Problem 53 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(x_{1}, x_{2} \ldots x_{n}\) be \(n\) observations, and let \(\bar{x}\) be their arithmetic mean and \(\sigma^{2}\) be their variance. Statement-1: Variance of \(2 x_{1}, 2 x_{2} \ldots 2 x_{n}\) is \(4 \sigma^{2}\). Statement-2: Arithmetic mean of \(2 x_{1}, 2 x_{2} \ldots 2 x_{n}\) 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

Answer

(D) Statement-1 is true, statement- 2 is false

Step 1: Understanding Variance

Variance is a measure of how much the values of a data set deviate from the mean. If each observation is multiplied by a constant, the variance is multiplied by the square of that constant. Thus, for the observations \(x_1, x_2, \ldots, x_n\) with variance \(\sigma^2\), the variance of \(2x_1, 2x_2, \ldots, 2x_n\) is given by: \[ \text{Variance}(2x_1, 2x_2, \ldots, 2x_n) = 2^2 \times \sigma^2 = 4\sigma^2 \] Therefore, Statement-1 is true.

Step 2: Calculating the Arithmetic Mean

The arithmetic mean of observations \(x_1, x_2, \ldots, x_n\) is \(\bar{x}\). If each observation is multiplied by a constant, the mean is also multiplied by the same constant, not its square. Therefore, the mean of \(2x_1, 2x_2, \ldots, 2x_n\) is: \[ \text{Mean}(2x_1, 2x_2, \ldots, 2x_n) = 2 \times \bar{x} \] Thus, Statement-2 is false, as it incorrectly claims the mean is \(4 \bar{x}\).

Step 3: Evaluating the Correct Explanation

Since Statement-2 is false, it cannot be a correct explanation for Statement-1, nor can it provide any justification for the truth of the statement about variance.

Key Concepts

Variance

Variance is a fundamental concept in statistics that helps us understand how data points differ from the average value in a dataset. It's like getting a sense of the data's spread. The formula to calculate variance (\(\sigma^2\)) involves taking the mean of the squared deviations from the arithmetic mean. This gives us a quantitative measure of variability, making it easier to compare different datasets.
When each observation in a dataset is multiplied by a constant, the variance is affected significantly. More precisely, if each data point is multiplied by a constant \(c\), the variance becomes \(c^2\sigma^2\). For example, if all observations are doubled (i.e., multiplied by 2), the variance increases by a factor of the square of 2, which is 4. Hence, for the given problem, the variance of \(2x_1, 2x_2, \ldots, 2x_n\) becomes \(4\sigma^2\). This explains why Statement-1 in the exercise is true.

Arithmetic Mean

The arithmetic mean, often just called the average, is a central value for a given set of numbers. It's calculated by summing all observations and dividing by the number of observations: \(\bar{x} = \frac{x_1 + x_2 + \ldots + x_n}{n}\). The arithmetic mean provides a single value that represents the entire dataset.
When all observations in a dataset are multiplied by a constant, the arithmetic mean of the new dataset is simply the original mean multiplied by that constant. So, if every observation is doubled, the new mean is \(2\bar{x}\). In the problem, multiplying each data point by 2 leads to an arithmetic mean of \(2\bar{x}\) rather than \(4\bar{x}\), making Statement-2 incorrect, as opposed to what was suggested.

Data Set Deviations

Understanding how individual data points differ from the dataset's mean is crucial in statistics. These differences are known as deviations. For any given point, the deviation from the mean is expressed as \((x_i - \bar{x})\), where \(x_i\) represents individual observations.
By squaring each deviation, we not only account for both positive and negative values, making them all positive, but also highlight larger deviations. Squared deviations are central to computing variance. However, data transformations, like scaling each data point, change these deviations proportionally. While the mean shifts by the constant of transformation, variance reacts to the square of the constant.
For clarity and better understanding, analyzing deviations extends beyond the arithmetic mean, allowing us to assess the dataset’s overall variability and spread. Thus, deviations are more than differences—they are the data’s fingerprint concerning how well it adheres to a central tendency.