The standard deviation of first 11 nutural numbers is
(i) 2
(ii) `2sqrt2`
(iii) 3
(iv) `sqrt10`

Options

A. 2
B. `2sqrt2`
C. 3
D. `sqrt10`

Answer

To find the standard deviation of the first 11 natural numbers, we will follow these steps:

### Step 1: Calculate the sum of the first 11 natural numbers.
The formula for the sum of the first \( n \) natural numbers is given by:

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

For \( n = 11 \):

\[ S_{11} = \frac{11(11 + 1)}{2} = \frac{11 \times 12}{2} = \frac{132}{2} = 66 \]

### Step 2: Calculate the sum of the squares of the first 11 natural numbers.
The formula for the sum of the squares of the first \( n \) natural numbers is:

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

For \( n = 11 \):

\[ S_{11^2} = \frac{11(11 + 1)(2 \times 11 + 1)}{6} = \frac{11 \times 12 \times 23}{6} \]

Calculating this step-by-step:

1. Calculate \( 11 \times 12 = 132 \)
2. Calculate \( 2 \times 11 + 1 = 23 \)
3. Now, calculate \( 132 \times 23 = 3036 \)
4. Finally, divide by 6:

\[ S_{11^2} = \frac{3036}{6} = 506 \]

### Step 3: Calculate the mean of the first 11 natural numbers.
The mean \( \mu \) is given by:

\[ \mu = \frac{S_n}{n} = \frac{66}{11} = 6 \]

### Step 4: Calculate the variance.
The variance \( \sigma^2 \) is given by the formula:

\[ \sigma^2 = \frac{S_{n^2}}{n} - \left(\frac{S_n}{n}\right)^2 \]

Substituting the values we have:

\[ \sigma^2 = \frac{506}{11} - \left(6\right)^2 \]

Calculating \( \frac{506}{11} \):

\[ \frac{506}{11} = 46 \]

Now calculate the variance:

\[ \sigma^2 = 46 - 36 = 10 \]

### Step 5: Calculate the standard deviation.
The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{10} \]

### Conclusion
Thus, the standard deviation of the first 11 natural numbers is:

\[ \sqrt{10} \]

The correct answer is option (iv) \( \sqrt{10} \).

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