If the variance of first n even natural numbers is 133, then the value of n is equal to

Options

A. 19
B. 24
C. 21
D. 20

Answer

To solve the problem of finding the value of \( n \) such that the variance of the first \( n \) even natural numbers is 133, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the first \( n \) even natural numbers**:
The first \( n \) even natural numbers are \( 2, 4, 6, \ldots, 2n \).

2. **Calculate the mean (\( \bar{x} \)) of these numbers**:
The mean can be calculated as:
\[ \bar{x} = \frac{\text{Sum of the first } n \text{ even numbers}}{n} = \frac{2 + 4 + 6 + \ldots + 2n}{n} \]
The sum of the first \( n \) even numbers is:
\[ 2 + 4 + 6 + \ldots + 2n = 2(1 + 2 + 3 + \ldots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1) \]
Therefore, the mean is:
\[ \bar{x} = \frac{n(n + 1)}{n} = n + 1 \]

3. **Calculate the variance (\( \sigma^2 \))**:
The formula for variance is:
\[ \sigma^2 = \frac{\sum (x_i^2)}{n} - \bar{x}^2 \]
We need to calculate \( \sum (x_i^2) \):
\[ \sum (x_i^2) = 2^2 + 4^2 + 6^2 + \ldots + (2n)^2 = 4(1^2 + 2^2 + 3^2 + \ldots + n^2) = 4 \cdot \frac{n(n + 1)(2n + 1)}{6} \]
Thus,
\[ \sum (x_i^2) = \frac{2n(n + 1)(2n + 1)}{3} \]

4. **Substituting into the variance formula**:
Now substituting back into the variance formula:
\[ \sigma^2 = \frac{\frac{2n(n + 1)(2n + 1)}{3}}{n} - (n + 1)^2 \]
Simplifying this gives:
\[ \sigma^2 = \frac{2(n + 1)(2n + 1)}{3} - (n + 1)^2 \]
\[ = (n + 1)\left(\frac{2(2n + 1)}{3} - (n + 1)\right) \]
\[ = (n + 1)\left(\frac{2(2n + 1) - 3(n + 1)}{3}\right) \]
\[ = (n + 1)\left(\frac{4n + 2 - 3n - 3}{3}\right) \]
\[ = (n + 1)\left(\frac{n - 1}{3}\right) \]

5. **Setting the variance equal to 133**:
We know the variance is given as 133, so we set up the equation:
\[ \frac{(n + 1)(n - 1)}{3} = 133 \]
Multiplying both sides by 3:
\[ (n + 1)(n - 1) = 399 \]
This simplifies to:
\[ n^2 - 1 = 399 \implies n^2 = 400 \implies n = 20 \]

### Final Answer:
Thus, the value of \( n \) is \( 20 \).