Question

The probability distribution of a discrete random variable \( X \) is given as \[ \begin{array}{|c|c|c|c|c|c|} \hline X & 1 & 2 & 4 & 2A & 3A & 5A \\ \hline P(X) & \frac{1}{2} & \frac{1}{5} & \frac{3}{25} & K & \frac{1}{25} & \frac{1}{25} \\ \hline \end{array} \] Then the value of \( A \) if \( E(X) = 2.94 \) is

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

Hint:

When working with probability distributions, always ensure that the sum of all probabilities equals 1, and use the expected value formula to solve for unknowns.

Answer 1

The Correct Option is A

We are given the probability distribution of the random variable \( X \), and we need to find the value of \( A \) if the expected value \( E(X) = 2.94 \). The expected value \( E(X) \) of a discrete random variable is given by: \[ E(X) = \sum X_i P(X_i) \] Where \( X_i \) are the values of the random variable and \( P(X_i) \) are the corresponding probabilities. Substitute the given values into the formula: \[ E(X) = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{5} + 4 \cdot \frac{3}{25} + 2A \cdot K + 3A \cdot \frac{1}{25} + 5A \cdot \frac{1}{25} \] The total probability must be equal to 1, so: \[ \frac{1}{2} + \frac{1}{5} + \frac{3}{25} + K + \frac{1}{25} + \frac{1}{25} = 1 \] Simplify this equation to find the value of \( K \). Once you have \( K \), substitute it into the expected value equation and solve for \( A \). After simplifying the calculations, you will find: \[ A = 3 \] Thus, the correct answer is \( \boxed{3} \).