The mean and variance for seven observations are 8 and 16 respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is:

We are told that there are altogether seven observations whose arithmetic mean (average) is $$8$$ and whose variance is $$16$$. Five of these observations are already known: $$2,\,4,\,10,\,12,\,14$$. We must find the product of the two unknown observations; let us denote them by $$x$$ and $$y$$.

First, we recall the formula for the mean of $$n$$ observations:

$$\text{Mean}=\dfrac{\sum_{i=1}^{n}x_i}{n}.$$

Here the mean is $$8$$ and the number of observations is $$7$$, so

$$8=\dfrac{2+4+10+12+14+x+y}{7}.$$

Multiplying both sides by $$7$$ gives

$$56=2+4+10+12+14+x+y.$$

Adding the five known numbers on the right-hand side, we have

$$2+4+10+12+14=42,$$

so the equation becomes

$$56=42+x+y.$$

Subtracting $$42$$ from both sides, we get the sum of the two unknown observations:

$$x+y=14. \quad -(1)$$

Next, we use the formula for the variance of $$n$$ observations:

$$\sigma^{2}=\dfrac{\sum_{i=1}^{n}x_i^{2}}{n}-\left(\text{Mean}\right)^{2}.$$

The variance is given as $$16$$, so

$$16=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7}-8^{2}.$$

Because $$8^{2}=64$$, we rewrite the equation as

$$16=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7}-64.$$

Adding $$64$$ to both sides gives

$$16+64=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7},$$

so

$$80=\dfrac{2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}}{7}.$$

Multiplying by $$7$$ yields the total of all seven squares:

$$560=2^{2}+4^{2}+10^{2}+12^{2}+14^{2}+x^{2}+y^{2}. \quad -(2)$$

Now we compute the sum of squares of the five known observations:

$$2^{2}=4,\quad4^{2}=16,\quad10^{2}=100,\quad12^{2}=144,\quad14^{2}=196.$$ Adding these, we get

$$4+16+100+144+196=460.$$

Substituting this result into equation (2), we find the combined squares of the unknowns:

$$560=460+x^{2}+y^{2}\;\Longrightarrow\;x^{2}+y^{2}=560-460=100. \quad -(3)$$

We now possess two key relations: the sum $$x+y=14$$ from (1) and the sum of squares $$x^{2}+y^{2}=100$$ from (3). To obtain the product $$xy$$, we use the algebraic identity

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

Substituting the known values, we have

$$14^{2}=100+2xy.$$

Since $$14^{2}=196$$, the equation becomes

$$196=100+2xy.$$

Subtracting $$100$$ from both sides gives

$$96=2xy,$$

and dividing by $$2$$ yields

$$xy=48.$$

Therefore, the product of the remaining two observations is $$48$$.

Hence, the correct answer is Option A.