Question 1(a)
1(a) [Maximum mark: 4]
A random variable \( X \) has the distribution \( \text{Po}(25) \). Use the normal approximation to the Poisson distribution to find \( P(X > 30) \).
▶️Answer/Explanation
Solution:
– Approximate \( X \sim N(25, 25) \).
– Apply continuity correction: \( P(X > 30) = P(X \geq 30.5) \).
– Standardize: \( Z = \frac{30.5 – 25}{\sqrt{25}} = 1.1 \).
– Find probability: \( 1 – \Phi(1.1) = 0.136 \) (3 sf).
Answer: \( 0.136 \)
Question 1(b)
1(b) [Maximum mark: 2]
A random variable \( Y \) has the distribution \( \text{B}(100, p) \) where \( p < 0.05 \). Use the Poisson approximation to the binomial distribution to write down an expression, in terms of \( p \), for \( P(Y < 3) \).
▶️Answer/Explanation
Solution:
– Approximate \( Y \sim \text{Po}(100p) \).
– Calculate cumulative probability: \( P(Y < 3) = e^{-100p} \left(1 + 100p + \frac{(100p)^2}{2!}\right) \).
Answer: \( e^{-100p} \left(1 + 100p + 5000p^2\right) \)
Question 2(a)
2(a) [Maximum mark: 3]
The length, in minutes, of mathematics lectures has mean \( \mu \) and standard deviation 8.3. A random sample of 85 lectures had a total length of 4590 minutes. Calculate a 95% confidence interval for \( \mu \).
▶️Answer/Explanation
Solution:
– Sample mean: \( \bar{x} = \frac{4590}{85} = 54 \).
– Standard error: \( \frac{8.3}{\sqrt{85}} \approx 0.900 \).
– 95% CI: \( 54 \pm 1.96 \times 0.900 \).
– Interval: \( (52.2, 55.8) \) minutes (3 sf).
Answer: \( 52.2 \) to \( 55.8 \)
Question 2(b)
2(b) [Maximum mark: 2]
The length of history lectures has mean \( m \) and standard deviation \( s \). A 95% confidence interval for \( m \) using a sample of 100 lectures had a width of 2.8 minutes. Find \( s \).
▶️Answer/Explanation
Solution:
– Width of CI: \( 2 \times 1.96 \times \frac{s}{\sqrt{100}} = 2.8 \).
– Solve for \( s \): \( s = \frac{2.8 \times 10}{2 \times 1.96} = 7.14 \) (3 sf).
Answer: \( 7.14 \)
Question 3(a)(i)
3(a)(i) [Maximum mark: 1]
A student calculates \( P(X = 8) = 0.0439 \) and claims this is sufficient to reject the null hypothesis at the 5% level. Explain why this is incorrect.
▶️Answer/Explanation
Solution:
– The test requires the tail probability \( P(X \geq 8) \), not just \( P(X = 8) \).
Answer: The comparison should be based on the cumulative probability, not a single value.
Question 3(a)(ii)
3(a)(ii) [Maximum mark: 5]
Carry out the hypothesis test for the researcher’s claim that boys are equally likely to choose either color cube, given that 8 out of 10 boys chose green.
▶️Answer/Explanation
Solution:
– Hypotheses: \( H_0: p = 0.5 \), \( H_1: p \neq 0.5 \).
– Calculate \( P(X \geq 8) = 0.0439 + P(X=9) + P(X=10) = 0.0547 \).
– Since \( 0.0547 > 0.05 \), do not reject \( H_0 \).
Conclusion: Insufficient evidence to reject the null hypothesis.
Question 3(b)
3(b) [Maximum mark: 1]
Explain why a Type I error cannot have occurred in the test.
▶️Answer/Explanation
Solution:
– A Type I error occurs when \( H_0 \) is wrongly rejected. Here, \( H_0 \) was not rejected.
Answer: The null hypothesis was not rejected, so no Type I error could occur.
Question 3(c)
3(c) [Maximum mark: 2]
Find the probability of a Type I error in a similar test with 10 boys.
▶️Answer/Explanation
Solution:
– Probability of extreme outcomes under \( H_0 \): \( P(X=9) + P(X=10) = 0.0107 \).
Answer: \( 0.0107 \)
Question 4(a)
4(a) [Maximum mark: 1]
Write down null and alternative hypotheses to test whether the mean height of trees exceeds 4.23 metres.
