1. [Maximum mark: 4]
Accidents at two factories occur randomly and independently. On average, the numbers of accidents per month are 3.1 at factory A and 1.7 at factory B. Find the probability that the total number of accidents in the two factories during a 2-month period is more than 3.
▶️Answer/Explanation
Solution:
Combine the average rates for both factories over 2 months:
\(\lambda = (3.1 + 1.7) \times 2 = 9.6\)
Use the Poisson distribution to find \(P(X > 3)\):
\(1 – e^{-9.6} \left(1 + 9.6 + \frac{9.6^2}{2} + \frac{9.6^3}{6}\right) = 0.986\) (3 sf)
2. [Maximum mark: 4]
The time, in minutes, taken by students to complete a test has the distribution \(N(125, 36)\).
(a) Find the probability that the mean time taken to complete the test by a random sample of 40 students is less than 123 minutes.
(b) Explain whether it was necessary to use the Central Limit theorem in the solution to part (a).
▶️Answer/Explanation
(a) Solution:
Standardize the sample mean:
\(z = \frac{123 – 125}{6/\sqrt{40}} = -2.108\)
\(P(z < -2.108) = 1 – \Phi(2.108) = 0.0175\) (3 sf)
(b) Explanation:
No, because the population is already normally distributed.
3. [Maximum mark: 2]
The graph of the probability density function of a random variable \(X\) is symmetrical about the line \(x=4\). Given that \(\text{P}(X<5)=\frac{20}{27}\), find \(\text{P}(3<X<5)\).
▶️Answer/Explanation
Solution:
Due to symmetry about \(x=4\):
\(\text{P}(3 < X < 5) = 2 \times \left(\frac{20}{27} – \frac{1}{2}\right) = \frac{13}{27}\) (or 0.481)
4. [Maximum mark: 6]
100 randomly chosen adults each throw a ball once. The length, \(l\) metres, of each throw is recorded. The results are summarised below:
\(n = 100\), \(\Sigma l = 3820\), \(\Sigma l^2 = 182200\)
Calculate a 94% confidence interval for the population mean length of throws by adults.
▶️Answer/Explanation
Solution:
Sample mean: \(\bar{l} = \frac{3820}{100} = 38.2\)
Sample variance: \(s^2 = \frac{182200 – 3820^2/100}{99} = 366.424\)
Standard error: \(\frac{s}{\sqrt{n}} = \frac{19.142}{\sqrt{100}} = 1.9142\)
94% CI: \(38.2 \pm 1.881 \times 1.9142 = (34.6, 41.8)\) (3 sf)
5. [Maximum mark: 7]
On average, 1 in 75,000 adults has a certain genetic disorder.
(a) Use a suitable approximating distribution to find the probability that, in a random sample of 10,000 people, at least 1 has the genetic disorder.
(b) In a random sample of \(n\) people, where \(n\) is large, the probability that no-one has the genetic disorder is more than 0.9. Find the largest possible value of \(n\).
▶️Answer/Explanation
(a) Solution:
Poisson approximation: \(\lambda = \frac{10000}{75000} = \frac{2}{15}\)
\(P(X \geq 1) = 1 – e^{-2/15} = 0.125\) (3 sf)
(b) Solution:
\(e^{-n/75000} > 0.9\)
\(-n/75000 > \ln(0.9)\)
\(n < 7902.04\)
Largest integer \(n = 7902\)
6. [Maximum mark: 6]
The probability density function, \(f\), of a random variable \(X\) is given by:
\(f(x) = \begin{cases} k(6x – x^2) & 0 \leq x \leq 6 \\ 0 & \text{otherwise} \end{cases}\)
where \(k\) is a constant. State the value of \(E(X)\) and show that \(\text{Var}(X) = \frac{9}{5}\).
▶️Answer/Explanation
Solution:
For PDF: \(\int_0^6 k(6x – x^2)dx = 1 \Rightarrow k = \frac{1}{36}\)
\(E(X) = 3\) (by symmetry)
\(E(X^2) = \int_0^6 \frac{1}{36}x^2(6x – x^2)dx = 10.8\)
\(\text{Var}(X) = 10.8 – 3^2 = 1.8 = \frac{9}{5}\)
7. [Maximum mark: 10]
The masses, in kilograms, of large and small sacks of flour have the distributions \(N(55, 3^2)\) and \(N(27, 2.5^2)\) respectively.
(a) Some sacks are loaded onto a boat. The maximum load of flour that the boat can carry safely is 340 kg. Find the probability that the boat can carry safely 3 randomly chosen large sacks of flour and 6 randomly chosen small sacks of flour.
(b) Find the probability that the mass of a randomly chosen large sack of flour is greater than the total mass of two randomly chosen small sacks of flour.
▶️Answer/Explanation
(a) Solution:
Total mass distribution: \(N(3 \times 55 + 6 \times 27, 3 \times 3^2 + 6 \times 2.5^2) = N(327, 64.5)\)
\(P(T < 340) = \Phi\left(\frac{340-327}{\sqrt{64.5}}\right) = \Phi(1.619) = 0.947\) (3 sf)
(b) Solution:
Difference distribution: \(N(55 – 2 \times 27, 3^2 + 2 \times 2.5^2) = N(1, 21.5)\)
\(P(D > 0) = \Phi\left(\frac{0-1}{\sqrt{21.5}}\right) = \Phi(0.216) = 0.586\) (3 sf)
8. [Maximum mark: 11]
At a certain large school it was found that the proportion of students not wearing correct uniform was 0.15. The school sent a letter to parents asking them to ensure that their children wear the correct uniform. The school now wishes to test whether the proportion not wearing correct uniform has been reduced.
(a) It is suggested that a random sample of the students in Grade 12 should be used for the test. Give a reason why this would not be an appropriate sample.
(b) State suitable null and alternative hypotheses.
(c) Use a binomial distribution to find the probability of a Type I error. You must justify your answer fully.
(d) In fact 4 students out of the 50 are not wearing correct uniform. State the conclusion of the test, explaining your answer.
(e) State, with a reason, which of the errors, Type I or Type II, may have been made.
▶️Answer/Explanation
(a) Answer: Not representative of all students in the school.
(b) Hypotheses:
\(H_0: p = 0.15\), \(H_1: p < 0.15\)
(c) Solution:
Critical region: \(X \leq 3\) (since \(P(X \leq 4) = 0.112 > 0.05\))
\(P(\text{Type I}) = P(X \leq 3) = 0.0460\) (3 sf)
(d) Conclusion:
4 is outside critical region, so no evidence proportion has decreased (accept \(H_0\)).
(e) Possible error:
Type II error (since \(H_0\) was not rejected).