Question

Find the expected value and variance of \( X \) for the following p.m.f: \[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline P(X) & 0.2 & 0.3 & 0.1 & 0.15 & 0.25 \\ \hline \end{array} \]

Hint:

To calculate the variance, use the formula \( \text{Variance} = E(X^2) - (E(X))^2 \), where \( E(X^2) \) is the expected value of \( X^2 \).

Answer 1

The expected value \( E(X) \) is given by: \[ E(X) = \sum x \cdot P(X=x) = (-2)(0.2) + (-1)(0.3) + (0)(0.1) + (A)(0.15) + (B)(0.25) \] \[ E(X) = -0.4 - 0.3 + 0 + 0.15 + 0.5 = -0.05. \] The expected value of \( X^2 \) is: \[ E(X^2) = \sum x^2 \cdot P(X=x) = (-2)^2(0.2) + (-1)^2(0.3) + (0)^2(0.1) + (A)^2(0.15) + (B)^2(0.25) \] \[ E(X^2) = 0.8 + 0.3 + 0 + 0.15 + 1 = 2.25. \] The variance \( \text{Var}(X) \) is given by: \[ \text{Var}(X) = E(X^2) - (E(X))^2 = 2.25 - (-0.05)^2 = 2.25 - 0.0025 = 2.2475. \] Thus, the variance is \( 2.2475 \).