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

Answer

To calculate the mean and standard deviation of the first n natural numbers, we will follow these steps:

### Step 1: Calculate the Mean
The mean (average) of the first n natural numbers can be calculated using the formula:

\[ \bar{x} = \frac{\sum_{r=1}^{n} r}{n} \]

We know that the sum of the first n natural numbers is given by the formula:

\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]

Substituting this into the mean formula:

\[ \bar{x} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

### Step 2: Calculate the Standard Deviation
The formula for the variance (σ²) is:

\[ \sigma^2 = \frac{\sum_{r=1}^{n} r^2}{n} - \bar{x}^2 \]

We need to find \(\sum_{r=1}^{n} r^2\), which is given by the formula:

\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]

Now substituting this into the variance formula:

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

This simplifies to:

\[ \sigma^2 = \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4} \]

To combine these fractions, we need a common denominator, which is 12:

\[ \sigma^2 = \frac{2(n + 1)(2n + 1)}{12} - \frac{3(n + 1)^2}{12} \]

Combining the fractions:

\[ \sigma^2 = \frac{2(n + 1)(2n + 1) - 3(n + 1)^2}{12} \]

Factoring out \((n + 1)\):

\[ \sigma^2 = \frac{(n + 1)(2(2n + 1) - 3(n + 1))}{12} \]

Expanding the terms inside the parentheses:

\[ 2(2n + 1) - 3(n + 1) = 4n + 2 - 3n - 3 = n - 1 \]

Thus, we have:

\[ \sigma^2 = \frac{(n + 1)(n - 1)}{12} \]

Finally, the standard deviation (σ) is the square root of the variance:

\[ \sigma = \sqrt{\frac{(n + 1)(n - 1)}{12}} = \frac{\sqrt{n^2 - 1}}{2\sqrt{3}} \]

### Final Results
- Mean of the first n natural numbers: \(\bar{x} = \frac{n + 1}{2}\)
- Standard deviation of the first n natural numbers: \(\sigma = \frac{\sqrt{n^2 - 1}}{2\sqrt{3}}\)