Let \(n \geq 3\). A list of numbers \(0

The mean, or average, is a measure of central tendency. It represents the arithmetic average of a set of numbers. To find the mean of a list of numbers, sum all the numbers and divide by the count of numbers in the list. For example, if we have a list of numbers \(x_1, x_2, \ldots, x_n\), the mean \(\mu\) is given by: \[ \mu = \frac{1}{n} \sum_{i=1}^n x_i \] Here, \(\mu\) represents the mathematical notation for the mean, \(\sum\) symbolizes the summation, and \(n\) is the total number of elements in the list. This formula tells us that to calculate the mean, you sum all elements \(x_i\) from \(i=1\) to \(n\), and then divide the total by \(n\). As seen in the provided problem, the original mean \(\mu\) of the list \(x_1, x_2, \ldots, x_n\) was calculated using this same formula, resulting in \[ \mu = \frac{1}{n} \sum_{i=1}^{n} x_i \] When a new list was formed, the mean \(\hat{\mu}\) was calculated to be the same as the original mean \(\mu\). The new mean remains unchanged since the overall sum and distribution were redefined without changing the total sum. This emphasizes that the mean is sensitive to the values but not their rearrangement.

Standard deviation measures the amount of variation or dispersion from the mean. A low standard deviation means that the data points tend to be close to the mean, whereas a high standard deviation means that the data points are spread out over a wider range. To calculate the standard deviation \(\sigma\) of a list \(x_1, x_2, \ldots, x_n\), you use the formula: \[ \sigma = \sqrt{\frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2} \] Here, \(\sigma\) represents the standard deviation, \(\mu\) is the mean of the list, and \(x_i\) are the individual elements in the list. The expression \(\(x_i - \mu\)^2\) represents the squared deviation from the mean for each element, and the summation \(\sum\) calculates the total of all squared deviations, which is then averaged and square-rooted. In our given problem, the new list \(y_1, y_2, \ldots, y_n\) leads to a new standard deviation \(\hat{\sigma}\). The new list's elements are dismantled and combined in a way that results in a larger spread, particularly through the addition of \(x_1\) and \(x_n\), making \(\hat{\sigma} > \sigma\). This transformation increases the overall deviation from the mean, reflecting higher spread or variability.

Variance is another measure of dispersion, closely related to standard deviation. It is the square of the standard deviation and tells us the average degree to which each number is different from the mean. The variance \(\text{Var}(X)\) of a list \(x_1, x_2, \ldots, x_n\) can be calculated as: \[ \text{Var}(X) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2 \] Here, variance captures how far each number in the list is from the mean \(\mu\), squared, and averaged out over the number of data points \(n\). Statement-2 in the problem correctly explains that variance is independent of the choice of origin. This means that if we shift all data points by a constant amount, the variance remains unchanged because the relative distances between points remain identical. This distinguishes it from mean, which changes with the shift in the origin. Thus, while standard deviation \(\hat{\sigma}\) increases in the new list due to larger spread, the fundamental principle that variance does not depend on a choice of origin informs us the structure of spread internally has changed, not the external shift.