Question

If \( X \) is a random variable such that \( P(X = -2) = P(X = -1) = P(X = 2) = P(X = 1) = \frac{1{6} \), and \( P(X = 0) = \frac{1}{3} \), then the mean of \( X \) is}

• A. \( \frac{5}{3} \)
• B. \( 1 \)
• C. \( 0 \)
• D. \( \frac{3}{5} \)

Hint:

To find the mean of a random variable, calculate the weighted sum of all its possible values, where the weights are the respective probabilities.

Answer 1

The Correct Option is C

The expected value (mean) of a random variable \( X \) is defined as: \[ E(X) = \sum_{i} x_i P(X = x_i), \] where \( x_i \) represents the possible values of \( X \), and \( P(X = x_i) \) is the probability of each value. Step 1: List all possible values of \( X \). The random variable \( X \) can take the values \( \{-2, -1, 0, 1, 2\} \), with the following corresponding probabilities: \[ P(X = -2) = P(X = -1) = P(X = 1) = P(X = 2) = \frac{1}{6}, \quad P(X = 0) = \frac{1}{3}. \] Step 2: Calculate \( E(X) \). Substitute these values and probabilities into the expected value formula: \[ E(X) = (-2)P(X = -2) + (-1)P(X = -1) + 0P(X = 0) + 1P(X = 1) + 2P(X = 2). \] Substituting the respective probabilities: \[ E(X) = (-2) \times \frac{1}{6} + (-1) \times \frac{1}{6} + 0 \times \frac{1}{3} + 1 \times \frac{1}{6} + 2 \times \frac{1}{6}. \] Simplifying the terms: \[ E(X) = \frac{-2}{6} + \frac{-1}{6} + 0 + \frac{1}{6} + \frac{2}{6}. \] Combining like terms: \[ E(X) = \frac{-2 - 1 + 1 + 2}{6} = \frac{0}{6} = 0. \] Final Answer: \[ \boxed{0} \]