IB Mathematics AHL 4.16 Confidence intervals for the mean of a normal population.AI HL Paper 2- Exam Style Questions- New Syllabus
Alex is engaged in a computer game where they target spaceships and battleships. The frequency of hitting spaceships per minute follows a Poisson distribution with a mean of \( 4.2 \). The frequency of hitting battleships per minute follows a Poisson distribution with a mean of \( 2.3 \). Each hit occurs independently of others.
(a) (i) Compute the probability that Alex hits at most 10 spaceships in 2 minutes.
(a) (ii) Compute the probability that Alex hits more than 10 spaceships and battleships combined in one minute. [5]
Each hit on a spaceship awards Alex 3 points, and each hit on a battleship awards 5 points. Let \( T \) represent the total points earned in one minute.
(b) (i) Determine \( \mathbb{E}(T) \).
(b) (ii) Determine \( \text{Var}(T) \). [3]
(c) Provide one reason why the distribution of \( T \) cannot be Poisson. [1]
Alex plans to play the game for one hour.
(d) Apply the central limit theorem to find the probability that Alex’s average score per minute exceeds 25. [4]
▶️ Answer/Explanation
Alex’s spaceship hits per minute \( S \sim \text{Poisson}(4.2) \) and battleship hits per minute \( B \sim \text{Poisson}(2.3) \), with independence.
(a) (i) At most 10 spaceships in 2 minutes. Over 2 minutes: \( S_{(2)} \sim \text{Poisson}(8.4) \). \( P(S_{(2)} \le 10) = \sum_{k=0}^{10} e^{-8.4} \frac{8.4^k}{k!} \approx 0.774 \)
(a) (ii) More than 10 ships in total in 1 minute. Since \( S + B \sim \text{Poisson}(6.5) \): \( P(S + B > 10) = 1 – \sum_{k=0}^{10} e^{-6.5} \frac{6.5^k}{k!} \approx 0.0668 \)
(b) (i) Expected score per minute. \( T = 3S + 5B \). \( \mathbb{E}(T) = 3 \times 4.2 + 5 \times 2.3 = 24.1 \)
(b) (ii) Variance of score per minute. \( \text{Var}(T) = 3^2 \times 4.2 + 5^2 \times 2.3 = 37.8 + 57.5 = 95.3 \)
(c) Why is \( T \) not Poisson? A Poisson distribution has mean equal to variance, but here \( \mathbb{E}(T) = 24.1 \ne 95.3 = \text{Var}(T) \). Also, \( T = 3S + 5B \) cannot take all integer values.
(d) Probability that mean score per minute exceeds 25 over 1 hour. Over 60 minutes: \( \overline{T} \sim N \big( 24.1, \frac{95.3}{60} \big), \quad \sigma_{\overline{T}} = \sqrt{\frac{95.3}{60}} \approx 1.261 \) \( P(\overline{T} > 25) = P \big( Z > \frac{25 – 24.1}{1.261} \big) = P(Z > 0.714) \approx 0.238 \)