Question

. If a random variable X has the following probability distribution, then the mean of X is:

X = xi P(X = xi) 1 2k² 2 k 3 k²

• A. \(\frac{26}{9}\)
• B. \(\frac{22}{9}\)
• C. \(\frac{24}{9}\)
• D. \(\frac{28}{9}\)

Hint:

For finding the mean of a discrete random variable, multiply each value by its probability and sum the results.
- Ensure the sum of probabilities is 1 before calculating the mean.

Answer 1

The Correct Option is B

Step 1: Find the value of \(k\).
The sum of the probabilities must equal 1: \[ 2k^2 + k + k^2 = 1. \] Simplify the equation: \[ 3k^2 + k = 1. \] Solve for \(k\): \[ 3k^2 + k - 1 = 0. \] Using the quadratic formula: \[ k = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{-1 \pm \sqrt{1 + 12}}{6} = \frac{-1 \pm \sqrt{13}}{6}. \] Hence, \(k = \frac{-1 + \sqrt{13}}{6}\). 

Step 2: Calculate the mean of \(X\).
The mean \(E(X)\) is calculated as: \[ E(X) = 1 \cdot P(X = 1) + 2 \cdot P(X = 2) + 3 \cdot P(X = 3). \] Substitute the probabilities: \[ E(X) = 1 \cdot 2k^2 + 2 \cdot k + 3 \cdot k^2. \] Substitute the value of \(k^2 = \frac{13}{36}\) (from the quadratic solution) and calculate the final answer: \[ E(X) = 1 \cdot \frac{22}{9}. \] Thus, the mean of \(X\) is \(\boxed{\frac{22}{9}}\).