1. **State the problem:** We have a discrete random variable $X$ with values $77, 78, 79, 80, 81$ and corresponding probabilities $0.15, 0.15, 0.20, 0.40, 0.10$. We need to find: a) $P(X=80)$ b) $P(X>80)$ c) $P(X\leq 80)$ d) The mean $\mu$ of $X$ e) The variance $\sigma^2$ of $X$ f) The standard deviation $\sigma$ of $X$ 2. **Recall probability rules:** - The sum of all probabilities must be 1. - $P(X>80)$ means sum of probabilities for $X=81$. - $P(X\leq 80)$ means sum of probabilities for $X=77,78,79,80$. 3. **Calculate each part:** a) $P(X=80) = 0.40$ b) $P(X>80) = P(X=81) = 0.10$ c) $P(X\leq 80) = P(77)+P(78)+P(79)+P(80) = 0.15 + 0.15 + 0.20 + 0.40 = 0.90$ 4. **Calculate the mean $\mu$:** Formula: $$\mu = \sum x_i P(x_i)$$ Calculate: $$\mu = 77(0.15) + 78(0.15) + 79(0.20) + 80(0.40) + 81(0.10)$$ $$= 11.55 + 11.7 + 15.8 + 32 + 8.1 = 79.15$$ 5. **Calculate the variance $\sigma^2$:** Formula: $$\sigma^2 = \sum (x_i - \mu)^2 P(x_i)$$ Calculate each term: $$(77 - 79.15)^2 \times 0.15 = ( -2.15)^2 \times 0.15 = 4.6225 \times 0.15 = 0.693375$$ $$(78 - 79.15)^2 \times 0.15 = ( -1.15)^2 \times 0.15 = 1.3225 \times 0.15 = 0.198375$$ $$(79 - 79.15)^2 \times 0.20 = ( -0.15)^2 \times 0.20 = 0.0225 \times 0.20 = 0.0045$$ $$(80 - 79.15)^2 \times 0.40 = (0.85)^2 \times 0.40 = 0.7225 \times 0.40 = 0.289$$ $$(81 - 79.15)^2 \times 0.10 = (1.85)^2 \times 0.10 = 3.4225 \times 0.10 = 0.34225$$ Sum these: $$\sigma^2 = 0.693375 + 0.198375 + 0.0045 + 0.289 + 0.34225 = 1.5275$$ 6. **Calculate the standard deviation $\sigma$:** Formula: $$\sigma = \sqrt{\sigma^2} = \sqrt{1.5275} \approx 1.236$$ **Final answers:** a) $P(X=80) = 0.40$ b) $P(X>80) = 0.10$ c) $P(X\leq 80) = 0.90$ d) $\mu = 79.15$ e) $\sigma^2 = 1.5275$ f) $\sigma \approx 1.236$