Find the variance of first 50 even natural numbers.

Answer

To find the variance of the first 50 even natural numbers, we can follow these steps:

### Step 1: Identify the first 50 even natural numbers
The first 50 even natural numbers are:
\[ 2, 4, 6, \ldots, 100 \]

### Step 2: Calculate the mean (X̄)
The mean of the first n even natural numbers can be calculated using the formula:
\[ \text{Mean} = \frac{\text{Sum of the first n even natural numbers}}{n} \]
The sum of the first n even natural numbers can be calculated as:
\[ \text{Sum} = 2 + 4 + 6 + \ldots + 2n = 2(1 + 2 + 3 + \ldots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]
For n = 50:
\[ \text{Sum} = 50 \cdot 51 = 2550 \]
Thus, the mean is:
\[ \text{Mean} = \frac{2550}{50} = 51 \]

### Step 3: Calculate the variance (σ²)
The variance is given by the formula:
\[ \sigma^2 = \frac{\sum_{i=1}^{n} x_i^2}{n} - \left( \bar{x} \right)^2 \]
Where \( x_i \) are the even natural numbers.

#### Step 3.1: Calculate \( \sum_{i=1}^{50} x_i^2 \)
The even natural numbers can be expressed as \( 2n \), where \( n = 1, 2, \ldots, 50 \). Therefore:
\[ x_i^2 = (2n)^2 = 4n^2 \]
So, we need to calculate:
\[ \sum_{n=1}^{50} (2n)^2 = 4 \sum_{n=1}^{50} n^2 \]
The formula for the sum of squares of the first n natural numbers is:
\[ \sum_{n=1}^{n} n^2 = \frac{n(n + 1)(2n + 1)}{6} \]
For n = 50:
\[ \sum_{n=1}^{50} n^2 = \frac{50 \cdot 51 \cdot 101}{6} = 42925 \]
Thus:
\[ \sum_{n=1}^{50} (2n)^2 = 4 \cdot 42925 = 171700 \]

#### Step 3.2: Substitute into the variance formula
Now we can substitute into the variance formula:
\[ \sigma^2 = \frac{171700}{50} - (51)^2 \]
Calculating \( \frac{171700}{50} \):
\[ \frac{171700}{50} = 3434 \]
Calculating \( (51)^2 \):
\[ (51)^2 = 2601 \]
So:
\[ \sigma^2 = 3434 - 2601 = 833 \]

### Final Answer
The variance of the first 50 even natural numbers is:
\[ \sigma^2 = 833 \]

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