What is the variance of the first 11 natural numbers ?

Options

A. 10
B. 11
C. 12
D. 13

Explanation

To find the variance of the first 11 natural numbers, we can follow these steps:

### Step 1: Identify the first 11 natural numbers
The first 11 natural numbers are:
\[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \]

### Step 2: Calculate the mean (average)
The mean \( \mu \) of the first \( n \) natural numbers can be calculated using the formula:
\[ \mu = \frac{\text{Sum of first } n \text{ natural numbers}}{n} \]
The sum of the first \( n \) natural numbers is given by:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
For \( n = 11 \):
\[ \text{Sum} = \frac{11(11 + 1)}{2} = \frac{11 \times 12}{2} = 66 \]
Thus, the mean is:
\[ \mu = \frac{66}{11} = 6 \]

### Step 3: Calculate the sum of squares of the first 11 natural numbers
The sum of squares of the first \( n \) natural numbers is given by:
\[ \text{Sum of squares} = \frac{n(n + 1)(2n + 1)}{6} \]
For \( n = 11 \):
\[ \text{Sum of squares} = \frac{11(11 + 1)(2 \times 11 + 1)}{6} = \frac{11 \times 12 \times 23}{6} \]
Calculating this step-by-step:
\[ = \frac{11 \times 12 \times 23}{6} = \frac{3036}{6} = 506 \]

### Step 4: Calculate the variance
The variance \( \sigma^2 \) is calculated using the formula:
\[ \sigma^2 = \frac{\text{Sum of squares}}{n} - \mu^2 \]
Substituting the values we have:
\[ \sigma^2 = \frac{506}{11} - 6^2 \]
Calculating \( \frac{506}{11} \):
\[ \frac{506}{11} = 46 \]
Now substituting back:
\[ \sigma^2 = 46 - 36 = 10 \]

### Final Answer
The variance of the first 11 natural numbers is:
\[ \boxed{10} \]

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