▶️Answer/Explanation
Solution:
– \( H_0: \mu = 4.23 \), \( H_1: \mu > 4.23 \).
Answer: \( H_0: \mu = 4.23 \), \( H_1: \mu > 4.23 \)
Question 4(b)
4(b) [Maximum mark: 3]
Given that the null hypothesis is rejected at the 5% level, find the possible values of the sample mean \( \bar{h} \).
▶️Answer/Explanation
Solution:
– Critical value: \( \bar{h} > 4.23 + 1.645 \times \frac{0.67}{\sqrt{200}} \).
– Calculate: \( \bar{h} > 4.31 \) (3 sf).
Answer: \( \bar{h} > 4.31 \)
Question 4(c)
4(c) [Maximum mark: 1]
Evaluate Ajit’s statement about the Central Limit Theorem.
▶️Answer/Explanation
Solution:
– The population is already normally distributed, so the CLT is not needed.
Answer: Incorrect, because the population distribution is given as normal.
Question 5(a)
5(a) [Maximum mark: 3]
Given \( E(X) = \ln 2 \) for a random variable \( X \) with PDF \( f(x) = \frac{1}{x^2} \) (\( a < x < b \)), show that \( b = 2a \).
▶️Answer/Explanation
Solution:
– Integrate: \( E(X) = \int_a^b \frac{1}{x} \, dx = \ln b – \ln a = \ln \frac{b}{a} \).
– Set equal to \( \ln 2 \): \( \frac{b}{a} = 2 \).
Answer: Hence \( b = 2a \).
Question 5(b)
5(b) [Maximum mark: 3]
Show that \( a = \frac{1}{2} \) for the same PDF.
▶️Answer/Explanation
Solution:
– Integrate PDF: \( \int_a^b \frac{1}{x^2} \, dx = 1 \).
– Substitute \( b = 2a \): \( \frac{1}{a} – \frac{1}{2a} = 1 \).
– Solve: \( a = \frac{1}{2} \).
Answer: Hence \( a = \frac{1}{2} \).
Question 5(c)
5(c) [Maximum mark: 3]
Find the median of \( X \) for the given PDF.
▶️Answer/Explanation
Solution:
– Set \( \int_{0.5}^m \frac{1}{x^2} \, dx = 0.5 \).
– Solve: \( m = \frac{2}{3} \) (3 sf).
Answer: \( \frac{2}{3} \)
Question 6
6 [Maximum mark: 7]
Calculate the probability that the profit on a batch of bread exceeds \$40, given costs and sales data.
▶️Answer/Explanation
Solution:
– Total cost \( T \sim N(99.45, 7.3629) \).
– Profit condition: \( P(T < 105) \).
– Standardize: \( Z = \frac{105 – 99.45}{\sqrt{7.3629}} \approx 2.045 \).
– Probability: \( \Phi(2.045) = 0.979 \) (3 sf).
Answer: \( 0.979 \)
Question 7(a)
7(a) [Maximum mark: 2]
For \( X \sim \text{Po}(2.4) \), find \( P(2 \leq X < 4) \).
▶️Answer/Explanation
Solution:
– Calculate: \( P(X=2) + P(X=3) = 0.261 + 0.209 = 0.470 \) (3 sf).
Answer: \( 0.470 \)
Question 7(b)
7(b) [Maximum mark: 3]
Find the probability that two independent values of \( X \) are both greater than 1.
▶️Answer/Explanation
Solution:
– \( P(X > 1) = 1 – P(X \leq 1) = 0.6916 \).
– Square for independence: \( 0.6916^2 = 0.478 \) (3 sf).
Answer: \( 0.478 \)
Question 7(c)(i)
7(c)(i) [Maximum mark: 3]
Find the set of values \( r \) for which \( P(X = r) < P(X = r+1) \).
▶️Answer/Explanation
Solution:
– Solve inequality: \( r + 1 < 2.4 \).
– Valid \( r \): \( 0, 1 \).
Answer: \( r = 0, 1 \)
Question 7(c)(ii)
7(c)(ii) [Maximum mark: 1]
Identify the value of \( r \) for which \( P(X = r) \) is greatest.
▶️Answer/Explanation
Solution:
– The maximum probability occurs at \( r = 2 \).
Answer: \( r = 2 \)