1. [Maximum mark: 3]
The heights, in metres, of a random sample of 10 mature trees of a certain variety are given below:
5.9, 6.5, 6.7, 5.9, 6.9, 6.0, 6.4, 6.2, 5.8, 5.8
Find unbiased estimates of the population mean and variance of the heights of all mature trees of this variety.
▶️Answer/Explanation
Unbiased estimate of mean: 6.21 m
Unbiased estimate of variance: 0.157 (3 sf)
2. [Maximum mark: 8]
A spinner has five sectors, each printed with a different colour. Susma and Sanjay both wish to test whether the spinner is biased so that it lands on red on fewer spins than it would if it were fair.
(a) Susma spins the spinner 40 times. She finds that it lands on red exactly 4 times. Use a binomial distribution to carry out the test at the 5% significance level.
(b) Sanjay also spins the spinner 40 times. He finds that it lands on red r times. Use a binomial distribution to find the largest value of r that lies in the rejection region for the test at the 5% significance level.
▶️Answer/Explanation
(a)
• Hypotheses: H₀: p = 0.2, H₁: p < 0.2
• P(X ≤ 4) = 0.0759 > 0.05
• Conclusion: Do not reject H₀ – insufficient evidence of bias
(b)
• P(X ≤ 3) = 0.0285 < 0.05
• Largest r in rejection region: 3
3. [Maximum mark: 7]
Drops of water fall randomly from a leaking tap at a constant average rate of 5.2 per minute.
(a) Find the probability that at least 3 drops fall during a randomly chosen 30-second period.
(b) Use a suitable approximating distribution to find the probability that at least 650 drops fall during a randomly chosen 2-hour period.
▶️Answer/Explanation
(a)
• λ = 2.6 (for 30 seconds)
• P(X ≥ 3) = 1 – P(X ≤ 2) = 0.482 (3 sf)
(b)
• Normal approximation: N(624, 624)
• With continuity correction: P(X ≥ 649.5)
• Probability = 0.154 (3 sf)
4. [Maximum mark: 8]
Each month a company sells X kg of brown sugar and Y kg of white sugar, where X and Y have the independent distributions N(2500, 120²) and N(3700, 130²) respectively.
(a) Find the mean and standard deviation of the total amount of sugar that the company sells in 3 randomly chosen months.
(b) The company makes a profit of $1.50 per kilogram of brown sugar sold and makes a loss of $0.20 per kilogram of white sugar sold. Find the probability that, in a randomly chosen month, the total profit is less than $3000.
▶️Answer/Explanation
(a)
• Mean = 18,600 kg
• Standard deviation = 306 kg (3 sf)
(b)
• Profit model: 1.5X – 0.2Y
• Mean profit = $3,010
• Variance = 33,076
• P(Profit < $3000) = 0.478 (3 sf)
5. [Maximum mark: 7]
A builders’ merchant sells stones of different sizes.
(a) The masses of size A stones have standard deviation 6 grams. The mean mass of a random sample of 200 size A stones is 45 grams. Find a 95% confidence interval for the population mean mass of size A stones.
(b) The masses of size B stones have standard deviation 11 grams. Using a random sample of size 200, an a% confidence interval for the population mean mass is found to have width 4 grams. Find a.
▶️Answer/Explanation
(a)
• 95% CI: (44.2, 45.8) grams
(b)
• z = 2.571
• Confidence level = 99.0% (3 sf)
6. [Maximum mark: 9]
The diagram shows the graph of the probability density function of a random variable X that takes values between -1 and 3 only. It is given that the graph is symmetrical about the line x = 1. Between x = -1 and x = 3 the graph is a quadratic curve.
The random variable S is such that E(S) = 2×E(X) and Var(S) = Var(X).
(a) On the grid below, sketch a quadratic graph for the probability density function of S.
The random variable T is such that E(T) = E(X) and Var(T) = ¼ Var(X).
(b) On the grid below, sketch a quadratic graph for the probability density function of T.
It is now given that:
f(x) = { (3/32)(3 + 2x – x²) -1 ≤ x ≤ 3,
0 otherwise }
(c) Given that P(1 – a < X < 1 + a) = 0.5, show that a³ – 12a + 8 = 0.
(d) Hence verify that 0.69 < a < 0.70.
▶️Answer/Explanation
(a)
• S has range [0,4] with peak at x=2
• Height at peak = 0.375 (3/8)
(b)
• T has range [0,2] with peak at x=1
• Height at peak = 0.75 (3/4)
(c)
• ∫ from (1-a) to (1+a) of f(x)dx = 0.5
• After integration and simplification: a³ – 12a + 8 = 0
(d)
• For a=0.69: -0.049 < 0
• For a=0.70: +0.043 > 0
• Sign change implies root in (0.69,0.70)
7. [Maximum mark: 10]
In the past Laxmi’s time, in minutes, for her journey to college had mean 32.5 and standard deviation 3.1. After a change in her route, Laxmi wishes to test whether the mean time has decreased.
(a) She notes her journey times for a random sample of 50 journeys and finds that the sample mean is 31.8 minutes. Carry out a hypothesis test, at the 8% significance level, of whether Laxmi’s mean journey time has decreased.
(b) Later Laxmi carries out a similar test with the same hypotheses, at the 8% significance level, using another random sample of size 50. Given that the population mean is now 31.5, find the probability of a Type II error.
▶️Answer/Explanation
(a)
• z = -1.597
• Critical value = -1.406
• Conclusion: Reject H₀ – evidence of decreased mean time
(b)
• Critical value = 31.88 minutes
• P(Type II error) = 0.191 (3 sf)