Missing Observation from Mean

To find the true product, we locate the student's mathematical error and re-evaluate the steps correctly:

1. Verify Step 1:
The sum of all 5 observations is:
$$5 \times 6 = 30$$
Subtracting the three known observations (4, 7, and 8) isolates the sum of the remaining two variables:
$$4 + 7 + 8 + x + y = 30 \implies 19 + x + y = 30 \implies x + y = 11$$
Step 1 is completely correct.

2. Verify Step 2:
The standard formula for variance is given by:
$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$
Substituting the given parameters ($$\sigma^2 = 2$$, $$n = 5$$, and $$\bar{x} = 6$$) yields:
$$\frac{4^2 + 7^2 + 8^2 + x^2 + y^2}{5} - 6^2 = 2$$
Step 2 is completely correct.

3. Find the mistake in Step 3:
Let us calculate the sum of the squares of the known observations:
$$4^2 + 7^2 + 8^2 = 16 + 49 + 64 = 129$$
Now let us correctly isolate the sum of squares term from the variance equation in Step 2:
$$\frac{129 + x^2 + y^2}{5} - 36 = 2$$
$$\frac{129 + x^2 + y^2}{5} = 2 + 36 = 38$$
$$129 + x^2 + y^2 = 38 \times 5 = 190$$
The student incorrectly calculated $$38 \times 5 = 180$$ instead of 190. This is where the error occurred.

4. Calculate the true value of $$x^2 + y^2$$:
Using the correct total sum of squares from our fix:
$$x^2 + y^2 = 190 - 129 = 61$$

5. Compute the true product $$xy$$:
Substitute the correct value into the algebraic identity:
$$(x + y)^2 = x^2 + y^2 + 2xy$$
$$(11)^2 = 61 + 2xy$$
$$121 = 61 + 2xy$$
$$2xy = 121 - 61 = 60$$
$$xy = 30$$

Thus, the corrected product of the remaining two observations is 30.

The difference between the calculated and the true value of $$xy$$ is $$|35 - 30 | = 5$$

We have ten observations: $$2,4,6,8,10,12,14,16,18,x$$ whose mean is given to be $$11$$.

1. Determine the unknown observation $$x$$.
If the mean of the ten numbers is $$11$$, then their total sum must be
$$\text{Sum}= (\text{mean}) \times (\text{number of observations}) = 11 \times 10 = 110.$$

The sum of the first nine known observations is an arithmetic-progression sum:
$$2+4+6+\dots+18 = \frac{9}{2}\,(2+18)=\frac{9}{2}\times20 = 90.$$

Hence
$$x = 110 - 90 = 20.$$

The complete data set is now $$\{2,4,6,8,10,12,14,16,18,20\}.$$

2. Compute the variance.
For a set of $$n$$ observations $$(x_1,x_2,\dots,x_n)$$ with mean $$\bar{x}$$, the variance is
$$\sigma^{2}= \frac{1}{n}\sum_{i=1}^{n}(x_i-\bar{x})^{2}.$$

Here, $$n=10$$ and $$\bar{x}=11$$. Calculate each deviation $$x_i-\bar{x}$$ and then square it:

$$\begin{aligned} 2-11 &= -9 &\Rightarrow&\; (-9)^2 = 81\\ 4-11 &= -7 &\Rightarrow&\; (-7)^2 = 49\\ 6-11 &= -5 &\Rightarrow&\; (-5)^2 = 25\\ 8-11 &= -3 &\Rightarrow&\; (-3)^2 = 9\\ 10-11 &= -1 &\Rightarrow&\; (-1)^2 = 1\\ 12-11 &= 1 &\Rightarrow&\; 1^2 = 1\\ 14-11 &= 3 &\Rightarrow&\; 3^2 = 9\\ 16-11 &= 5 &\Rightarrow&\; 5^2 = 25\\ 18-11 &= 7 &\Rightarrow&\; 7^2 = 49\\ 20-11 &= 9 &\Rightarrow&\; 9^2 = 81\\ \end{aligned}$$

Add these squared deviations:
$$81+49+25+9+1+1+9+25+49+81 = 330.$$

Finally, divide by $$n=10$$ to get the variance:
$$\sigma^{2}= \frac{330}{10}=33.$$

Therefore, the variance of the given data is $$33$$.

Option A which is: $$33$$