Standard deviation of first 'n' natural numbers is

Options

A. `(n+1)/(2)`
B. `sqrt((n(n+1))/(2))`
C. `sqrt((n^(2) - 1)/(12))`
D. n

Answer

To find the standard deviation of the first 'n' natural numbers, we can follow these steps:

### Step 1: Identify the first 'n' natural numbers
The first 'n' natural numbers are:
\[ 1, 2, 3, \ldots, n \]

### Step 2: Calculate the mean (\(\bar{x}\))
The mean of the first 'n' natural numbers can be calculated using the formula:
\[ \bar{x} = \frac{\sum_{i=1}^{n} i}{n} \]
The sum of the first 'n' natural numbers is given by:
\[ \sum_{i=1}^{n} i = \frac{n(n + 1)}{2} \]
Thus, the mean is:
\[ \bar{x} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

### Step 3: Calculate the sum of squares of the first 'n' natural numbers
The sum of the squares of the first 'n' natural numbers is given by:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 4: Use the formula for standard deviation
The formula for standard deviation (\(\sigma\)) is:
\[ \sigma = \sqrt{\frac{\sum_{i=1}^{n} i^2}{n} - \left(\frac{\sum_{i=1}^{n} i}{n}\right)^2} \]
Substituting the values we calculated:
\[ \sigma = \sqrt{\frac{\frac{n(n + 1)(2n + 1)}{6}}{n} - \left(\frac{n + 1}{2}\right)^2} \]
This simplifies to:
\[ \sigma = \sqrt{\frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4}} \]

### Step 5: Simplify the expression
To simplify:
1. Find a common denominator, which is 12:
\[ \sigma = \sqrt{\frac{2(n + 1)(2n + 1) - 3(n + 1)^2}{12}} \]
2. Expanding the terms:
\[ = \sqrt{\frac{(n + 1)(4n + 2 - 3(n + 1))}{12}} \]
3. This simplifies to:
\[ = \sqrt{\frac{(n + 1)(n - 1)}{12}} \]

### Final Result
Thus, the standard deviation of the first 'n' natural numbers is:
\[ \sigma = \frac{\sqrt{n^2 - 1}}{2\sqrt{3}} \]