Find the mean and variance for the following data
(i)Find n natural numbers.
(ii)First 10 multiple of 3.

Answer

To solve the problem of finding the mean and variance for the given data, we will break it down into two parts as outlined in the question.

### Part (i): Mean and Variance of the first n Natural Numbers

1. **Identify the Data**:
The first n natural numbers are: \( 1, 2, 3, \ldots, n \).

2. **Calculate the Mean**:
The mean \( \bar{x} \) is calculated using the formula:
\[ \bar{x} = \frac{\text{Sum of all observations}}{\text{Total number of observations}} = \frac{1 + 2 + 3 + \ldots + n}{n} \]
The sum of the first n natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
Therefore, the mean becomes:
\[ \bar{x} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

3. **Calculate the Variance**:
The variance \( \sigma^2 \) is calculated using the formula:
\[ \sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2 \]
The sum of the squares of the first n natural numbers is given by:
\[ \sum x_i^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Thus, the variance becomes:
\[ \sigma^2 = \frac{\frac{n(n + 1)(2n + 1)}{6}}{n} - \left(\frac{n + 1}{2}\right)^2 \]
Simplifying this gives:
\[ \sigma^2 = \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4} \]
Finding a common denominator (12):
\[ \sigma^2 = \frac{2(n + 1)(2n + 1) - 3(n + 1)^2}{12} \]
Expanding and simplifying:
\[ \sigma^2 = \frac{(n + 1)(2n + 1 - 3(n + 1))}{12} = \frac{(n + 1)(n - 2)}{12} \]

### Part (ii): Mean and Variance of the First 10 Multiples of 3

1. **Identify the Data**:
The first 10 multiples of 3 are: \( 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 \).

2. **Calculate the Mean**:
The mean \( \bar{x} \) is calculated as:
\[ \bar{x} = \frac{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}{10} \]
The sum is:
\[ 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30 = 165 \]
Therefore, the mean is:
\[ \bar{x} = \frac{165}{10} = 16.5 \]

3. **Calculate the Variance**:
The variance \( \sigma^2 \) is calculated using:
\[ \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} \]
First, we calculate \( x_i - \bar{x} \) for each \( x_i \):
- For \( 3 \): \( 3 - 16.5 = -13.5 \)
- For \( 6 \): \( 6 - 16.5 = -10.5 \)
- For \( 9 \): \( 9 - 16.5 = -7.5 \)
- For \( 12 \): \( 12 - 16.5 = -4.5 \)
- For \( 15 \): \( 15 - 16.5 = -1.5 \)
- For \( 18 \): \( 18 - 16.5 = 1.5 \)
- For \( 21 \): \( 21 - 16.5 = 4.5 \)
- For \( 24 \): \( 24 - 16.5 = 7.5 \)
- For \( 27 \): \( 27 - 16.5 = 10.5 \)
- For \( 30 \): \( 30 - 16.5 = 13.5 \)

Now, squaring these results:
- \( (-13.5)^2 = 182.25 \)
- \( (-10.5)^2 = 110.25 \)
- \( (-7.5)^2 = 56.25 \)
- \( (-4.5)^2 = 20.25 \)
- \( (-1.5)^2 = 2.25 \)
- \( (1.5)^2 = 2.25 \)
- \( (4.5)^2 = 20.25 \)
- \( (7.5)^2 = 56.25 \)
- \( (10.5)^2 = 110.25 \)
- \( (13.5)^2 = 182.25 \)

Summing these squared values:
\[ 182.25 + 110.25 + 56.25 + 20.25 + 2.25 + 2.25 + 20.25 + 56.25 + 110.25 + 182.25 = 742.5 \]
Finally, the variance is:
\[ \sigma^2 = \frac{742.5}{10} = 74.25 \]

### Summary of Results
- For the first n natural numbers:
- Mean: \( \frac{n + 1}{2} \)
- Variance: \( \frac{(n + 1)(n - 2)}{12} \)

- For the first 10 multiples of 3:
- Mean: \( 16.5 \)
- Variance: \( 74.25 \)