1. **State the problem:** We have a discrete random variable $x$ with values $0,1,2,3,4,5$ and their corresponding probabilities $p(x) = 0.15, 0.25, 0.05, 0.05, 0.3, 0.2$. We need to find the mean $\mu$, variance $\sigma^2$, and standard deviation $\sigma$ of this distribution. 2. **Formulas:** - Mean (expected value): $$\mu = \sum x_i p(x_i)$$ - Variance: $$\sigma^2 = \sum (x_i - \mu)^2 p(x_i)$$ - Standard deviation: $$\sigma = \sqrt{\sigma^2}$$ 3. **Calculate the mean $\mu$:** $$\mu = 0(0.15) + 1(0.25) + 2(0.05) + 3(0.05) + 4(0.3) + 5(0.2)$$ $$\mu = 0 + 0.25 + 0.1 + 0.15 + 1.2 + 1.0 = 2.7$$ 4. **Calculate the variance $\sigma^2$:** First calculate each squared difference times probability: $$ (0 - 2.7)^2(0.15) = ( -2.7)^2(0.15) = 7.29 \times 0.15 = 1.0935 $$ $$ (1 - 2.7)^2(0.25) = ( -1.7)^2(0.25) = 2.89 \times 0.25 = 0.7225 $$ $$ (2 - 2.7)^2(0.05) = ( -0.7)^2(0.05) = 0.49 \times 0.05 = 0.0245 $$ $$ (3 - 2.7)^2(0.05) = (0.3)^2(0.05) = 0.09 \times 0.05 = 0.0045 $$ $$ (4 - 2.7)^2(0.3) = (1.3)^2(0.3) = 1.69 \times 0.3 = 0.507 $$ $$ (5 - 2.7)^2(0.2) = (2.3)^2(0.2) = 5.29 \times 0.2 = 1.058 $$ Sum these values: $$\sigma^2 = 1.0935 + 0.7225 + 0.0245 + 0.0045 + 0.507 + 1.058 = 3.41$$ 5. **Calculate the standard deviation $\sigma$:** $$\sigma = \sqrt{3.41} \approx 1.847$$ **Final answers:** $$\mu = 2.7$$ $$\sigma^2 = 3.41$$ $$\sigma \approx 1.847$$