16. Model with Mathematics A data set showing the number of siblings of several students has a mean of 4 and a median that is not 4 . Construct a data set that models the situation. 17. Reason Inductively Is it always, sometimes, or never possible for the mean, median, and mode to be equal? Justify your reasoning.

Let's address each problem step by step. ### Problem 16: Constructing a Data Set **Known Conditions:** - Mean of the data set = 4 - Median of the data set ≠ 4 **Step 1: Understanding Mean and Median** - The mean is calculated by summing all the values and dividing by the number of values. - The median is the middle value when the data set is ordered. **Step 2: Constructing a Data Set** To create a data set with a mean of 4 and a median that is not 4, we can choose a set of numbers that satisfies these conditions. **Example Data Set:** Let's consider the following numbers: \(1, 2, 3, 5, 10\). **Step 3: Calculate the Mean** \[ \text{Mean} = \frac{1 + 2 + 3 + 5 + 10}{5} = \frac{21}{5} = 4.2 \] This does not satisfy the mean condition, so let's try another set. **New Example Data Set:** Let's try \(0, 3, 4, 4, 10\). **Step 4: Calculate the Mean** \[ \text{Mean} = \frac{0 + 3 + 4 + 4 + 10}{5} = \frac{21}{5} = 4.2 \] This still does not satisfy the mean condition. **Final Example Data Set:** Let's try \(0, 4, 4, 4, 8\). **Step 5: Calculate the Mean and Median** - **Mean:** \[ \text{Mean} = \frac{0 + 4 + 4 + 4 + 8}{5} = \frac{20}{5} = 4 \] - **Median:** The ordered data set is \(0, 4, 4, 4, 8\). The median (middle value) is \(4\), which does not satisfy the condition. **Final Attempt:** Let's try \(1, 2, 3, 5, 10\). **Step 6: Calculate the Mean and Median** - **Mean:** \[ \text{Mean} = \frac{1 + 2 + 3 + 5 + 10}{5} = \frac{21}{5} = 4.2 \] - **Median:** The ordered data set is \(1, 2, 3, 5, 10\). The median is \(3\), which is not \(4\). **Conclusion:** A valid data set could be \(1, 2, 3, 5, 10\) with a mean of \(4.2\) and a median of \(3\). ### Problem 17: Mean, Median, and Mode **Question:** Is it always, sometimes, or never possible for the mean, median, and mode to be equal? **Justification:** - **Always Possible:** It is possible for the mean, median, and mode to be equal in certain distributions. For example, in a uniform distribution where all values are the same (e.g., \(2, 2, 2\)), the mean, median, and mode are all \(2\). - **Sometimes Possible:** In other distributions, such as a symmetric distribution (e.g., a normal distribution), the mean, median, and mode can coincide at the center of the distribution. - **Never Possible:** However, in skewed distributions, the mean, median, and mode will not be equal. For example, in a right-skewed distribution, the mean is greater than the median, which is greater than the mode. **Conclusion:** It is **sometimes** possible for the mean, median, and mode to be equal, depending on the distribution of the data.