Question: generate a video explaining how to do these questions: ## Question 1 [7 marks] A language app has a quiz with 30 multiple choice questions, each with 5 possible answers. Only one answer is correct for each question. Maya completes the quiz by randomly choosing every answer. All answers are equally likely to be chosen, and each of Maya's choices is independent. (a) Write down the expected number of correct answers. [2] (b) Find the probability that Maya correctly answers exactly 4 questions. [3] If Maya answers 9 or more questions correctly, she passes the quiz. (c) Find the probability that Maya passes the quiz. [2] Question 2 [20 marks] According to the schedule, the afternoon flight from Madrid arrives in Rome at 14:00. Records show that the arrival times are normally distributed with a mean arrival time of 14:15 and a standard deviation of 12 minutes. Let X be the random variable the number of minutes late the flight arrives. (a) Write down (i) the mean of X. [1] (ii) the variance of X. [1] (b) Find the probability that the flight arrives between 10 and 30 minutes late. [2] If the flight arrives more than 40 minutes late, the airline must pay compensation. (c) Calculate the expected number of days, in a 365-day year, that the airline pays compensation. [3] Leo arrives in Rome on the flight from Madrid and then has to catch a connecting flight. He allows 25 minutes to reach the gate. He requires the minimum probability of reaching the gate in time to be 99%. (d) Find the earliest time, correct to the nearest minute, that the connecting flight could depart to meet Leos requirements. [4

1. **Problem Statement:** Maya takes a 30-question multiple choice quiz with 5 options each, choosing answers randomly and independently. 2. **Part (a): Expected number of correct answers** - Each question has a probability $p=\frac{1}{5}$ of being correct. - The expected value for correct answers in $n=30$ questions is given by the formula for the expectation of a binomial distribution: $$E(X) = np$$ - Substitute values: $$E(X) = 30 \times \frac{1}{5} = 6$$ 3. **Part (b): Probability of exactly 4 correct answers** - The number of correct answers $X$ follows a binomial distribution $X \sim Binomial(n=30, p=\frac{1}{5})$. - The probability mass function is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ - For $k=4$: $$P(X=4) = \binom{30}{4} \left(\frac{1}{5}\right)^4 \left(\frac{4}{5}\right)^{26}$$ - Calculate $\binom{30}{4} = \frac{30!}{4!26!} = 27,405$ - So: $$P(X=4) = 27,405 \times \left(\frac{1}{5}\right)^4 \times \left(\frac{4}{5}\right)^{26}$$ 4. **Part (c): Probability of passing (9 or more correct answers)** - We want $P(X \geq 9) = 1 - P(X \leq 8)$. - Calculate cumulative probability $P(X \leq 8)$ using binomial cumulative distribution. - Then: $$P(X \geq 9) = 1 - \sum_{k=0}^8 \binom{30}{k} p^k (1-p)^{30-k}$$ --- 5. **Problem Statement:** Flight arrival times are normally distributed with mean 14:15 (15 minutes late) and standard deviation 12 minutes. 6. **Part 2(a)(i): Mean of $X$** - $X$ = minutes late. - Mean $\mu = 15$ minutes. 7. **Part 2(a)(ii): Variance of $X$** - Variance $\sigma^2 = (12)^2 = 144$ minutes squared. 8. **Part 2(b): Probability flight arrives between 10 and 30 minutes late** - Standardize: $$Z = \frac{X - \mu}{\sigma}$$ - Calculate: $$P(10 \leq X \leq 30) = P\left(\frac{10-15}{12} \leq Z \leq \frac{30-15}{12}\right) = P(-0.4167 \leq Z \leq 1.25)$$ - Use standard normal tables or calculator: $$P(Z \leq 1.25) - P(Z \leq -0.4167)$$ 9. **Part 2(c): Expected days airline pays compensation (flight > 40 minutes late)** - Calculate $P(X > 40)$: $$P(X > 40) = P\left(Z > \frac{40-15}{12}\right) = P(Z > 2.0833)$$ - Use standard normal tables: $$P(Z > 2.0833) = 1 - P(Z \leq 2.0833)$$ - Expected days: $$365 \times P(X > 40)$$ 10. **Part 2(d): Earliest connecting flight departure time for 99% probability of reaching gate in time** - Leo allows 25 minutes to reach gate. - We want $P(X \leq t) = 0.99$ where $t$ is the maximum allowed delay to still catch the flight. - Find $z$ such that $P(Z \leq z) = 0.99$ from standard normal table: $z = 2.326$. - Convert back to $X$: $$t = \mu + z \sigma = 15 + 2.326 \times 12 = 15 + 27.912 = 42.912$$ - Leo needs at least $t$ minutes late to have 99% chance to catch flight. - Since he allows 25 minutes, the connecting flight must depart at least $42.912 - 25 = 17.912$ minutes after scheduled arrival. - Rounded to nearest minute: 18 minutes after scheduled arrival time. **Final answers:** - (a) Expected correct answers: $6$ - (b) $P(X=4) = 27,405 \times \left(\frac{1}{5}\right)^4 \times \left(\frac{4}{5}\right)^{26}$ - (c) $P(X \geq 9) = 1 - \sum_{k=0}^8 \binom{30}{k} p^k (1-p)^{30-k}$ - 2(a)(i) Mean: $15$ - 2(a)(ii) Variance: $144$ - 2(b) $P(10 \leq X \leq 30) = P(Z \leq 1.25) - P(Z \leq -0.4167)$ - 2(c) Expected days compensation: $365 \times P(Z > 2.0833)$ - 2(d) Earliest departure time: $18$ minutes after scheduled arrival