1. **State the problem:** We have a random variable $X$ representing the number of three-point shots made by a basketball player with the distribution: $$\begin{array}{c|cccc} X & 0 & 1 & 2 & 3 \\ P(X) & 0.10 & 0.25 & 0.40 & 0.25 \end{array}$$ We need to find the mean ($\mu$), variance ($\sigma^2$), and standard deviation ($\sigma$). 2. **Mean formula:** $$\mu = \sum x P(x)$$ This means multiply each value of $X$ by its probability and sum all. 3. **Calculate mean:** $$\mu = 0 \times 0.10 + 1 \times 0.25 + 2 \times 0.40 + 3 \times 0.25$$ $$\mu = 0 + 0.25 + 0.80 + 0.75 = 1.80$$ 4. **Variance formula:** $$\sigma^2 = \sum (x - \mu)^2 P(x)$$ We find the squared difference between each $x$ and the mean, multiply by $P(x)$, then sum. 5. **Calculate each term for variance:** $$x - \mu: 0 - 1.8 = -1.8, \quad 1 - 1.8 = -0.8, \quad 2 - 1.8 = 0.2, \quad 3 - 1.8 = 1.2$$ $$ (x - \mu)^2: (-1.8)^2 = 3.24, \quad (-0.8)^2 = 0.64, \quad 0.2^2 = 0.04, \quad 1.2^2 = 1.44$$ $$ (x - \mu)^2 P(x): 3.24 \times 0.10 = 0.324, \quad 0.64 \times 0.25 = 0.16, \quad 0.04 \times 0.40 = 0.016, \quad 1.44 \times 0.25 = 0.36$$ 6. **Sum for variance:** $$\sigma^2 = 0.324 + 0.16 + 0.016 + 0.36 = 0.86$$ 7. **Standard deviation formula:** $$\sigma = \sqrt{\sigma^2}$$ 8. **Calculate standard deviation:** $$\sigma = \sqrt{0.86} \approx 0.927$$ **Final answers:** - Mean $\mu = 1.80$ - Variance $\sigma^2 = 0.86$ - Standard deviation $\sigma \approx 0.927$