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
Mean: \( \frac{5.9 + 6.5 + 6.7 + 5.9 + 6.9 + 6.0 + 6.4 + 6.2 + 5.8 + 5.8}{10} = 6.21 \) m
Variance: \( \frac{1}{9} \left( \sum x^2 – \frac{(\sum x)^2}{10} \right) = \frac{1}{9} (387.05 – \frac{62.1^2}{10}) = 0.157 \) (3 sf)
2. [Maximum mark: 8]
A spinner has five sectors, each printed with a different colour.
(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 to determine whether the spinner is biased so that it lands on red fewer times than if it were fair.
(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: \( H_0: p = 0.2 \), \( H_1: p < 0.2 \)
– Calculation: \( P(X ≤ 4) = 0.0759 \) (using B(40,0.2))
– Conclusion: Since 0.0759 > 0.05, do not reject \( H_0 \). There is not enough evidence to suggest the spinner is biased against red. (b)
– Calculation: \( P(X ≤ 3) = 0.0285 \)
– Since
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: 5.2 × 0.5 = 2.6
– Probability: \( P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – e^{-2.6}(1 + 2.6 + \frac{2.6^2}{2}) = 0.482 \) (3 sf) (b)
– Normal approximation: \( X \sim N(624, 624) \) (where μ = 5.2 × 120 = 624)
– Continuity correction: \( P(X ≥ 649.5) \)
– z-score: \( \frac{649.5 – 624}{\sqrt{624}} = 1.021 \)
– Probability: \( 1 – Φ(1.021) = 0.154 \) (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: \( 3 × (2500 + 3700) = 18,600 \) kg
– Variance: \( 3 × (120² + 130²) = 93,900 \)
– Standard deviation: \( \sqrt{93,900} = 306 \) kg (3 sf) (b)
– Expected profit: \( 1.5 × 2500 – 0.2 × 3700 = 3,010 \)
– Variance: \( 1.5² × 120² + 0.2² × 130² = 33,076 \)
– z-score: \( \frac{3000 – 3010}{\sqrt{33076}} = -0.055 \)
– Probability: \( Φ(-0.055) = 0.478 \) (3 sf)
5. [Maximum mark: 7]
(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)
– Confidence interval: \( 45 ± 1.96 × \frac{6}{\sqrt{200}} \)
– Result: \( (44.2, 45.8) \) grams (3 sf) (b)
– Equation: \( 2 × z × \frac{11}{\sqrt{200}} = 4 \) → \( z = 2.571 \)
– Confidence level: \( Φ(2.571) = 0.9949 \) → \( a = 99.0 \)% (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.
(a) The random variable S is such that E(S) = 2×E(X) and Var(S) = Var(X). On the grid below, sketch a quadratic graph for the probability density function of S.
(b) The random variable T is such that E(T) = E(X) and Var(T) = ¼ Var(X). On the grid below, sketch a quadratic graph for the probability density function of T.
It is now given that: \[ f(x) = \begin{cases} \frac{3}{32}(3 + 2x – x^2) & -1 \leq x \leq 3, \\ 0 & \text{otherwise}. \end{cases} \]
(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)
– Curve from x=0 to x=4, symmetrical about x=2 – Highest point at (2, 0.375) (b)
– Curve from x=0 to x=2, symmetrical about x=1 – Highest point at (1, 0.75) (c)
– \(\frac{3}{32} \int_{1-a}^{1+a} (3 + 2x – x^2) dx = 0.5\) – After integration: \(\frac{3}{32} [3x + x^2 – \frac{x^3}{3}]_{1-a}^{1+a} = 0.5\) – Substituting limits and simplifying leads to: \(a^3 – 12a + 8 = 0\) (d)
– For a=0.69: \(0.69^3 – 12×0.69 + 8 ≈ 0.049 > 0\) – For a=0.70: \(0.70^3 – 12×0.70 + 8 ≈ -0.057 < 0\) – Since continuous function changes sign, root is between 0.69 and 0.70
7. [Maximum mark: 10]
In the past Laxmi’s journey time to college had mean 32.5 minutes and standard deviation 3.1 minutes. After a route change:
(a) She tests whether the mean time has decreased using a sample of 50 journeys with mean 31.8 minutes. Carry out a hypothesis test at the 8% significance level.
(b) Later she carries out a similar test with the same hypotheses at the 8% significance level using another sample of size 50. Given that the population mean is now 31.5 minutes, find the probability of a Type II error.
▶️Answer/Explanation
(a)
– Hypotheses: \( H_0: μ = 32.5 \), \( H_1: μ < 32.5 \)
– z-test: \( \frac{31.8 – 32.5}{3.1/\sqrt{50}} = -1.597 \)
– Critical value: -1.406 (for 8% significance)
– Conclusion: Since -1.597 < -1.406, reject \( H_0 \). There is evidence that the mean journey time has decreased. (b)
– Critical value: \( 32.5 – 1.406 × \frac{3.1}{\sqrt{50}} = 31.88 \)
– Type II error probability: \( P(\bar{X} > 31.88 | μ = 31.5) \)
– z-score: \( \frac{31.88 – 31.5}{3.1/\sqrt{50}} = 0.866 \)
– Probability: \( 1 – Φ(0.866) = 0.193 \) (3 sf)