Question

Consider the following statements:
Statement - I: The variance of the first n even natural numbers is \( \frac{n^{2}-1}{4} \)
Statement - II: The difference between the variance of the first 20 even natural numbers and their mean is 112
Which of the following is correct?

• A. Both the statements I and II are true
• B. Both the statements I and II are false
• C. Statement I is false and Statement II is true
• D. Statement I is true and Statement II is false

Hint:

Test with a small number like \( n = 2 \) (numbers are 2 and 4): \[ \text{Mean} = 3, \quad \text{Variance} = \frac{(2-3)^2 + (4-3)^2}{2} = 1 \] Formula from statement I gives: \( \frac{2^2-1}{4} = 0.75 \neq 1 \). This instantly proves statement I is false!

Answer 1

The Correct Option is C

Concept: The first \( n \) even natural numbers are \( 2, 4, 6, \dots, 2n \).

• The mean of the first \( n \) even natural numbers is \( \mu = n + 1 \).

• The variance of the first \( n \) consecutive natural numbers is \( \frac{n^2-1}{12} \). Since multiplying each data point by a constant \( k \) scales the variance by \( k^2 \), multiplying by 2 changes the variance to: \[ \text{Variance} = 2^2 \times \frac{n^2-1}{12} = \frac{n^2-1}{3} \]

Step 1: Evaluating Statement I.
As derived above, the variance of the first \( n \) even numbers is \( \frac{n^2-1}{3} \). The statement claims it is \( \frac{n^2-1}{4} \), which is incorrect. Thus, Statement I is false.

Step 2: Evaluating Statement II.
For the first \( n = 20 \) even natural numbers: \[ \text{Mean} = n + 1 = 20 + 1 = 21 \] \[ \text{Variance} = \frac{20^2 - 1}{3} = \frac{400 - 1}{3} = \frac{399}{3} = 133 \] Calculating the difference between variance and mean: \[ \text{Difference} = 133 - 21 = 112 \] This matches Statement II perfectly. Hence, Statement II is true.