Problem 36
Question
Let \(x_{1}, x_{2}, \ldots, x_{n}\) be \(n\) observations, and let \(\bar{x}\) be their arithmetic mean and \(\sigma^{2}\) be the variance Statement-1: Variance of \(2 x_{1}, 2 x_{2}, \ldots, 2 x_{n}\) is \(4 \sigma^{2}\). Statement-2: Arithmetic mean \(2 x_{1}, 2 x_{2}, \ldots, 2 x_{n}\) is \(4 \bar{x}\). (A) Statement- 1 is false, Statement- 2 is true. (B) Statement-1 is true, statement- 2 is true; statement- 2 is a correct explanation for Statement-1. (C) Statement-1 is true, statement-2 is true; statement- 2 is not a correct explanation for Statement-1. (D) Statement- 1 is true, statement- 2 is false.
Step-by-Step Solution
Verified Answer
(D) Statement-1 is true, Statement-2 is false.
1Step 1: Understand the Definitions
The arithmetic mean \(\bar{x}\) is calculated as \(\bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n}\). The variance \(\sigma^2\) is calculated as \(\sigma^2 = \frac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \cdots + (x_n - \bar{x})^2}{n}\).
2Step 2: Calculate the New Arithmetic Mean
If the observations are \(2x_1, 2x_2, \ldots, 2x_n\), the new arithmetic mean is \(\frac{(2x_1) + (2x_2) + \cdots + (2x_n)}{n} = 2\bar{x}\). Statement-2 claims this is \(4\bar{x}\), which is incorrect. Therefore, Statement-2 is false.
3Step 3: Calculate the New Variance
For the new observations \(2x_1, 2x_2, \ldots, 2x_n\), the variance is \(\sigma'^2 = \frac{(2x_1 - 2\bar{x})^2 + (2x_2 - 2\bar{x})^2 + \cdots + (2x_n - 2\bar{x})^2}{n}\). Simplifying: \(\sigma'^2 = 4\frac{(x_1 - \bar{x})^2 + \cdots + (x_n - \bar{x})^2}{n} = 4\sigma^2\). Statement-1 is true.
4Step 4: Evaluate Statements
Based on Step 2 and Step 3, Statement-1 is true and Statement-2 is false. Therefore, the correct choice is (D): Statement-1 is true, Statement-2 is false.
Key Concepts
Arithmetic MeanVariance CalculationFactor Effect on Variance
Arithmetic Mean
The arithmetic mean is one of the most commonly used measures of central tendency. To find the arithmetic mean of a set of numbers, you simply add up all the numbers in the set and then divide by the number of values. This gives you an average value that represents the center of the data.
Mathematically, the arithmetic mean \( \bar{x} \) of a set of observations \( x_1, x_2, \ldots, x_n \) is calculated as:
\[ \bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} \]
In the original exercise, when all observations are multiplied by 2, the arithmetic mean also gets multiplied by the same factor. Thus, the new arithmetic mean becomes \( 2\bar{x} \). It's important to note that multiplying each data point by a constant factor will multiply the arithmetic mean by the same factor. However, Statement-2 in our exercise mistakenly states it would be \( 4\bar{x} \), which is incorrect.
Mathematically, the arithmetic mean \( \bar{x} \) of a set of observations \( x_1, x_2, \ldots, x_n \) is calculated as:
\[ \bar{x} = \frac{x_1 + x_2 + \cdots + x_n}{n} \]
In the original exercise, when all observations are multiplied by 2, the arithmetic mean also gets multiplied by the same factor. Thus, the new arithmetic mean becomes \( 2\bar{x} \). It's important to note that multiplying each data point by a constant factor will multiply the arithmetic mean by the same factor. However, Statement-2 in our exercise mistakenly states it would be \( 4\bar{x} \), which is incorrect.
Variance Calculation
Variance is a measure of how data points vary from the mean. It assesses the spread of the data set by squaring the differences from the mean and averaging them. A higher variance means that the data points are spread out over a larger range of values.
The formula for variance \( \sigma^2 \) of a set of observations \( x_1, x_2, \ldots, x_n \) with mean \( \bar{x} \) is:
\[ \sigma^2 = \frac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \cdots + (x_n - \bar{x})^2}{n} \]
When each observation is multiplied by a constant, the variance is affected quadratically. That is, if each observation is multiplied by 2, the new variance becomes \( 4\sigma^2 \), as verified in Statement-1 of the exercise. This demonstrates how variance dramatically scales with factors of the observations, being sensitive to modifications in the data.
The formula for variance \( \sigma^2 \) of a set of observations \( x_1, x_2, \ldots, x_n \) with mean \( \bar{x} \) is:
\[ \sigma^2 = \frac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \cdots + (x_n - \bar{x})^2}{n} \]
When each observation is multiplied by a constant, the variance is affected quadratically. That is, if each observation is multiplied by 2, the new variance becomes \( 4\sigma^2 \), as verified in Statement-1 of the exercise. This demonstrates how variance dramatically scales with factors of the observations, being sensitive to modifications in the data.
Factor Effect on Variance
Understanding the effect of multiplying a factor to data points is crucial when dealing with variance. If each observation in the data set is scaled by a multiplication factor, the arithmetic mean is directly multiplied by that factor, as discussed earlier.
However, for variance, we must consider the square of that factor. If the scaling factor is \( k \), the variance of the new data becomes \( k^2 \times \sigma^2 \). This quadratic relationship is key because it illustrates that variance not only considers the distribution in linear terms (like the mean does) but also amplifies the spread depending on the square of changes applied.
This means that even small changes, when magnified through a square, can result in significant differences in variance, providing deeper insight into data variability and stability. In the context of the original exercise, multiplying each observation by 2 affected the variance by a factor of 4 \( (2^2) \), affirming Statement-1's correctness.
However, for variance, we must consider the square of that factor. If the scaling factor is \( k \), the variance of the new data becomes \( k^2 \times \sigma^2 \). This quadratic relationship is key because it illustrates that variance not only considers the distribution in linear terms (like the mean does) but also amplifies the spread depending on the square of changes applied.
This means that even small changes, when magnified through a square, can result in significant differences in variance, providing deeper insight into data variability and stability. In the context of the original exercise, multiplying each observation by 2 affected the variance by a factor of 4 \( (2^2) \), affirming Statement-1's correctness.
Other exercises in this chapter
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