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

Short Answer

Expert verified
Option (2) is the correct answer: Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.

Step by step solution

01

Calculate the mean (μ) of the original list

The mean of the original list is given by \ \[ \mu = \frac{x_1 + x_2 + \ldots + x_n}{n} \]
02

Calculate the variance (σ^2) of the original list

The variance of the original list is given by \ \[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n}(x_i - \mu)^2 \]
03

Calculate the mean (\hat{\mu}) of the new list

The mean of the new list \ \[ y_1, y_2, y_3, \ldots, y_{n-1}, y_n \] can be calculated as follows: \ \[ \hat{\mu} = \frac{y_1 + y_2 + \ldots + y_n}{n} \] Given \ \[ y_1 = 0, y_2 = x_2, \ldots, y_{n-1} = x_{n-1}, y_n = x_1 + x_n \] So, \ \[ \hat{\mu} = \frac{0 + x_2 + \ldots + x_{n-1} + x_1 + x_n}{n} \] Since \ \[ x_1 + x_n = x_1 + x_n \] Thus the mean of the new list \ \[ \hat{\mu} = \mu \]
04

Calculate the variance (\hat{\sigma}^2) of the new list

The variance of the new list can be calculated as follows: \ \[ \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (y_i - \hat{\mu})^2 \] Given \ \[ y_1 = 0, y_2 = x_2, \ldots, y_{n-1} = x_{n-1}, y_n = x_1 + x_n \] and \ \[ \hat{\mu} = \mu \]we have: \ \[ y_1 - \hat{\mu} = 0 - \mu = -\mu \] and the remaining terms are similar to the original terms. However, the term \ \[ (y_n - \hat{\mu}) = (x_1 + x_n - \mu) \] adds more deviation. Therefore, \ \[ \hat{\sigma} > \sigma \]
05

Choose the Correct Statement

Statement-1: \ \[ \sigma < \ hat{\sigma}\] and\[\ mu = \hat{\mu}\]. Is correct based on our calculation. Statement-2: \ Variance is independent of the choice of origin while the mean is dependent on the choice of origin, for the statement \

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

mean calculation
The mean, often denoted by \(\frac{\text{sum of values}}{\text{number of values}}\), represents the central value of a dataset. It's the average of all numbers. For any list of numbers, we calculate the mean by adding up all the numbers and then dividing by the count of numbers in the list. For example, if the list is \[x_1, x_2, \ldots, x_n \], the mean is found using \[ \mu = \frac{x_1 + x_2 + \cdots + x_n}{n} \]. This equation applies to the original list of numbers. By substituting the values into this formula, students can calculate the average. Understanding how to compute the mean helps in grasping how data points are centered around a value.
standard deviation
Standard deviation is a measure of how spread out the numbers in a dataset are. It tells us how much the values deviate from the mean. To calculate it, we first find the variance, which is the average of the squared differences from the mean. For the original list of numbers \[{x_1, x_2, \ldots, x_n}\], the formula for variance (\(\baan{\b{\refersentry}{test}^2}\)) is \[ \frac{1}{n} \sum_{i=1}^{n}(x_i - \mu)^2 \]. The square root of the variance then gives the standard deviation, so \( \sigma = \sqrt{\baan{\b{\refersentry}{test}^2}}\). This value indicates the dispersion of data points. If the standard deviation is high, it means the values are spread out a lot from the mean. If it's low, the values are closer to the mean. Understanding standard deviation is crucial for determining the consistency of data.
variance
Variance is a fundamental concept in statistics that measures the degree of spread in a set of numbers. It is calculated by taking the mean of squared differences from the mean of the dataset. Variance helps us understand the variability of data points. For our original dataset \[x_1, x_2, \ldots, x_n \], the formula for variance, denoted as \(\baan{\b{\refersentry}{test}^2} \), is \(\frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2 \). Once we know the variance, we can calculate the standard deviation by taking its square root, i.e., \(\baan{\b{\refersentry}{test}} = \sqrt{\baan{\b{\refersentry}{test}^2}}\). Variance is less intuitive but still critical. It quantifies the degree to which each number in the set differs from the mean, providing a powerful way to understand data dispersion. It's important to note that variance is independent of the choice of origin, meaning it does not change if we shift the entire dataset by the same amount.

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