1. [Maximum mark: 3]
A random variable \( X \) has the distribution \( N(410, 400) \). Find the probability that the mean of a random sample of 36 values of \( X \) is less than 405.
▶️Answer/Explanation
Step 1: Standard error: \( \sigma_{\bar{X}} = \frac{20}{\sqrt{36}} = 3.333 \)
Step 2: \( Z \)-score: \( Z = \frac{405 – 410}{3.333} = -1.5 \)
Step 3: \( P(\bar{X} < 405) = \Phi(-1.5) = 0.0668 \)
2. [Maximum mark: 4]
In a survey of 300 adults, 134 exercised regularly. The upper bound of a confidence interval is 0.487. Find the confidence level \( \alpha \) (to nearest integer).
▶️Answer/Explanation
Step 1: \( \hat{p} = \frac{134}{300} = 0.4467 \)
Step 2: Solve \( 0.4467 + z \sqrt{\frac{0.4467 \times 0.5533}{300}} = 0.487 \) → \( z \approx 1.405 \)
Step 3: Confidence level = \( 2\Phi(1.405) – 1 \approx 84\% \)
3. [Maximum mark: 8]
(a)(i) Find \( P(\geq 4 \text{ hits in 10 mins}) \) for \( \text{Po}(3) \).
(ii) Approximate \( P(< 40 \text{ hits in 3 hours}) \).
(b) Explain why day/night hits contradict Poisson assumption.
▶️Answer/Explanation
(a)(i): \( P(\geq 4) = 1 – e^{-3}\left(1 + 3 + \frac{9}{2} + \frac{27}{6}\right) = 0.353 \)
(a)(ii): Normal approximation \( N(54, 54) \). \( P(X < 39.5) = 0.0242 \).
(b): Poisson requires constant rate. Day/night rates differ (non-homogeneous).
4. [Maximum mark: 8]
(a) Find mean and variance of daily income for Chemical A.
(b) Find \( P(\text{Income A} > \text{Income B}) \).
▶️Answer/Explanation
(a): Mean = \( 25.75 \), Variance = \( 36 \)
(b): \( D = 2.5X – 3.25Y \sim N(-11.3, 137.5) \). \( P(D > 0) = 0.168 \)
5. [Maximum mark: 5]
Test if mean enquiries increased from \( \lambda = 0.31 \), given 5 enquiries in 5 minutes (2.5% significance).
▶️Answer/Explanation
Step 1: \( H_0: \lambda = 1.55 \), \( H_1: \lambda > 1.55 \)
Step 2: \( P(X \geq 5) = 0.021 \)
Step 3: \( 0.021 < 0.025 \) → Reject \( H_0 \). Evidence supports increase.
6. [Maximum mark: 8]
(a) Show \( p \leq \frac{5}{23} \).
(b) Find \( P(b < X < 6 – a) \).
(c) Given \( b = 2 \), \( p = \frac{5}{27} \), find \( a \).
▶️Answer/Explanation
(a): \( p + 1.3p \leq 0.5 \) → \( p \leq \frac{5}{23} \)
(b): Symmetry → \( P(b < X < 6 – a) = 3.6p \)
(c): Solve \( \frac{1}{36} \int_{a}^{2} (6x – x^2) dx = \frac{5}{27} \) → \( a = 1 \)
7. [Maximum mark: 12]
(a) Test \( H_0: \mu = 0.5 \) vs \( H_1: \mu < 0.5 \) (5% significance).
(b) Find probability of Type II error if \( \mu = 0.4 \).
▶️Answer/Explanation
(a): Test statistic \( Z = -1.27 \). Do not reject \( H_0 \).
(b): Critical value \( = 0.448 \). Type II error \( = 0.064 \)