1. [Maximum mark: 4]
The lengths, in millimetres, of a random sample of 12 rods made by a certain machine are as follows:
200, 201, 198, 202, 200, 199, 199, 201, 197, 202, 200, 199.
(a) Find unbiased estimates of the population mean and variance.
(b) Give a statistical reason why these estimates may not be reliable.
▶️Answer/Explanation
(a) Unbiased estimate of the population mean:
\(\text{Mean} = \frac{200 + 201 + 198 + 202 + 200 + 199 + 199 + 201 + 197 + 202 + 200 + 199}{12} = 199.833 \, \text{mm}\)
Unbiased estimate of the population variance:
\(\text{Variance} = \frac{1}{11} \left( \sum x^2 – \frac{(\sum x)^2}{12} \right) = 2.33 \, \text{mm}^2\) (3 sf).
(b) The sample size is small (only 12 rods), so the estimates may not be reliable due to limited data.
2. [Maximum mark: 5]
Harry has a five-sided spinner with sectors coloured blue, green, red, yellow, and black. Harry thinks the spinner may be biased. He plans to carry out a hypothesis test with the following hypotheses:
\( H_0 \): P(spinner lands on blue) = \( \frac{1}{5} \)
\( H_1 \): P(spinner lands on blue) ≠ \( \frac{1}{5} \).
Harry spins the spinner 300 times. It lands on blue on 45 spins. Use a suitable approximation to carry out Harry’s test at the 5% significance level.
▶️Answer/Explanation
Solution:
– Under \( H_0 \), the distribution is \( B(300, 0.2) \). Approximate with \( N(60, 48) \).
– Test statistic: \( z = \frac{45.5 – 60}{\sqrt{48}} = -2.093 \).
– Critical value at 5% significance (two-tailed): \( \pm 1.96 \).
– Since \( |z| = 2.093 > 1.96 \), reject \( H_0 \).
Conclusion: There is evidence at the 5% level that the spinner is biased.
3. [Maximum mark: 5]
A random sample of 500 households in a certain town was chosen. Using this sample, a confidence interval for the proportion \( p \) of all households in that town that owned two or more cars was found to be \( 0.355 < p < 0.445 \).
Find the confidence level of this confidence interval. Give your answer correct to the nearest integer.
▶️Answer/Explanation
Solution:
– Estimated proportion \( \hat{p} = \frac{0.355 + 0.445}{2} = 0.4 \).
– Margin of error: \( 0.445 – 0.4 = 0.045 \).
– Solve \( z \sqrt{\frac{0.4 \times 0.6}{500}} = 0.045 \) → \( z = 2.054 \).
– Corresponding confidence level: \( 96\% \) (nearest integer).
4. [Maximum mark: 6]
In the past, the time taken by students to complete a certain challenge had a mean of 25.5 minutes and a standard deviation of 5.2 minutes. A new challenge is devised, and it is expected that students will take, on average, less than 25.5 minutes to complete this challenge. A random sample of 40 students is chosen, and their mean time for the new challenge is found to be 23.7 minutes.
(a) Assuming the standard deviation of the time for the new challenge is 5.2 minutes, test at the 1% significance level whether the population mean time for the new challenge is less than 25.5 minutes.
(b) State, with a reason, whether it is possible that a Type I error was made in the test in part (a).
▶️Answer/Explanation
(a)
– Hypotheses: \( H_0: \mu = 25.5 \), \( H_1: \mu < 25.5 \).
– Test statistic: \( z = \frac{23.7 – 25.5}{5.2 / \sqrt{40}} = -2.189 \).
– Critical value at 1% significance: \( -2.326 \).
– Since \( -2.189 > -2.326 \), do not reject \( H_0 \).
Conclusion: No evidence that the mean time has decreased.
(b) No, because \( H_0 \) was not rejected, so a Type I error (false rejection of \( H_0 \)) could not occur.
5. [Maximum mark: 9]
The heights of buildings in a large city are normally distributed with a mean of 18.3 m and a standard deviation of 2.5 m.
(a) Find the probability that the total height of 5 randomly chosen buildings in the city is more than 95 m.
(b) Find the probability that the difference between the heights of two randomly chosen buildings in the city is less than 1 m.
▶️Answer/Explanation
(a)
– Total height distribution: \( N(5 \times 18.3, 5 \times 2.5^2) = N(91.5, 31.25) \).
– \( P(X > 95) = P\left(Z > \frac{95 – 91.5}{\sqrt{31.25}}\right) = P(Z > 0.626) = 0.266 \).
(b)
– Difference distribution: \( N(0, 2 \times 2.5^2) = N(0, 12.5) \).
– \( P(|D| < 1) = P(-1 < D < 1) = \Phi(0.283) – \Phi(-0.283) = 0.223 \).
6. [Maximum mark: 11]
In a game, a ball is rolled down a slope and along a track until it stops. The distance \( X \) (in metres) travelled by the ball is modelled by the probability density function:
\( f(x) = \begin{cases} -k(x-1)(x-3) & \text{for } 1 \leq x \leq 3, \\ 0 & \text{otherwise.} \end{cases} \)
(a) Without calculation, explain why \( E(X) = 2 \).
(b) Show that \( k = \frac{3}{4} \).
(c) Find \( \text{Var}(X) \).
(d) One turn consists of rolling the ball 3 times and noting the largest value of \( X \) obtained. If this largest value is greater than 2.5, the player scores a point. Find the probability that on a particular turn the player scores a point.
▶️Answer/Explanation
(a) The PDF is symmetric about \( x = 2 \), so the mean \( E(X) = 2 \).
(b) Integrate \( f(x) \) over \( [1, 3] \) and set equal to 1:
\( \int_{1}^{3} -k(x-1)(x-3) \, dx = 1 \) → \( k = \frac{3}{4} \).
(c) \( \text{Var}(X) = E(X^2) – [E(X)]^2 = \frac{21}{5} – 4 = 0.2 \).
(d) \( P(X > 2.5) = \frac{5}{32} \). Probability of scoring a point: \( 1 – \left(1 – \frac{5}{32}\right)^3 = 0.399 \).
7. [Maximum mark: 10]
(a) Two ponds, A and B, each contain a large number of fish. It is known that 2.4% of fish in pond A are carp and 1.8% of fish in pond B are carp. Random samples of 50 fish from pond A and 60 fish from pond B are selected.
(i) Use Poisson approximations to find the probability that the samples contain at least 2 carp from pond A and at least 2 carp from pond B.
(ii) Find the probability that the samples contain at least 4 carp altogether.
(b) The random variables \( X \) and \( Y \) have distributions \( \text{Po}(\lambda) \) and \( \text{Po}(\mu) \), respectively. Given that \( P(X=0) = [P(Y=0)]^2 \) and \( P(X=2) = k[P(Y=1)]^2 \), find the value of \( k \).
▶️Answer/Explanation
(a)(i)
– \( \lambda_A = 50 \times 0.024 = 1.2 \), \( \lambda_B = 60 \times 0.018 = 1.08 \).
– \( P(\geq 2 \text{ carp in A}) \times P(\geq 2 \text{ carp in B}) = 0.0991 \).
(a)(ii)
– Total carp: \( \text{Po}(2.28) \). \( P(\geq 4) = 0.197 \).
(b) Solve \( e^{-\lambda} = e^{-2\mu} \) and \( \frac{\lambda^2 e^{-\lambda}}{2} = k (\mu e^{-\mu})^2 \) → \( k = 2 \).