Let x₁, x₂, ..., xₙ be n observations, and let x̄ be their arithmetic mean and σ² be the variance. S

Q: Why does the variance scale by 4 and not 2?

Because variance measures the *squared* deviation. When you double the distance of each point from the mean, the square of that distance quadruples. Think of it this way: if you stretch a rubber band to twice its length, the potential energy (which is proportional to the square of the extension) increases by a factor of 4.

Q: Is the mean always linear?

Yes, the mean is a linear operator. For any constant $k$, $E(kX) = kE(X)$. It simply shifts the center of gravity of your data set. It does not care about the 'spread' or the 'square' of the data points.

Q: Can I use a counter-example to prove Statement 2 is false?

Absolutely! Take the set $\{1, 2\}$. The mean is $\frac{1+2}{2} = 1.5$. If you scale the set by 2, you get $\{2, 4\}$. The new mean is $\frac{2+4}{2} = 3$. Since $3 \neq 4 \times 1.5$, Statement 2 is clearly false. Counter-examples are a powerful tool in your JEE arsenal!

Question

Let $x_1, x_2, \dots, x_n$ be n observations, and let $\bar{x}$ be their arithmetic mean and $\sigma^2$ be the variance. 
Statement-1 : Variance of $2x_1, 2x_2, \dots, 2x_n$ is $4\sigma^2$ 
Statement-2 : Arithmetic mean $2x_1, 2x_2, \dots, 2x_n$ is $4\bar{x}$.

Options:

(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.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast, infinite number line. You have a collection of $n$ data points, $x_1, x_2, \dots, x_n$, scattered like stars in the night sky.

At the heart of this cluster lies the arithmetic mean, $\bar{x}$, the center of gravity of your data. Surrounding this mean, we have the variance, $\sigma^2$, which quantifies the dispersion of these points.

Now, consider the transformation where we multiply every data point by a constant $k=2$. We must determine how this scaling affects our statistical parameters.

### The Linearity of the Mean

The arithmetic mean is defined as:

$$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$$

When we transform our data to $2x_1, 2x_2, \dots, 2x_n$, the new mean $\bar{x}_{new}$ is calculated as:

$$\bar{x}_{new} = \frac{1}{n} \sum_{i=1}^{n} (2x_i)$$

By the distributive property of summation, we extract the constant $2$:

$$\bar{x}_{new} = 2 	imes \left( \frac{1}{n} \sum_{i=1}^{n} x_i ight) = 2\bar{x}$$

The mean follows the scaling factor exactly. Therefore, any claim that the mean becomes $4\bar{x}$ is mathematically incorrect, as it erroneously applies quadratic logic to a linear operator.

### The Quadratic Nature of Variance

The variance is defined as the average of the squared deviations from the mean:

$$\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2$$

When we scale our observations by $2$, the new variance $\sigma^2_{new}$ becomes:

$$\sigma^2_{new} = \frac{1}{n} \sum_{i=1}^{n} (2x_i - 2\bar{x})^2$$

Factoring the $2$ out of the squared term yields:

$$\sigma^2_{new} = \frac{1}{n} \sum_{i=1}^{n} [2(x_i - \bar{x})]^2 = \frac{1}{n} \sum_{i=1}^{n} 4(x_i - \bar{x})^2$$

Extracting the constant $4$ from the summation, we arrive at:

$$\sigma^2_{new} = 4 	imes \left( \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 ight) = 4\sigma^2$$

The variance scales by the **square of the factor** because it is inherently a measure of squared deviations. Stretching the data stretches the deviations, and squaring those deviations results in a quadrupling effect.

### The Final Verdict

We have dissected the problem with surgical precision.

**Statement-1**, which claims the variance becomes $4\sigma^2$, is **true**.

**Statement-2**, which claims the mean becomes $4\bar{x}$, is **false**.

In the high-stakes arena of the JEE Advanced, clarity triumphs over confusion. Always break down transformations to their fundamental definitions to ensure your logic remains sound.

Select Answer:

Visualized Solution

Initial Observations and Parameters

Property of Mean under Scaling

Calculating the New Mean

Evaluating Statement-2

Property of Variance under Scaling

Calculating the New Variance

Evaluating Statement-1

Final Verdict

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Linearity of the Mean

The Quadratic Nature of Variance

The Final Verdict

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