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

Step 1: List the numbers.
The first $n$ even natural numbers are $2, 4, 6, \ldots, 2n$. Their mean is $n+1$.
Step 2: Build the variance.
The variance of the first $n$ natural numbers is $\dfrac{n^2-1}{12}$. Multiplying each value by 2 multiplies the variance by $2^2 = 4$, giving $\dfrac{n^2-1}{3}$.
Step 3: Judge Statement I.
The correct variance is $\dfrac{n^2-1}{3}$, but the statement says $\dfrac{n^2-1}{4}$. So Statement I is false.
Step 4: Compute for n = 20.
Mean $= 21$, and variance $= \dfrac{400-1}{3} = \dfrac{399}{3} = 133$.
Step 5: Find the difference.
Difference $= 133 - 21 = 112$, which matches Statement II. So Statement II is true.
Step 6: Choose the option.
Statement I false, Statement II true. \[ \boxed{\text{I false, II true}} \]