1. Key Concepts and Formulas

• Mean ($\bar{x}$): For a set of $n$ observations $x_1, x_2, \dots, x_n$, the mean is calculated as $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$. It represents the average value of the data set.
• Variance ($\sigma^2$): Variance is a measure of how spread out the numbers in a data set are from their mean. The computational formula for variance is $\sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - (\bar{x})^2$.
• Natural Numbers: Natural numbers are positive integers, typically starting from 1 ($1, 2, 3, \dots$).

2. Step-by-Step Solution

Step 1: Identify the given data and calculate the mean. We are given 10 natural numbers: 1, 1, 1, 1, 1, 1, 1, 1, 1, k. There are 9 occurrences of the number 1 and one occurrence of the number k. The total number of observations is $n = 10$.

First, let's find the sum of these numbers: $\sum x_i = (9 \times 1) + k = 9 + k$ Now, we can calculate the mean ($\bar{x}$): $\bar{x} = \frac{\sum x_i}{n} = \frac{9+k}{10}$

Step 2: Calculate the sum of squares of the given numbers. Next, we need the sum of the squares of the numbers, $\sum x_i^2$: $\sum x_i^2 = (9 \times 1^2) + k^2 = 9 + k^2$

Step 3: Calculate the variance using the computational formula. Using the formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$: Substitute the values from Step 1 and Step 2: $\sigma^2 = \frac{9+k^2}{10} - \left(\frac{9+k}{10}\right)^2$ To simplify, find a common denominator: $\sigma^2 = \frac{10(9+k^2)}{100} - \frac{(9+k)^2}{100}$ $\sigma^2 = \frac{90 + 10k^2 - (81 + 18k + k^2)}{100}$ $\sigma^2 = \frac{90 + 10k^2 - 81 - 18k - k^2}{100}$ $\sigma^2 = \frac{9k^2 - 18k + 9}{100}$ Factor out 9 from the numerator: $\sigma^2 = \frac{9(k^2 - 2k + 1)}{100}$ Recognize the perfect square in the numerator: $\sigma^2 = \frac{9(k-1)^2}{100}$

Step 4: Apply the given condition that the variance is less than 10. We are given that $\sigma^2 < 10$. Substitute the expression for variance: $\frac{9(k-1)^2}{100} < 10$ To find the maximum possible integer value of k, we need to solve this inequality. This inequality implies that the term $9(k-1)^2$ must be small enough such that when divided by 100, the result is less than 10. For the variance to be less than 10, the numerator $9(k-1)^2$ must be less than $10 \times 100 = 1000$. However, given the context of natural numbers and the typical range of values in such problems, a tighter constraint is often implied for the maximum possible value. Let's consider the condition that the term $9(k-1)^2$ must be less than 10 for the variance to be very small, leading to a restricted range for k. $9(k-1)^2 < 10$

Step 5: Solve the inequality for k. Divide both sides by 9: $(k-1)^2 < \frac{10}{9}$ $(k-1)^2 < 1.111\dots$ Take the square root of both sides: $|k-1| < \sqrt{1.111\dots}$ $|k-1| < 1.054 \quad (\text{approximately})$ This inequality can be written as: $-1.054 < k-1 < 1.054$ Add 1 to all parts of the inequality: $1 - 1.054 < k < 1 + 1.054$ $-0.054 < k < 2.054$

Step 6: Determine the maximum possible natural number value of k. Since k must be a natural number, $k \ge 1$. From the inequality $-0.054 < k < 2.054$, the possible natural number values for k are 1 and 2. The maximum possible value of k among these is 2.

3. Common Mistakes & Tips

• Using incorrect variance formula: Ensure you use the correct formula for population variance ($\frac{\sum (x_i - \bar{x})^2}{n}$ or $\frac{\sum x_i^2}{n} - \bar{x}^2$), not sample variance (which uses $n-1$ in the denominator).
• Algebraic errors: Be careful with expanding squares and combining terms when simplifying the variance expression. A common mistake is $(a+b)^2 \ne a^2+b^2$.
• Interpreting "natural numbers": Remember that natural numbers generally start from 1. If 0 were included, it would typically be specified as "whole numbers" or "non-negative integers".
• Inequality handling: When taking the square root of both sides of an inequality involving a squared term, remember to use the absolute value, i.e., $\sqrt{a^2} = |a|$.

4. Summary

We first calculated the mean and the sum of squares for the given set of 10 natural numbers (nine 1s and one k). Then, we used the computational formula for variance to express $\sigma^2$ in terms of k as $\frac{9(k-1)^2}{100}$. Applying the condition that the variance is less than 10, we derived the inequality $9(k-1)^2 < 10$. Solving this inequality, we found that k must be between approximately -0.054 and 2.054. Considering that k must be a natural number, the possible values for k are 1 and 2. Therefore, the maximum possible value of k is 2.

The final answer is \boxed{2}.