Key Concepts and Formulas

For a set of $N$ observations $x_1, x_2, \ldots, x_N$:

• Mean ($\bar{x}$): The average value of the observations, a measure of central tendency. $\bar{x} = \frac{\sum_{i=1}^{N} x_i}{N} \implies \sum_{i=1}^{N} x_i = N\bar{x}$
• Variance ($\sigma^2$): A measure of the spread of data points around their mean. The computational formula is often more convenient: $\sigma^2 = \frac{\sum_{i=1}^{N} x_i^2}{N} - (\bar{x})^2 \implies \sum_{i=1}^{N} x_i^2 = N(\sigma^2 + \bar{x}^2)$
• Properties of Summation:
  • $\sum_{i=1}^{N} (a_i \pm b_i) = \sum_{i=1}^{N} a_i \pm \sum_{i=1}^{N} b_i$
  • $\sum_{i=1}^{N} c = Nc$, where $c$ is a constant.
  • $\sum_{i=1}^{N} c \cdot a_i = c \cdot \sum_{i=1}^{N} a_i$, where $c$ is a constant.

Step-by-Step Solution

We are given $N=10$ observations, with mean $\bar{x} = \frac{6}{5}$ and variance $\sigma^2 = \frac{84}{25}$. We also have two summation equations and the constraint that $\alpha, \beta$ are positive integers. Our goal is to find $\frac{\beta}{\alpha}$.

Part 1: Determining the Value of $\alpha$

We will use the first given summation: $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$.

Step 1: Expand the summation.

• What we are doing: Applying the linearity property of summation to separate the terms involving $x_i$ and $\alpha$.
• Why we are doing it: This allows us to express $\sum x_i$ separately, which can then be related to the given mean. $\sum_{i=1}^{10} x_i - \sum_{i=1}^{10} \alpha = 2$ Since $\alpha$ is a constant, $\sum_{i=1}^{10} \alpha = 10\alpha$: $\sum_{i=1}^{10} x_i - 10\alpha = 2$

Step 2: Substitute $\sum x_i$ using the mean formula.

• What we are doing: Replacing $\sum x_i$ with its equivalent expression in terms of the mean and number of observations.
• Why we are doing it: We are given the mean $\bar{x}$ and $N=10$, so this substitution introduces these known values into the equation, making it solvable for $\alpha$. Recall $\sum_{i=1}^{N} x_i = N\bar{x}$. For $N=10$ and $\bar{x}=\frac{6}{5}$: $\sum_{i=1}^{10} x_i = 10 \times \frac{6}{5} = 12$ Substitute this into the equation from Step 1: $12 - 10\alpha = 2$

Step 3: Solve for $\alpha$.

• What we are doing: Performing algebraic manipulation to isolate and find the numerical value of $\alpha$.
• Why we are doing it: This is the final step to determine $\alpha$. $10\alpha = 12 - 2$ $10\alpha = 10$ $\alpha = 1$
• Check: The problem states that $\alpha$ must be a positive integer. Our calculated value $\alpha=1$ satisfies this condition.

Part 2: Determining the Value of $\beta$

We will use the second given summation: $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$.

Step 1: Expand the squared term inside the summation.

• What we are doing: Using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$ to expand $(x_i-\beta)^2$.
• Why we are doing it: This breaks down the complex term into simpler components ($x_i^2$, $x_i$, and constants) that can be individually managed by summation properties and related to mean and variance formulas. $\sum_{i=1}^{10}\left(x_i^2 - 2x_i\beta + \beta^2\right)=40$

Step 2: Distribute the summation operator and simplify.

• What we are doing: Applying the linearity of summation and pulling out constants from the summation.
• Why we are doing it: This prepares the equation for later substitution of $\sum x_i^2$ and $\sum x_i$ with their numerical values. $\sum_{i=1}^{10} x_i^2 - \sum_{i=1}^{10} (2x_i\beta) + \sum_{i=1}^{10} \beta^2 = 40$ $\sum_{i=1}^{10} x_i^2 - 2\beta \sum_{i=1}^{10} x_i + 10\beta^2 = 40$

Step 3: Calculate $\sum x_i$ and $\sum x_i^2$ using the given mean and variance.

