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
The expected value (mean) of a random variable \( X \) is calculated using the formula: \[ E(X) = \sum_{i} x_i P(X = x_i), \] where \( x_i \) are the possible values of \( X \) and \( P(X = x_i) \) are their respective probabilities.
Step 1: Identify the possible values and probabilities for \( X \). The set of possible values for \( X \) is \( \{-2, -1, 0, 1, 2\} \), with the following probability distribution: \[ P(X = -2) = P(X = -1) = P(X = 1) = P(X = 2) = \frac{1}{6}, \quad P(X = 0) = \frac{1}{3}. \]
Step 2: Compute \( E(X) \) by applying the formula with the identified values and probabilities: \[ E(X) = (-2)P(X = -2) + (-1)P(X = -1) + 0P(X = 0) + 1P(X = 1) + 2P(X = 2). \] Substitute the probabilities: \[ E(X) = (-2) \times \frac{1}{6} + (-1) \times \frac{1}{6} + 0 \times \frac{1}{3} + 1 \times \frac{1}{6} + 2 \times \frac{1}{6}. \] Simplify the terms: \[ E(X) = \frac{-2}{6} + \frac{-1}{6} + 0 + \frac{1}{6} + \frac{2}{6}. \] Combine the terms: \[ E(X) = \frac{-2 - 1 + 1 + 2}{6} = \frac{0}{6} = 0. \]
Final Answer: \[ \boxed{0} \]