IB Mathematics AHL 4.15 A linear combination of n independent normal AI HL Paper 2- Exam Style Questions- New Syllabus
The city of Verona has an adult population of four million people. It is assumed that the weights of adults in Verona can be modelled by a normal distribution with mean \(72 \, \text{kg}\) and standard deviation \(10 \, \text{kg}\).
(a) If 10 adults in Verona are chosen independently and at random, find the probability that more than 3 of them have a weight greater than \(85 \, \text{kg}\). [4 marks]
Clara runs a travel agency in Verona. The elevator to her office is designed to hold a maximum of 8 people.
(b) Write down a probability distribution that models the total weight of 8 adults chosen independently and at random from Verona. [3 marks]
The total weight of 8 adults exceeds \(w\) on less than 1% of all occasions that 8 adults enter the elevator.
(c) Find the value of \(w\). [2 marks]
A newspaper claims that 42% of the adults in Verona who go on holiday choose to go abroad. Clara believes that this is an overestimation of the true number. During the past month, Clara found that 67 of her clients chose a holiday abroad, and 133 chose a holiday that was not abroad.
(d) Clara decides to perform a test using the binomial distribution on her data for the population proportion, \(p\), that go on holiday abroad.
(i) State two assumptions that Clara makes in order to conduct the test.
(ii) Write down the null and the alternative hypotheses for Clara’s test, in terms of \(p\).
(iii) Using the data from Clara’s sample, perform the test at a 5% significance level to determine whether Clara’s belief is reasonable. [8 marks]
▶️ Answer/Explanation
(a)
We are told adult weights in Verona follow \[ X \sim N(72, 10^2). \] We want the probability that more than 3 out of 10 adults weigh above 85 kg.
First, find the single probability: \[ P(X > 85) = P\!\left( Z > \frac{85-72}{10} \right) = P(Z > 1.3) = 0.0968. \]
Let \( Y \) = number of people out of 10 weighing above 85 kg. \[ Y \sim B(10, 0.0968). \] Then, \[ P(Y \geq 4) = 1 – P(Y \leq 3). \] Using the binomial probabilities gives \[ P(Y \geq 4) = 0.0114. \] Final Answer (a): The probability is about **1.14%**.
(b)
The total weight of 8 adults is a sum of independent normal random variables. If \( X \sim N(72, 10^2) \), then \[ W \sim N(8 \times 72, 8 \times 10^2). \] So, \[ W \sim N(576, 800). \] where mean = 576 kg and variance = 800, giving \( SD = 28.28 \).
Final Answer (b): \[ W \sim N(576, 800). \]
(c)
We want the weight \( w \) such that in less than 1% of cases, \[ P(W > w) = 0.01. \] Using inverse normal: \[ P(Z > z) = 0.01 \quad \Rightarrow \quad z = 2.33. \] So, \[ w = 576 + 2.33 \times 28.28 = 642 \, \text{kg (approx)}. \] Final Answer (c): \( w \approx 642 \, \text{kg} \).
(d) (i)
Two assumptions Clara must make in her test:
- Her clients are a random sample of the Verona population.
- Each person takes at most one holiday a year.
- Choices are independent of each other.
- She is the only agent for her clients.
Final Answer (d)(i): Any two of the above.
(d) (ii)
The newspaper claims 42% of holidays are abroad. Clara thinks this is too high. Hypotheses: \[ H_0: p = 0.42, \quad H_1: p < 0.42. \]
(d) (iii)
Clara surveyed 200 clients: 67 abroad, 133 not abroad. Let \( Q \) be the number of clients choosing abroad: \[ Q \sim B(200, 0.42). \] Observed value: \( Q = 67 \). We compute the probability: \[ P(Q \leq 67) = 0.00850. \] Since \( 0.0085 < 0.05 \), the result is significant. This means the observed proportion (67/200 = 0.335) is unusually low compared with 0.42.
Conclusion: There is evidence to support Clara’s belief that fewer than 42% of adults choose to go abroad.