Calculate the mean and standard deviation of first `n` natural numbers.

Explanation

To calculate the mean and standard deviation of the first `n` natural numbers, we will 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
The mean (average) of a set of numbers is given by the formula:
\[ \text{Mean} ( \bar{x} ) = \frac{\text{Sum of all observations}}{\text{Number of observations}} \]
For the first `n` natural numbers, the sum can be calculated using the formula for the sum of the first `n` natural numbers:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
Thus, the mean can be calculated as:
\[ \bar{x} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

### Step 3: Calculate the Variance
The variance is defined as:
\[ \text{Variance} ( \sigma^2 ) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]
This can be expanded to:
\[ \sigma^2 = \frac{1}{n} \left( \sum_{i=1}^{n} x_i^2 - n \bar{x}^2 \right) \]
We need to find \( \sum_{i=1}^{n} x_i^2 \), which is given by the formula:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Now substituting this into the variance formula:
\[ \sigma^2 = \frac{1}{n} \left( \frac{n(n + 1)(2n + 1)}{6} - n \left( \frac{n + 1}{2} \right)^2 \right) \]
Calculating \( n \left( \frac{n + 1}{2} \right)^2 \):
\[ n \left( \frac{n + 1}{2} \right)^2 = n \cdot \frac{(n + 1)^2}{4} = \frac{n(n + 1)^2}{4} \]
Substituting back into the variance formula:
\[ \sigma^2 = \frac{1}{n} \left( \frac{n(n + 1)(2n + 1)}{6} - \frac{n(n + 1)^2}{4} \right) \]
Finding a common denominator (which is 12):
\[ \sigma^2 = \frac{1}{n} \left( \frac{2n(n + 1)(2n + 1)}{12} - \frac{3n(n + 1)^2}{12} \right) \]
This simplifies to:
\[ \sigma^2 = \frac{1}{n} \cdot \frac{n(n + 1)}{12} \left( 2(2n + 1) - 3(n + 1) \right) \]
Simplifying the expression inside the parentheses:
\[ 2(2n + 1) - 3(n + 1) = 4n + 2 - 3n - 3 = n - 1 \]
Thus, we have:
\[ \sigma^2 = \frac{(n + 1)(n - 1)}{12} \]

### Step 4: Calculate the Standard Deviation
The standard deviation is the square root of the variance:
\[ \sigma = \sqrt{\sigma^2} = \sqrt{\frac{(n^2 - 1)}{12}} \]

### Final Results
- Mean of the first `n` natural numbers:
\[ \bar{x} = \frac{n + 1}{2} \]
- Standard deviation of the first `n` natural numbers:
\[ \sigma = \sqrt{\frac{n^2 - 1}{12}} \]