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: The variance formula.
Variance is mean of squares minus square of the mean: $\sigma^2 = \frac{\sum x_i^2}{n} - \bar x^2$.

Step 2: List the numbers.
The first $n$ even numbers are $2, 4, \ldots, 2n$.

Step 3: Find the mean.
\[ \bar x = \frac{2(1 + 2 + \cdots + n)}{n} = \frac{2 \cdot \frac{n(n+1)}{2}}{n} = n+1 \]

Step 4: Find the mean of squares.
\[ \frac{\sum x_i^2}{n} = \frac{4 \cdot \frac{n(n+1)(2n+1)}{6}}{n} = \frac{2(n+1)(2n+1)}{3} \]

Step 5: Subtract.
\[ \sigma^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2 = (n+1)\cdot\frac{4n+2 - 3n - 3}{3} \]

Step 6: Simplify.
\[ \sigma^2 = (n+1)\cdot\frac{n-1}{3} = \frac{n^2 - 1}{3} \] \[ \boxed{ \frac{n^2 - 1}{3} } \]