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 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: \[ \boxed{0} \]
List-I | List-II |
---|---|
(A) 4î − 2ĵ − 4k̂ | (I) A vector perpendicular to both î + 2ĵ + k̂ and 2î + 2ĵ + 3k̂ |
(B) 4î − 4ĵ + 2k̂ | (II) Direction ratios are −2, 1, 2 |
(C) 2î − 4ĵ + 4k̂ | (III) Angle with the vector î − 2ĵ − k̂ is cos⁻¹(1/√6) |
(D) 4î − ĵ − 2k̂ | (IV) Dot product with −2î + ĵ + 3k̂ is 10 |