1. **Problem 1: Lottery Tickets** We have 5000 tickets sold at 1 each. - 1 ticket wins 1000 - 2 tickets win 500 each - 10 tickets win 100 each - The rest win 0 Let $X$ be the net gain from buying one ticket, so $X = \text{prize} - 1$. 2. **Constructing the probability distribution of $X$** - $P(X=999) = \frac{1}{5000}$ (win 1000, net gain 999) - $P(X=499) = \frac{2}{5000}$ (win 500, net gain 499) - $P(X=99) = \frac{10}{5000}$ (win 100, net gain 99) - $P(X=-1) = 1 - \frac{1+2+10}{5000} = \frac{4987}{5000}$ (no win, net loss 1) 3. **Expected value $E(X)$** $$E(X) = \sum x_i P(x_i) = 999 \times \frac{1}{5000} + 499 \times \frac{2}{5000} + 99 \times \frac{10}{5000} + (-1) \times \frac{4987}{5000}$$ Calculate: $$E(X) = \frac{999}{5000} + \frac{998}{5000} + \frac{990}{5000} - \frac{4987}{5000} = \frac{999 + 998 + 990 - 4987}{5000} = \frac{0}{5000} = 0$$ Interpretation: On average, the net gain is 0, meaning the game is fair. 4. **Variance $Var(X)$** Formula: $$Var(X) = E(X^2) - [E(X)]^2$$ Calculate $E(X^2)$: $$E(X^2) = 999^2 \times \frac{1}{5000} + 499^2 \times \frac{2}{5000} + 99^2 \times \frac{10}{5000} + (-1)^2 \times \frac{4987}{5000}$$ Calculate each term: $$999^2 = 998001, \quad 499^2 = 249001, \quad 99^2 = 9801, \quad (-1)^2 = 1$$ So: $$E(X^2) = \frac{998001}{5000} + \frac{2 \times 249001}{5000} + \frac{10 \times 9801}{5000} + \frac{4987}{5000} = \frac{998001 + 498002 + 98010 + 4987}{5000} = \frac{1,583,000}{5000} = 316.6$$ Since $E(X) = 0$, $$Var(X) = 316.6$$ 5. **Standard deviation $\sigma$** $$\sigma = \sqrt{Var(X)} = \sqrt{316.6} \approx 17.8$$ --- 6. **Problem 2: Borachio's Tires** Given $X$ with values and probabilities: | x | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---| | P(x) | 0.48 | 0.36 | 0.12 | 0.04 | ? | Sum probabilities to find missing: $$0.48 + 0.36 + 0.12 + 0.04 + P(5) = 1 \Rightarrow P(5) = 1 - 1 = 0$$ 7. **a. Probability $X > 3$** $$P(X > 3) = P(4) + P(5) = 0.04 + 0 = 0.04$$ 8. **b. Probability $X \leq 2$** $$P(X \leq 2) = P(1) + P(2) = 0.48 + 0.36 = 0.84$$ 9. **c. Mean $\mu$** $$\mu = E(X) = \sum x_i P(x_i) = 1 \times 0.48 + 2 \times 0.36 + 3 \times 0.12 + 4 \times 0.04 + 5 \times 0 = 0.48 + 0.72 + 0.36 + 0.16 + 0 = 1.72$$ Interpretation: On average, Borachio produces 1.72 blemished tires per day. 10. **Standard deviation $\sigma$** Calculate $E(X^2)$: $$E(X^2) = 1^2 \times 0.48 + 2^2 \times 0.36 + 3^2 \times 0.12 + 4^2 \times 0.04 + 5^2 \times 0 = 0.48 + 1.44 + 1.08 + 0.64 + 0 = 3.64$$ Variance: $$Var(X) = E(X^2) - [E(X)]^2 = 3.64 - (1.72)^2 = 3.64 - 2.9584 = 0.6816$$ Standard deviation: $$\sigma = \sqrt{0.6816} \approx 0.8258$$ --- 11. **Problem 3: Discrete Random Variable $X$** Given: | x | 13 | 18 | 20 | 24 | 27 | |---|----|----|----|----|----| | P(x) | 0.22 | 0.25 | 0.20 | 0.17 | 0.16 | 12. **a. $P(20)$** $$P(20) = 0.20$$ 13. **b. $P(X > 18)$** Values greater than 18 are 20, 24, 27: $$P(X > 18) = 0.20 + 0.17 + 0.16 = 0.53$$ 14. **c. $P(X \leq 18)$** Values 13 and 18: $$P(X \leq 18) = 0.22 + 0.25 = 0.47$$ 15. **d. Mean $\mu$** $$\mu = \sum x_i P(x_i) = 13 \times 0.22 + 18 \times 0.25 + 20 \times 0.20 + 24 \times 0.17 + 27 \times 0.16$$ Calculate: $$= 2.86 + 4.5 + 4 + 4.08 + 4.32 = 19.76$$ 16. **e. Variance $\sigma^2$** Calculate $E(X^2)$: $$E(X^2) = 13^2 \times 0.22 + 18^2 \times 0.25 + 20^2 \times 0.20 + 24^2 \times 0.17 + 27^2 \times 0.16$$ Calculate each term: $$169 \times 0.22 = 37.18$$ $$324 \times 0.25 = 81$$ $$400 \times 0.20 = 80$$ $$576 \times 0.17 = 97.92$$ $$729 \times 0.16 = 116.64$$ Sum: $$E(X^2) = 37.18 + 81 + 80 + 97.92 + 116.64 = 412.74$$ Variance: $$Var(X) = E(X^2) - \mu^2 = 412.74 - (19.76)^2 = 412.74 - 390.34 = 22.4$$ 17. **f. Standard deviation $\sigma$** $$\sigma = \sqrt{22.4} \approx 4.73$$