1. **State the problem:** Given a discrete probability distribution with values of $x$ and their probabilities $P(x)$, we want to calculate the mean $\mu$ and the variance $\sigma^2$. 2. **Recall formulas:** - Mean: $$\mu = \sum x P(x)$$ - Variance: $$\sigma^2 = \sum (x - \mu)^2 P(x)$$ 3. **Calculate the mean $\mu$:** $$\mu = 0 \times 0.30 + 1 \times 0.28 + 2 \times 0.25 + 3 \times 0.17$$ $$\mu = 0 + 0.28 + 0.50 + 0.51 = 1.29$$ 4. **Calculate each $(x - \mu)$:** - For $x=0$: $0 - 1.29 = -1.29$ - For $x=1$: $1 - 1.29 = -0.29$ - For $x=2$: $2 - 1.29 = 0.71$ - For $x=3$: $3 - 1.29 = 1.71$ 5. **Calculate each $(x - \mu)^2$:** - $(-1.29)^2 = 1.6641$ - $(-0.29)^2 = 0.0841$ - $(0.71)^2 = 0.5041$ - $(1.71)^2 = 2.9241$ 6. **Calculate each $(x - \mu)^2 P(x)$:** - $1.6641 \times 0.30 = 0.49923$ - $0.0841 \times 0.28 = 0.02355$ - $0.5041 \times 0.25 = 0.12603$ - $2.9241 \times 0.17 = 0.49710$ 7. **Sum these to find variance $\sigma^2$:** $$\sigma^2 = 0.49923 + 0.02355 + 0.12603 + 0.49710 = 1.14591$$ **Final answers:** - Mean $\mu = 1.29$ - Variance $\sigma^2 = 1.15$ (rounded to two decimals)