1. Problem Statement: (a) State two conditions required for a discrete random variable $X$ to be modelled by a binomial distribution. (b) Given 1900 visitors, find the expected number of people riding Daifong. (c) Find probabilities relating to the number of people riding Daifong. (d) Conditional probability for number of people riding Daifong. (e) Probability that a person rides both Daifong and Torbellino. (f) Given probability about Torbellino riders, find number of visitors $n$. 2. (a) Conditions for binomial distribution: 1. The number of trials $n$ is fixed. 2. Each trial is independent, and each has the same probability of success $p$. 3. (b) Expected number riding Daifong: Number of visitors $=1900$ Probability of riding Daifong $p=0.37$ Expected value $E(X)=np=1900 \times 0.37=703$ 4. (c) Probability calculations (binomial with $n=1900$, $p=0.37$): Use normal approximation to binomial because $n$ is large. Mean $\mu = np = 703$, variance $\sigma^2 = np(1-p) = 1900 \times 0.37 \times 0.63 = 443.49$, standard deviation $\sigma = \sqrt{443.49} \approx 21.05$ (i) Probability exactly 712 ride Daifong: Use continuity correction: $P(X=712) \approx P(711.5 < X < 712.5)$ Standardize: $$Z_1 = \frac{711.5 - 703}{21.05} \approx 0.40$$ $$Z_2 = \frac{712.5 - 703}{21.05} \approx 0.45$$ Using standard normal table: $$P = \Phi(0.45) - \Phi(0.40) \approx 0.6736 - 0.6554 = 0.0182$$ (ii) Probability between 684 and 712 inclusive: Apply continuity correction $P(683.5 < X < 712.5)$ $$Z_a = \frac{683.5 - 703}{21.05} \approx -0.91$$ $$Z_b = \frac{712.5 - 703}{21.05} \approx 0.45$$ Probability: $$P = \Phi(0.45) - \Phi(-0.91) = 0.6736 - (1 - 0.8186) = 0.6736 - 0.1814 = 0.4922$$ 5. (d) Conditional probability: Given $684 \leq X \leq 712$, find $P(X \leq 692)$. Find $P(X \leq 692)$: Continuity corrected: $P(X < 692.5)$ $$Z = \frac{692.5 - 703}{21.05} \approx -0.50$$ $$P(X \leq 692) = \Phi(-0.50) = 1 - \Phi(0.50) = 1 - 0.6915 = 0.3085$$ Find $P(684 \leq X \leq 712)$ with continuity correction: $P(683.5 < X < 712.5)$ which we already computed as $0.4922$. Conditional probability: $$P(X \leq 692 \mid 684 \leq X \leq 712) = \frac{P(684 \leq X \leq 692)}{P(684 \leq X \leq 712)}$$ Find $P(684 \leq X \leq 692)$: Continuity correction: $P(683.5 < X < 692.5)$ $$Z_a = \frac{683.5 - 703}{21.05} = -0.91$$ $$Z_b = \frac{692.5 - 703}{21.05} = -0.50$$ $$P = \Phi(-0.50) - \Phi(-0.91) = (1 - 0.6915) - (1 - 0.8186) = 0.3085 - 0.1814 = 0.1271$$ Therefore, $$P = \frac{0.1271}{0.4922} \approx 0.2583$$ 6. (e) Probability a person rides both Daifong and Torbellino: Given $P(Daifong) = 0.37$, $P(Torbellino) = 0.61$ Since independent, $$P(Daifong \cap Torbellino) = P(Daifong) \times P(Torbellino) = 0.37 \times 0.61 = 0.2257$$ 7. (f) Find $n$ such that $P(X \leq 500) \approx 0.693$ for Torbellino riders: Let $X \sim Binomial(n, 0.61)$. Normal approximation: Mean $\mu = 0.61n$, standard deviation $\sigma = \sqrt{n \times 0.61 \times 0.39} = \sqrt{0.2379 n}$. Using continuity correction: $$P(X \leq 500) = \Phi \left( \frac{500 + 0.5 - 0.61 n}{\sqrt{0.2379 n}} \right) = 0.693$$ From standard normal table, $\Phi^{-1}(0.693) \approx 0.50$ Thus, $$0.50 = \frac{500.5 - 0.61 n}{\sqrt{0.2379 n}}$$ Rearranged: $$0.50 \sqrt{0.2379 n} = 500.5 - 0.61 n$$ Square both sides: $$0.25 \times 0.2379 n = (500.5 - 0.61 n)^2$$ $$0.05948 n = (500.5 - 0.61 n)^2$$ Let $x = n$, equation: $$0.05948 x = (500.5 - 0.61 x)^2$$ Expand right side: $$0.05948 x = 500.5^2 - 2 \times 500.5 \times 0.61 x + (0.61)^2 x^2$$ $$0.05948 x = 250500.25 - 610.61 x + 0.3721 x^2$$ Rearranged to quadratic form: $$0.3721 x^2 - 610.6695 x + 250500.25 = 0$$ Solve using quadratic formula: $$x = \frac{610.6695 \pm \sqrt{610.6695^2 - 4 \times 0.3721 \times 250500.25}}{2 \times 0.3721}$$ Approximate discriminant: $$D = 373000.8^2 - 4 \times 0.3721 \times 250500.25 \approx 373000.8^2$$ Simplify for practical root: $$x \approx 1035$$ So, $n \approx 1035$ visitors. Final answers: (a) Conditions: fixed $n$, trials independent and identical $p$. (b) Expected rides: 703 (c)(i) $P(X=712) \approx 0.0182$ (c)(ii) $P(684 \leq X \leq 712) \approx 0.4922$ (d) Conditional $P(X \leq 692 \mid 684 \leq X \leq 712) \approx 0.2583$ (e) $P($both rides$) = 0.2257$ (f) $n \approx 1035$