This problem involves calculating the mean and variance of a dataset to find the values of two unknown observations. We'll use the fundamental formulas for mean and variance and systematically solve for the unknowns.

1. Key Concepts and Formulas

   • Mean ($\bar{x}$): The average of all data points. For a dataset $x_1, x_2, \ldots, x_N$, the mean is given by: $\bar{x} = \frac{\sum_{i=1}^{N} x_i}{N}$ where $\sum x_i$ is the sum of all observations and $N$ is the total number of observations.

   • Variance ($\sigma^2$): A measure of the spread of data points around the mean. A computationally convenient 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, $N$ is the total number of observations, and $\bar{x}$ is the mean.

2. Step-by-Step Solution

   We are given the data: $65, 68, 58, 44, 48, 45, 60, \alpha, \beta, 60$. The total number of observations is $N = 10$. The mean of the data is $\bar{x} = 56$. The variance of the data is $\sigma^2 = 66.2$. We are also given the condition $\alpha > \beta$.

   Step 1: Use the mean formula to find the sum of $\alpha$ and $\beta$. First, we sum the known numerical observations in the dataset. This simplifies the equation for the mean. $\sum_{\text{known}} x_i = 65 + 68 + 58 + 44 + 48 + 45 + 60 + 60 = 448$ Now, apply the mean formula, substituting the sum of known values, $\alpha$, $\beta$, $N$, and the given mean $\bar{x}$: $\bar{x} = \frac{\left( \sum_{\text{known}} x_i \right) + \alpha + \beta}{N}$ $56 = \frac{448 + \alpha + \beta}{10}$ Multiply both sides by 10 to clear the denominator: $56 \times 10 = 448 + \alpha + \beta$ $560 = 448 + \alpha + \beta$ Subtract 448 from both sides to solve for $\alpha + \beta$: $\alpha + \beta = 560 - 448$ $\alpha + \beta = 112 \quad \text{(Equation 1)}$ This gives us the sum of the two unknown values.

   Step 2: Use the variance formula to find the sum of squares of $\alpha$ and $\beta$. To use the variance formula, we need the sum of squares of all observations. We begin by calculating the sum of squares of the known numerical observations: $\sum_{\text{known}} x_i^2 = 65^2 + 68^2 + 58^2 + 44^2 + 48^2 + 45^2 + 60^2 + 60^2$ $\sum_{\text{known}} x_i^2 = 4225 + 4624 + 3364 + 1936 + 2304 + 2025 + 3600 + 3600$ $\sum_{\text{known}} x_i^2 = 25678$ Now, apply the variance formula, substituting the sum of squares of known values, $\alpha^2$, $\beta^2$, $N$, the given variance $\sigma^2$, and the given mean $\bar{x}$: $\sigma^2 = \frac{\left( \sum_{\text{known}} x_i^2 \right) + \alpha^2 + \beta^2}{N} - (\bar{x})^2$ $66.2 = \frac{25678 + \alpha^2 + \beta^2}{10} - (56)^2$ Calculate the square of the mean: $(56)^2 = 3136$ Substitute this value back into the variance equation: $66.2 = \frac{25678 + \alpha^2 + \beta^2}{10} - 3136$ Add 3136 to both sides of the equation: $66.2 + 3136 = \frac{25678 + \alpha^2 + \beta^2}{10}$ $3202.2 = \frac{25678 + \alpha^2 + \beta^2}{10}$ Multiply both sides by 10: $3202.2 \times 10 = 25678 + \alpha^2 + \beta^2$ $32022 = 25678 + \alpha^2 + \beta^2$ Subtract 25678 from both sides to solve for $\alpha^2 + \beta^2$: $\alpha^2 + \beta^2 = 32022 - 25678$ $\alpha^2 + \beta^2 = 6344$

   Step 3: (Optional) Find $\alpha$ and $\beta$ to verify consistency. While the question only asks for $\alpha^2 + \beta^2$, it's good practice to find $\alpha$ and $\beta$ to ensure they are real and satisfy the condition $\alpha > \beta$. We have:

   1. $\alpha + \beta = 112$
   2. $\alpha^2 + \beta^2 = 6344$ Using the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$: $(112)^2 = 6344 + 2\alpha\beta$ $12544 = 6344 + 2\alpha\beta$ $2\alpha\beta = 12544 - 6344$ $2\alpha\beta = 6200$ $\alpha\beta = 3100$ Now we have the sum and product of $\alpha$ and $\beta$. They are the roots of the quadratic equation $t^2 - (\alpha+\beta)t + \alpha\beta = 0$: $t^2 - 112t + 3100 = 0$ Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{112 \pm \sqrt{(-112)^2 - 4(1)(3100)}}{2(1)}$ $t = \frac{112 \pm \sqrt{12544 - 12400}}{2}$ $t = \frac{112 \pm \sqrt{144}}{2}$ $t = \frac{112 \pm 12}{2}$ This yields two solutions: $t_1 = \frac{112 + 12}{2} = \frac{124}{2} = 62$ $t_2 = \frac{112 - 12}{2} = \frac{100}{2} = 50$ Given the condition $\alpha > \beta$, we have $\alpha = 62$ and $\beta = 50$. Both are real numbers, confirming the consistency of our calculations. For these values: $\alpha^2 + \beta^2 = 62^2 + 50^2 = 3844 + 2500 = 6344$.

3. Common Mistakes & Tips

   • Arithmetic Precision: Summing many numbers and their squares can easily lead to calculation errors. Double-check all additions and squaring operations.
   • Formula Application: Ensure you use the correct formulas for mean and variance. The variance formula $\sigma^2 = \frac{\sum x_i^2}{N} - (\bar{x})^2$ is generally more efficient than the definition involving deviations from the mean.
   • Algebraic Manipulation: Be careful with isolating terms and performing operations on both sides of the equation, especially when dealing with multiplication and division.
   • Systematic Approach: Break down the problem into smaller, manageable steps (e.g., sum knowns, sum squares of knowns, apply mean, apply variance). This reduces complexity and helps in error detection.

4. Summary

   By applying the definition of the mean, we first determined the sum of the two unknown values, $\alpha + \beta = 112$. Subsequently, by utilizing the variance formula and carefully calculating the sum of squares of the known data points, we established an equation that directly yielded the value of $\alpha^2 + \beta^2 = 6344$. An optional verification step confirmed that real values for $\alpha$ and $\beta$ (62 and 50 respectively) exist and satisfy the given conditions.

5. Final Answer

   The final answer is $\boxed{56}$.