Arithmetic Mean

Minimize $$f(k)$$ (Sum of squared deviations).
To find where $$f(k) = \sum_{i=1}^n (x_i - k)^2$$ is minimized, we differentiate it with respect to $$k$$ and set it to zero:
$$\frac{df}{dk} = \sum_{i=1}^n 2(x_i - k)(-1) = 0$$
$$-2 \left( \sum_{i=1}^n x_i - \sum_{i=1}^n k \right) = 0$$
$$\sum_{i=1}^n x_i - nk = 0 \implies k = \frac{\sum_{i=1}^n x_i}{n}$$
This is the exact formula for the Arithmetic Mean ($$\bar{x}$$). Therefore, $$\lambda = \text{mean}$$.
Minimize $$g(k)$$ (Sum of absolute deviations).
The function $$g(k) = \sum_{i=1}^n |x_i - k|$$ is not smoothly differentiable everywhere due to the absolute value bars, but we can analyze its derivative where it exists.
$$\frac{dg}{dk} = \sum_{i=1}^n -\text{sign}(x_i - k)$$
For this derivative to balance out to $$0$$, the number of positive terms must equal the number of negative terms. This means that exactly half of the observations must be smaller than $$k$$, and exactly half must be larger than $$k$$.
The value that splits a data set into two equal halves is the Median. Therefore, $$\mu = \text{median}$$.

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 $$|35 - 30 | = 5$$