What is the variance of first 11 natural numbers?

Options

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

Answer

To find the variance of the first 11 natural numbers, we will 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 sum of the first 11 natural numbers.
We can use the formula for the sum of the first n natural numbers:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
Here, \( n = 11 \):
\[ \text{Sum} = \frac{11 \times (11 + 1)}{2} = \frac{11 \times 12}{2} = \frac{132}{2} = 66 \]

### Step 3: Calculate the mean of the first 11 natural numbers.
The mean is calculated by dividing the sum by the total number of values:
\[ \text{Mean} = \frac{\text{Sum}}{n} = \frac{66}{11} = 6 \]

### Step 4: Calculate the variance.
The variance is calculated using the formula:
\[ \text{Variance} = \frac{\sum (x_i - \text{Mean})^2}{n} \]
Where \( x_i \) are the individual numbers. We will calculate \( (x_i - \text{Mean})^2 \) for each \( x_i \):

- For \( 1: (1 - 6)^2 = (-5)^2 = 25 \)
- For \( 2: (2 - 6)^2 = (-4)^2 = 16 \)
- For \( 3: (3 - 6)^2 = (-3)^2 = 9 \)
- For \( 4: (4 - 6)^2 = (-2)^2 = 4 \)
- For \( 5: (5 - 6)^2 = (-1)^2 = 1 \)
- For \( 6: (6 - 6)^2 = (0)^2 = 0 \)
- For \( 7: (7 - 6)^2 = (1)^2 = 1 \)
- For \( 8: (8 - 6)^2 = (2)^2 = 4 \)
- For \( 9: (9 - 6)^2 = (3)^2 = 9 \)
- For \( 10: (10 - 6)^2 = (4)^2 = 16 \)
- For \( 11: (11 - 6)^2 = (5)^2 = 25 \)

### Step 5: Sum the squared differences.
Now, we sum these squared differences:
\[ 25 + 16 + 9 + 4 + 1 + 0 + 1 + 4 + 9 + 16 + 25 = 110 \]

### Step 6: Divide by the number of values to find the variance.
Now we divide the total by the number of values (11):
\[ \text{Variance} = \frac{110}{11} = 10 \]

### Final Answer:
The variance of the first 11 natural numbers is **10**.

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