The random variable X takes the values -2, 0 and 4 only. It is given that \(P(X = -2) = 2p\), \(P(X = 0) = p\) and \(P(X = 4) = 3p\).
- Find \(p\).
- Find \(\overline{E}(X)\) and \(\text{Var}(X)\).
Solution
(i) The probabilities must sum to 1: \(2p + p + 3p = 1\). Simplifying gives \(6p = 1\), so \(p = \frac{1}{6}\).
(ii) To find \(\overline{E}(X)\), use the formula \(\overline{E}(X) = \sum x_i P(x_i)\):
\(\overline{E}(X) = (-2) \times \frac{2}{6} + 0 \times \frac{1}{6} + 4 \times \frac{3}{6} = \frac{-4}{6} + \frac{12}{6} = \frac{4}{3}\).
To find \(\text{Var}(X)\), use \(\text{Var}(X) = \sum x_i^2 P(x_i) - (\overline{E}(X))^2\):
\(\text{Var}(X) = 4 \times \frac{2}{6} + 0 + 16 \times \frac{3}{6} - \left(\frac{4}{3}\right)^2\).
\(\text{Var}(X) = \frac{8}{6} + \frac{48}{6} - \frac{16}{9} = \frac{56}{6} - \frac{16}{9} = \frac{68}{9}\).
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