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
Solution:
Standardize using \( Z = \frac{\bar{X} – \mu}{\sigma/\sqrt{n}} \):
\( Z = \frac{405 – 410}{\sqrt{400}/\sqrt{36}} = \frac{-5}{20/6} = -1.5 \).
Find \( P(Z < -1.5) = \Phi(-1.5) = 1 – \Phi(1.5) \).
From tables, \( \Phi(1.5) = 0.9332 \), so \( P(Z < -1.5) = 0.0668 \).
2. [Maximum mark: 4]
In a survey of 300 randomly chosen adults, 134 said they exercised regularly. The upper bound of a confidence interval for the proportion exercising regularly was 0.487. Find the confidence level \( a\% \).
▶️Answer/Explanation
Solution:
Let \( \hat{p} = \frac{134}{300} \approx 0.4467 \).
The upper bound is \( \hat{p} + z \sqrt{\frac{\hat{p}(1-\hat{p})}{300}} = 0.487 \).
Solve for \( z \): \( z \approx 1.405 \).
Find \( \Phi(1.405) \approx 0.92 \).
The confidence level is \( 2 \times 0.92 – 1 = 0.84 \) or \( 84\% \).
3(a)(i). [Maximum mark: 3]
Assume a website receives 0.3 hits per minute (Poisson). Find the probability of at least 4 hits in 10 minutes.
▶️Answer/Explanation
Solution:
For 10 minutes, \( \lambda = 0.3 \times 10 = 3 \).
\( P(X \geq 4) = 1 – P(X \leq 3) \).
Calculate \( P(X \leq 3) \) using Poisson formula:
\( e^{-3}(1 + 3 + \frac{9}{2} + \frac{27}{6}) \approx 0.647 \).
Thus, \( P(X \geq 4) = 1 – 0.647 = 0.353 \).
3(a)(ii). [Maximum mark: 4]
Use a suitable approximating distribution to find the probability of fewer than 40 hits in 3 hours.
▶️Answer/Explanation
Solution:
For 3 hours (180 minutes), \( \lambda = 0.3 \times 180 = 54 \).
Approximate with \( N(54, 54) \).
Use continuity correction: \( P(X < 40) \approx P(Y \leq 39.5) \).
Standardize: \( Z = \frac{39.5 – 54}{\sqrt{54}} \approx -1.973 \).
\( P(Z < -1.973) \approx 0.0242 \).
3(b)(i). [Maximum mark: 1]
Explain why the owner’s belief in a Poisson distribution is contradicted if hits vary by time of day.
▶️Answer/Explanation
Solution:
The Poisson distribution assumes a constant rate, but the varying hit rates (day vs. night) violate this assumption.
3(b)(ii). [Maximum mark: 2]
Specify separate Poisson distributions for day-time and night-time hits.
▶️Answer/Explanation
Solution:
Day-time (12 hours): \( \text{Po}(0.4) \) per minute.
Night-time (12 hours): \( \text{Po}(0.2) \) per minute.
4(a). [Maximum mark: 2]
Find the mean and variance of daily income for chemical A, given \( X \sim N(10.3, 5.76) \) and income is \$2.50 per kg.
▶️Answer/Explanation
Solution:
Mean income: \( 10.3 \times 2.50 = 25.75 \).
Variance: \( 5.76 \times (2.50)^2 = 36 \).
4(b). [Maximum mark: 6]
Find the probability that the income from chemical A exceeds that of chemical B on a randomly chosen day.
▶️Answer/Explanation
Solution:
Income for A: \( N(25.75, 36) \).
Income for B: \( N(37.05, 101.506) \).
Difference \( D = A – B \sim N(-11.3, 137.506) \).
Find \( P(D > 0) \):
Standardize: \( Z = \frac{0 – (-11.3)}{\sqrt{137.506}} \approx 0.964 \).
\( P(Z > 0.964) = 1 – \Phi(0.964) \approx 0.168 \).
5. [Maximum mark: 5]
Test whether the mean number of enquiries per minute has increased from 0.31 (Poisson) using a 5-minute sample with 5 enquiries (2.5% significance level).
▶️Answer/Explanation
Solution:
\( H_0: \lambda = 0.31 \times 5 = 1.55 \), \( H_1: \lambda > 1.55 \).
\( P(X \geq 5) = 1 – P(X \leq 4) \).
Calculate \( P(X \leq 4) \) using Poisson:
\( e^{-1.55}(1 + 1.55 + \frac{1.55^2}{2} + \frac{1.55^3}{6} + \frac{1.55^4}{24}) \approx 0.979 \).
\( P(X \geq 5) = 0.021 < 0.025 \). Reject \( H_0 \).
6(a). [Maximum mark: 1]
Show that \( p \leq \frac{5}{23} \) for a symmetrical random variable \( X \) with given probabilities.
▶️Answer/Explanation
Solution:
Total probability must be \( \leq 1 \):
\( p + \frac{13}{10}p \leq 0.5 \) (due to symmetry).
Solve: \( \frac{23}{10}p \leq 0.5 \) → \( p \leq \frac{5}{23} \).
6(b). [Maximum mark: 2]
Find \( P(b < X < 6 – a) \) in terms of \( p \) for a symmetrical \( X \).
▶️Answer/Explanation
Solution:
By symmetry, \( P(b < X < 6 – a) = 2 \times \frac{13}{10}p = \frac{26}{10}p = \frac{13}{5}p \).
6(c). [Maximum mark: 5]
Given \( b = 2 \) and \( p = \frac{5}{27} \), find \( a \) for the pdf \( f(x) = \frac{1}{36}(6x – x^2) \).
▶️Answer/Explanation
Solution:
Integrate \( f(x) \) from \( a \) to 2 and set equal to \( \frac{5}{27} \):
\( \int_a^2 \frac{6x – x^2}{36} dx = \frac{5}{27} \).
Solve: \( \frac{1}{36} \left[ 3x^2 – \frac{x^3}{3} \right]_a^2 = \frac{5}{27} \).
Simplify to \( a^3 – 9a^2 + 8 = 0 \).
Factor: \( (a – 1)(a^2 – 8a – 8) = 0 \). Valid solution: \( a = 1 \).
7(a). [Maximum mark: 7]
Test \( H_0: \mu = 0.5 \) vs \( H_1: \mu < 0.5 \) using sample data (\( n = 50 \), \( \Sigma x = 23.0 \), \( \Sigma x^2 = 13.02 \)) at 5% significance.
▶️Answer/Explanation
Solution:
Sample mean \( \bar{x} = \frac{23}{50} = 0.46 \).
Unbiased variance: \( s^2 = \frac{50}{49} \left( \frac{13.02}{50} – 0.46^2 \right) \approx 0.0498 \).
Test statistic: \( t = \frac{0.46 – 0.5}{\sqrt{0.0498/50}} \approx -1.268 \).
Compare to \( z_{0.05} = -1.645 \). Since \( -1.268 > -1.645 \), do not reject \( H_0 \).
7(b). [Maximum mark: 5]
Find the probability of a Type II error if \( \mu = 0.4 \) (true mean) in a similar test.
▶️Answer/Explanation
Solution:
Critical value under \( H_0 \): \( 0.5 – 1.645 \times \sqrt{0.0498/50} \approx 0.448 \).
Under \( H_1 \): \( \mu = 0.4 \), same variance.
Find \( P(\bar{X} > 0.448) \):
Standardize: \( Z = \frac{0.448 – 0.4}{\sqrt{0.0498/50}} \approx 1.524 \).
\( P(Z > 1.524) \approx 0.064 \).