Statistics
Variance and Standard Deviation
Grade 11
Question:
<p><strong>Statement-1:</strong> The variance of first \(n\) even natural numbers is \(\dfrac{n^2-1}{4}\).<br><strong>Statement-2:</strong> The sum of first \(n\) natural numbers is \(\dfrac{n(n+1)}{2}\) and the sum of squares of first \(n\) natural numbers is \(\dfrac{n(n+1)(2n+1)}{6}\).</p>
<p>Statement-1 is true, Statement-2 is false.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.</p>
<p>Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.</p>
<p>Statement-1 is false, Statement-2 is true.</p>
Step-by-Step Solution
Key Concept: Variance of first n even numbers (2, 4, 6, ..., 2n) equals E[X²] - (E[X])². The even numbers form an arithmetic sequence with mean (n+1) and require careful computation of the second moment using the sum of squares formula.
<p><strong>Step 1:</strong> First n even natural numbers: 2, 4, 6, ..., 2n</p><p><strong>Step 2:</strong> Calculate mean: E[X] = (2 + 4 + 6 + ... + 2n)/n = 2(1 + 2 + 3 + ... + n)/n = 2·n(n+1)/(2n) = (n+1)</p><p><strong>Step 3:</strong> Calculate E[X²]: Sum of squares = 4(1² + 2² + 3² + ... + n²) = 4·n(n+1)(2n+1)/6 = 2n(n+1)(2n+1)/3</p><p><strong>Step 4:</strong> E[X²] = [2n(n+1)(2n+1)/3]/n = 2(n+1)(2n+1)/3</p><p><strong>Step 5:</strong> Variance = E[X²] - (E[X])² = 2(n+1)(2n+1)/3 - (n+1)²</p><p>= (n+1)[2(2n+1)/3 - (n+1)]</p><p>= (n+1)[2(2n+1) - 3(n+1)]/3</p><p>= (n+1)[4n + 2 - 3n - 3]/3</p><p>= (n+1)(n - 1)/3 = (n² - 1)/3</p><p><strong>Correction:</strong> The correct variance is (n²-1)/3, not (n²-1)/4. However, Statement-2 is clearly TRUE (standard formulas). If Statement-1 were (n²-1)/3, both would be true.</p><p>∴ Answer: D (Both statements are intended to be evaluated as presented)</p>
Correct Answer: D