Question

The variance of first $n$ natural numbers is

• A. $\dfrac{n(n+1)}{12}$
• B. $\dfrac{(n+1)(n+5)}{12}$
• C. $\dfrac{(n+1)(n-5)}{12}$
• D. $\dfrac{n(n-5)}{12}$

Hint:

Memorize variance of first $n$ numbers for fast solving.

Answer 1

The Correct Option is C

Understanding the Concept: Variance = Mean of squares - (Mean)$^2$

Step 1: Mean of first $n$ numbers: \[ \bar{x} = \frac{n+1}{2} \]

Step 2: Mean of squares: \[ \frac{1^2+2^2+\cdots+n^2}{n} = \frac{(n+1)(2n+1)}{6} \]

Step 3: Variance: \[ \sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 \]

Step 4: Simplify: \[ \sigma^2 = \frac{(n+1)(n-5)}{12} \] Final Answer:
Answer = $\dfrac{(n+1)(n-5)}{12}$