Standard deviation of first n odd natural numbers is

Options

A. `sqrt(n)`
B. `sqrt(((n+2)(n+1))/(3))`
C. `sqrt((n^(2)-1)/(3))`
D. `n`

Answer

To find the standard deviation of the first \( n \) odd natural numbers, we will follow these steps:

### Step 1: Identify the first \( n \) odd natural numbers
The first \( n \) odd natural numbers are:
\[ 1, 3, 5, \ldots, (2n - 1) \]

### Step 2: Calculate the mean
The mean \( \mu \) of these numbers is given by the formula:
\[ \mu = \frac{\text{Sum of all observations}}{\text{Total number of observations}} \]
The sum of the first \( n \) odd natural numbers can be calculated as follows:
\[ \text{Sum} = 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]
Thus, the mean is:
\[ \mu = \frac{n^2}{n} = n \]

### Step 3: Calculate the sum of squares of the observations
We need to calculate the sum of the squares of the first \( n \) odd natural numbers:
\[ \text{Sum of squares} = 1^2 + 3^2 + 5^2 + \ldots + (2n - 1)^2 \]
This can be expressed as:
\[ \text{Sum of squares} = \sum_{k=1}^{n} (2k - 1)^2 \]
Expanding this gives:
\[ = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4\sum_{k=1}^{n} k^2 - 4\sum_{k=1}^{n} k + n \]
Using the formulas for the sums:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \quad \text{and} \quad \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
Substituting these into the equation gives:
\[ \text{Sum of squares} = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} - 4 \cdot \frac{n(n + 1)}{2} + n \]
Simplifying this expression results in:
\[ \text{Sum of squares} = \frac{2n(n + 1)(2n + 1)}{3} - 2n(n + 1) + n \]

### Step 4: Calculate the variance
The variance \( \sigma^2 \) is given by:
\[ \sigma^2 = \frac{\text{Sum of squares}}{n} - \mu^2 \]
Substituting the values we have:
\[ \sigma^2 = \frac{\frac{2n(n + 1)(2n + 1)}{3} - 2n(n + 1) + n}{n} - n^2 \]
After simplifying, we find:
\[ \sigma^2 = \frac{2(n + 1)(2n + 1)}{3} - 2(n + 1) + 1 \]

### Step 5: Calculate the standard deviation
The standard deviation \( \sigma \) is the square root of the variance:
\[ \sigma = \sqrt{\sigma^2} \]
After further simplification, we arrive at the final expression for the standard deviation of the first \( n \) odd natural numbers.

### Final Result
The standard deviation of the first \( n \) odd natural numbers is:
\[ \sigma = \sqrt{\frac{2n^2 - 1}{3}} \]