Question

The p.m.f. of a random variable \( X \) is: \[ P(X) = \frac{2x}{n(n+1)}, \quad x = 1, 2, 3, \ldots, n \] \[ P(X) = 0, \quad \text{Otherwise.} \] Then \( E(X) \) is:

• A. \( \frac{n+1}{3} \)
• B. \( \frac{2n+1}{3} \)
• C. \( \frac{n+2}{3} \)
• D. \( \frac{2n-1}{2} \)

Hint:

For discrete random variables, calculate \( E(X) \) by summing \( x \cdot P(X = x) \), and use known summation formulas for efficiency.

Answer 1

The Correct Option is B

The expected value \( E(X) \) is calculated as:\[E(X) = \sum_{x=1}^n x \cdot P(X = x).\]Substituting \( P(X = x) = \frac{2x}{n(n+1)} \) yields:\[E(X) = \sum_{x=1}^n x \cdot \frac{2x}{n(n+1)}.\]This simplifies to:\[E(X) = \frac{2}{n(n+1)} \sum_{x=1}^n x^2.\]Step 1: Apply the sum of squares formula.
The sum of the squares of the first \( n \) natural numbers is given by:\[\sum_{x=1}^n x^2 = \frac{n(n+1)(2n+1)}{6}.\]Substituting this into the expression for \( E(X) \):\[E(X) = \frac{2}{n(n+1)} \cdot \frac{n(n+1)(2n+1)}{6}.\]Further simplification results in:\[E(X) = \frac{2(2n+1)}{6}.\]Which reduces to:\[E(X) = \frac{2n+1}{3}.\] Final Answer:\[\boxed{\frac{2n+1}{3}}\]