Question

Standard deviation of first \( n \) odd natural numbers is:

• A. \( \sqrt{n} \)
• B. \( \sqrt{\frac{(n+2)(n+1)}{3}} \)
• C. \( \sqrt{\frac{n^2 - 1}{3}} \)
• D. \( n \)
• E. \( 2n \)

Hint:

The variance of an Arithmetic Progression with common difference \( d \) is \( \sigma^2 = \frac{(n^2-1)d^2}{12} \). For odd numbers, \( d=2 \). Thus, \( \sigma^2 = \frac{(n^2-1)4}{12} = \frac{n^2-1}{3} \).

Answer 1

The Correct Option is C

Concept: The first \( n \) odd natural numbers are \( 1, 3, 5, \dots, (2n-1) \). The standard deviation \( \sigma \) is the square root of the variance \( \sigma^2 \), which is defined as \( \frac{\sum x^2}{n} - (\bar{x})^2 \).

Step 1: Calculate the mean and sum of squares.
Sum of first \( n \) odd numbers: \( \sum x = n^2 \). Mean \( \bar{x} = \frac{n^2}{n} = n \). Sum of squares of first \( n \) odd numbers: \( \sum x^2 = \frac{n(4n^2 - 1)}{3} \).

Step 2: Calculate Variance and Standard Deviation.
\[ \sigma^2 = \frac{\sum x^2}{n} - (\bar{x})^2 = \frac{4n^2 - 1}{3} - n^2 \] \[ \sigma^2 = \frac{4n^2 - 1 - 3n^2}{3} = \frac{n^2 - 1}{3} \] \[ \sigma = \sqrt{\frac{n^2 - 1}{3}} \]