Question

The variance of the first 50 even natural numbers is:

• A. 833
• B. \( \frac{437}{4} \)
• C. \( \frac{833}{4} \)
• D. 437

Hint:

For arithmetic progressions, use the mean and variance formulas effectively: - Mean = \( \frac{\text{Sum of terms}}{\text{Number of terms}} \), - Variance = \( \text{Mean of squares} - (\text{Square of mean}) \).

Answer 1

The Correct Option is A

Step 1: Identify the first 50 even natural numbers. The first 50 even natural numbers are: \[ 2, 4, 6, \dots, 100. \] This is an arithmetic progression with:
- First term \( a = 2 \),
- Common difference \( d = 2 \),
- Number of terms \( n = 50 \).
Step 2: Find the sum and mean. The sum of an arithmetic progression is given by: \[ S = \frac{n}{2} \left( 2a + (n-1)d \right) = \frac{50}{2} \left( 2(B) + (50-1)2 \right) = 25(4 + 98) = 2550. \] The mean is: \[ \text{Mean} = \frac{S}{n} = \frac{2550}{50} = 51. \] Step 3: Find the sum of squares. The sum of squares is given by: \[ \sum i^2 = \sum (2i)^2 = 4 \sum i^2 = 4 \times \frac{n(n+1)(2n+1)}{6}. \] Using \( n = 50 \), we get: \[ 4 \times 50 \times 51 \times 101 / 6 = 171700. \] Step 4: Calculate the variance. Now, calculate \( E(X^2) \) and variance: \[ E(X^2) = \frac{171700}{50} = 3434, \] \[ \text{Variance} = E(X^2) - (\text{Mean})^2 = 3434 - 51^2 = 3434 - 2601 = 833. \] Therefore, the variance is \( 833 \).