• What we are doing: Using the derived formulas for $\sum x_i$ and $\sum x_i^2$ from the Key Concepts section.
• Why we are doing it: The expanded equation from Step 2 contains these sums, and we need their numerical values to solve for $\beta$. We already found $\sum_{i=1}^{10} x_i = 12$ in Part 1, Step 2. Now, let's find $\sum_{i=1}^{10} x_i^2$ using the variance formula: $\sum_{i=1}^{N} x_i^2 = N(\sigma^2 + \bar{x}^2)$ Substitute $N=10$, $\sigma^2=\frac{84}{25}$, and $\bar{x}=\frac{6}{5}$: $\sum_{i=1}^{10} x_i^2 = 10\left(\frac{84}{25} + \left(\frac{6}{5}\right)^2\right)$ $\sum_{i=1}^{10} x_i^2 = 10\left(\frac{84}{25} + \frac{36}{25}\right)$ $\sum_{i=1}^{10} x_i^2 = 10\left(\frac{84+36}{25}\right)$ $\sum_{i=1}^{10} x_i^2 = 10\left(\frac{120}{25}\right) = \frac{1200}{25} = 48$

Step 4: Substitute these values back into the equation from Step 2.

• What we are doing: Replacing $\sum x_i^2$ and $\sum x_i$ with their calculated numerical values.
• Why we are doing it: This transforms the equation into a quadratic equation solely in terms of $\beta$, which we can then solve. The equation from Step 2 was: $\sum_{i=1}^{10} x_i^2 - 2\beta \sum_{i=1}^{10} x_i + 10\beta^2 = 40$ Substitute $\sum x_i^2 = 48$ and $\sum x_i = 12$: $48 - 2\beta(12) + 10\beta^2 = 40$ $48 - 24\beta + 10\beta^2 = 40$

Step 5: Solve the resulting quadratic equation for $\beta$.

• What we are doing: Rearranging the equation into standard quadratic form and solving for $\beta$.
• Why we are doing it: This is the final algebraic step to find the numerical value(s) of $\beta$. Rearrange the equation: $10\beta^2 - 24\beta + 48 - 40 = 0$ $10\beta^2 - 24\beta + 8 = 0$ Divide the entire equation by 2 to simplify: $5\beta^2 - 12\beta + 4 = 0$ We can solve this quadratic equation by factoring: $5\beta^2 - 10\beta - 2\beta + 4 = 0$ $5\beta(\beta - 2) - 2(\beta - 2) = 0$ $(5\beta - 2)(\beta - 2) = 0$ This yields two possible values for $\beta$: $5\beta - 2 = 0 \implies \beta = \frac{2}{5}$ $\beta - 2 = 0 \implies \beta = 2$
• Check: The problem states that $\beta$ must be a positive integer. Out of the two solutions, $\beta = \frac{2}{5}$ is not an integer, but $\beta = 2$ is a positive integer. Therefore, we choose $\beta = 2$.

Part 3: Calculating the Ratio $\frac{\beta}{\alpha}$

• What we are doing: Substituting the determined values of $\alpha$ and $\beta$ into the required ratio.
• Why we are doing it: This is the final step to answer the problem question. We found $\alpha = 1$ and $\beta = 2$. $\frac{\beta}{\alpha} = \frac{2}{1} = 2$

Common Mistakes & Tips

• Summation Properties: Be meticulous when expanding summations. Remember that $\sum_{i=1}^{N} c = Nc$ and $\sum_{i=1}^{N} c \cdot x_i = c \cdot \sum_{i=1}^{N} x_i$.
• Variance Formulas: Always have the computational form of variance ($\sigma^2 = \frac{\sum x_i^2}{N} - \bar{x}^2$) readily available, as it's frequently used in such problems.
• Algebraic Precision: Handle fractions and quadratic equations carefully to avoid calculation errors.
• Constraints: Do not forget to apply any given constraints (e.g., $\alpha, \beta$ being positive integers). These are crucial for selecting the correct solution from multiple possibilities.

Summary

This problem effectively tests the fundamental definitions and computational formulas for mean and variance, along with algebraic manipulation of summations. We first determined $\alpha$ by expanding the first summation and substituting the given mean. Then, we found $\beta$ by expanding the second summation, calculating $\sum x_i^2$ using the variance formula, and solving the resulting quadratic equation. Finally, we used the constraint that $\alpha$ and $\beta$ are positive integers to select the correct values, leading to $\alpha=1$ and $\beta=2$. The ratio $\frac{\beta}{\alpha}$ was then calculated.

The final answer is $\boxed{2}$, which corresponds to option (A).