Question

The variance of the first $n$ even natural numbers is:

• A. $\frac{n^2 - 1}{12}$
• B. $\frac{n^2 - 1}{3}$
• C. $\frac{n^2 + 1}{3}$
• D. $\frac{n^2 - 1}{4}$

Hint:

Since the first $n$ natural numbers have a variance of $\frac{n^2-1}{12}$, scaling them by 2 (even numbers) scales the variance by $2^2 = 4$. Thus, $4 \times \frac{n^2-1}{12} = \frac{n^2-1}{3}$.

Answer 1

The Correct Option is B

Step 1: Concept
The variance ($\sigma^2$) of a set of observations $x_i$ is given by the formula $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$, where $\bar{x}$ is the mean.

Step 2: Meaning
The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$. We compute their mean and sum of squares.

Step 3: Analysis
Mean ($\bar{x}$): \[ \bar{x} = \frac{2 + 4 + \dots + 2n}{n} = \frac{2(1 + 2 + \dots + n)}{n} = \frac{2 \cdot \frac{n(n+1)}{2}}{n} = n+1 \] Sum of squares of the terms: \[ \sum x_i^2 = 2^2 + 4^2 + \dots + (2n)^2 = 4(1^2 + 2^2 + \dots + n^2) = 4 \cdot \frac{n(n+1)(2n+1)}{6} \] Now, calculating the variance: \[ \sigma^2 = \frac{4}{n} \cdot \frac{n(n+1)(2n+1)}{6} - (n+1)^2 \] \[ \sigma^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2 \] \[ \sigma^2 = (n+1) \left[ \frac{2(2n+1) - 3(n+1)}{3} \right] \] \[ \sigma^2 = (n+1) \left[ \frac{4n + 2 - 3n - 3}{3} \right] = (n+1) \left[ \frac{n-1}{3} \right] = \frac{n^2-1}{3} \]

Step 4: Conclusion
The variance of the first $n$ even natural numbers is $\frac{n^2-1}{3}$.

Final Answer: (B)