1. Key Concepts and Formulas

To solve this problem, we'll use the fundamental definitions and formulas for mean and variance in statistics, along with essential algebraic identities.

• Mean ($\bar{x}$): The average of a set of observations. It is calculated as the sum of all observations divided by the total number of observations. $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$
• Variance ($\sigma^2$): A measure of the spread of data points around the mean. A convenient computational formula for variance is: $\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$ where $\sum x_i^2$ is the sum of the squares of all observations.
• Algebraic Identities: We will use the following identities to relate sums, products, and sums of powers of two numbers, say $a$ and $b$:
  • $(a+b)^2 = a^2+b^2+2ab$
  • $a^3+b^3 = (a+b)^3 - 3ab(a+b)$

2. Step-by-Step Solution

Step 1: Define the unknown observations and state given information. We are given 5 observations. Let the two unknown observations be $x$ and $y$. The complete set of observations is: $1, 3, 5, x, y$. The total number of observations, $n = 5$. The given mean, $\bar{x} = 5$. The given variance, $\sigma^2 = 8$. Our objective is to find the value of $x^3 + y^3$.

Step 2: Utilize the mean information to find the sum of unknown observations ($x+y$). We use the formula for the mean to establish a relationship involving the sum of all observations. $\bar{x} = \frac{1+3+5+x+y}{n}$ Substitute the given mean value ($\bar{x}=5$) and total number of observations ($n=5$): $5 = \frac{9+x+y}{5}$ To solve for $x+y$, multiply both sides by 5: $25 = 9+x+y$ Subtract 9 from both sides: $x+y = 16 \quad \text{(Equation 1)}$ This equation gives us the sum of the two unknown observations.

Step 3: Utilize the variance information to find the sum of squares of unknown observations ($x^2+y^2$). The variance formula allows us to establish another relationship involving the squares of the unknown observations. First, calculate the sum of squares of the known observations: $1^2 + 3^2 + 5^2 = 1 + 9 + 25 = 35$ Now, apply the computational formula for variance: $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ Substitute the given variance ($\sigma^2=8$), the mean ($\bar{x}=5$), and the sum of squares of all observations (which is $35 + x^2 + y^2$): $8 = \frac{35+x^2+y^2}{5} - (5)^2$ Simplify the equation: $8 = \frac{35+x^2+y^2}{5} - 25$ Add 25 to both sides of the equation: $8+25 = \frac{35+x^2+y^2}{5}$ $33 = \frac{35+x^2+y^2}{5}$ To solve for $x^2+y^2$, multiply both sides by 5: $33 \times 5 = 35+x^2+y^2$ $195 = 35+x^2+y^2$ Subtract 35 from both sides: $x^2+y^2 = 195 - 35$ $x^2+y^2 = 160 \quad \text{(Equation 2)}$ This equation gives us the sum of the squares of the two unknown observations.

Step 4: Solve for the product of unknown observations ($xy$). We have the sum ($x+y$) from Equation 1 and the sum of squares ($x^2+y^2$) from Equation 2. We can find their product ($xy$) using the algebraic identity $(x+y)^2 = x^2+y^2+2xy$. $(x+y)^2 = x^2+y^2+2xy$ Substitute the values from Equation 1 and Equation 2: $(16)^2 = 160 + 2xy$ $256 = 160 + 2xy$ Subtract 160 from both sides: $2xy = 256 - 160$ $2xy = 96$ Divide by 2: $xy = 48$ Now we have both the sum ($x+y=16$) and the product ($xy=48$) of the two unknown observations. (The individual observations can be found by solving the quadratic equation $t^2 - (x+y)t + xy = 0$, which is $t^2 - 16t + 48 = 0$. This factors as $(t-4)(t-12)=0$, so the observations are 4 and 12.)

Step 5: Calculate the sum of cubes ($x^3+y^3$). Finally, we use the algebraic identity for the sum of cubes: $a^3+b^3 = (a+b)^3 - 3ab(a+b)$. $x^3+y^3 = (x+y)^3 - 3xy(x+y)$ Substitute the calculated values of $x+y=16$ and $xy=48$: $x^3+y^3 = (16)^3 - 3(48)(16)$ Perform the calculations: First, calculate $16^3$: $16^3 = 4096$. Next, calculate $3 \times 48 \times 16$: $3 \times 48 = 144$, and $144 \times 16 = 2304$. $x^3+y^3 = 4096 - 2304$ $x^3+y^3 = 1792$

3. Common Mistakes & Tips

• Arithmetic Precision: Statistical problems often involve multiple numerical calculations. A small error in addition, multiplication, or squaring can lead to an incorrect final answer. Double-check all steps.
• Correct Formula Application: Always ensure you are using the correct formulas for mean and variance. The computational formula for variance ($\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$) is generally more efficient than the definition involving deviations from the mean.
• Algebraic Identity Recall: Familiarity with fundamental algebraic identities, especially those involving sums and products of two variables, is crucial for simplifying expressions and efficiently solving for unknowns.

4. Summary

This problem effectively tests the application of basic statistical measures (mean and variance) in conjunction with algebraic manipulation. By systematically using the mean to determine the sum of the unknown observations and the variance to find the sum of their squares, we were able to deduce their product. The final step involved applying the algebraic identity for the sum of cubes to compute the required value. The solution highlights the importance of a structured approach and accurate calculations.

5. Final Answer

The sum of cubes of the remaining two observations is $\boxed{\text{1792}}$, which corresponds to option (A).