1. **State the problem:** We want to compute the expected value $E[X]$ and variance $\mathrm{Var}(X)$ of a discrete random variable $X$ with the given probability mass function (pmf): $$\begin{array}{c|ccccc} X & -2 & -1 & 0 & 1 & 2 \\ \hline P(X) & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{4}{15} & \frac{5}{15} \end{array}$$ 2. **Recall formulas:** - Expected value: $$E[X] = \sum_x x P(X=x)$$ - Variance: $$\mathrm{Var}(X) = E[X^2] - (E[X])^2$$ where $$E[X^2] = \sum_x x^2 P(X=x)$$ 3. **Calculate $E[X]$:** $$E[X] = (-2)\cdot\frac{1}{15} + (-1)\cdot\frac{2}{15} + 0\cdot\frac{3}{15} + 1\cdot\frac{4}{15} + 2\cdot\frac{5}{15}$$ $$= \frac{-2}{15} + \frac{-2}{15} + 0 + \frac{4}{15} + \frac{10}{15} = \frac{-2 - 2 + 0 + 4 + 10}{15} = \frac{10}{15} = \frac{2}{3}$$ 4. **Calculate $E[X^2]$:** $$E[X^2] = (-2)^2\cdot\frac{1}{15} + (-1)^2\cdot\frac{2}{15} + 0^2\cdot\frac{3}{15} + 1^2\cdot\frac{4}{15} + 2^2\cdot\frac{5}{15}$$ $$= 4\cdot\frac{1}{15} + 1\cdot\frac{2}{15} + 0 + 1\cdot\frac{4}{15} + 4\cdot\frac{5}{15} = \frac{4}{15} + \frac{2}{15} + 0 + \frac{4}{15} + \frac{20}{15} = \frac{30}{15} = 2$$ 5. **Calculate variance:** $$\mathrm{Var}(X) = E[X^2] - (E[X])^2 = 2 - \left(\frac{2}{3}\right)^2 = 2 - \frac{4}{9} = \frac{18}{9} - \frac{4}{9} = \frac{14}{9}$$ **Final answers:** - Expected value: $$E[X] = \frac{2}{3}$$ - Variance: $$\mathrm{Var}(X) = \frac{14}{9}$$