=64,$$

and substituting this into the previous equation gives

$$64+x+y=80.$$

So

$$x+y=80-64=16.$$

Next we use the information about variance. For a data set of size $$n,$$ the variance $$\sigma^{2}$$ is given by the formula

$$\sigma^{2}=\dfrac{\sum x_{i}^{2}}{n}-\mu^{2},$$

where $$\sum x_{i}^{2}$$ is the sum of the squares of the observations and $$\mu$$ is the mean.

Here $$\sigma^{2}=25,\; n=8 \text{ and } \mu=10.$$ Substituting these values, we get

$$25=\dfrac{\sum x_{i}^{2}}{8}-10^{2}.$$

Since $$10^{2}=100,$$ we have

$$25=\dfrac{\sum x_{i}^{2}}{8}-100.$$

Adding $$100$$ to both sides,

$$25+100=\dfrac{\sum x_{i}^{2}}{8},$$

so

$$\dfrac{\sum x_{i}^{2}}{8}=125.$$

Multiplying by $$8,$$

$$\sum x_{i}^{2}=125 \times 8=1000.$$

We now separate the squares of the known numbers:

$$3^{2}=9,\;7^{2}=49,\;9^{2}=81,\;12^{2}=144,\;13^{2}=169,\;20^{2}=400.$$

Adding these squares,

$$9+49+81+144+169+400=852.$$

Hence

$$x^{2}+y^{2}=1000-852=148.$$

We already found $$x+y=16.$$ To connect $$x+y$$ and $$x^{2}+y^{2},$$ we recall the algebraic identity

$$\bigl(x+y\bigr)^{2}=x^{2}+2xy+y^{2}.$$

Writing this identity with our known values,

$$16^{2}=x^{2}+2xy+y^{2}.$$

Since $$x^{2}+y^{2}=148,$$ the equation becomes

$$256=148+2xy.$$

Subtracting $$148$$ from both sides,

$$256-148=2xy,$$

so

$$2xy=108.$$

Finally, dividing by $$2,$$ we get

$$xy=\dfrac{108}{2}=54.$$

So, the answer is $$54$$.