Missing Observation from Variance

For $$N = 10$$ observations:

$$\bar{x}_{10} = 5 \implies \sum_{i=1}^{10} x_i = 10 \times 5 = 50$$

$$\sigma_{10}^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - (\bar{x}_{10})^2 = 7 \implies \frac{1}{10}\sum_{i=1}^{10} x_i^2 - 25 = 7 \implies \sum_{i=1}^{10} x_i^2 = 320$$

For the first 8 observations ($$N = 8$$):

$$\bar{x}_{8} = 4 \implies \sum_{i=1}^{8} x_i = 8 \times 4 = 32$$

$$\sigma_{8}^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - (\bar{x}_{8})^2 = 3.5 \implies \frac{1}{8}\sum_{i=1}^{8} x_i^2 - 16 = 3.5 \implies \sum_{i=1}^{8} x_i^2 = 156$$

Calculating relations for $$x_9$$ and $$x_{10}$$:

$$\sum_{i=1}^{10} x_i - \sum_{i=1}^{8} x_i = x_9 + x_{10} \implies 50 - 32 = x_9 + x_{10} \implies x_9 + x_{10} = 18 \quad \text{--- (1)}$$

$$\sum_{i=1}^{10} x_i^2 - \sum_{i=1}^{8} x_i^2 = x_9^2 + x_{10}^2 \implies 320 - 156 = x_9^2 + x_{10}^2 \implies x_9^2 + x_{10}^2 = 164 \quad \text{--- (2)}$$

$$18^2 = 164 + 2x_9x_{10} \implies 324 = 164 + 2x_9x_{10} \implies x_9x_{10} = 80 \quad \text{--- (3)}$$

Solving the quadratic equation $$t^2 - (x_9+x_{10})t + x_9x_{10} = 0$$:

$$t^2 - 18t + 80 = 0 \implies (t-8)(t-10) = 0 \implies t = 8, 10$$

Since $$x_9 < x_{10}$$: $$x_9 = 8, \quad x_{10} = 10$$

$$3x_9 + 2x_{10} = 3(8) + 2(10) = 24 + 20 = 44$$

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 $$|35 - 30 | = 5$$