1. [Maximum mark: 7]
The diameters, \( x \) millimetres, of a random sample of 200 discs made by a certain machine were recorded. The results are summarised below.
\( n = 200 \)
\( \Sigma x = 2520 \)
\( \Sigma x^2 = 31852 \)
(a) Calculate a 95% confidence interval for the population mean diameter.
(b) Jean chose 40 random samples and used each sample to calculate a 95% confidence interval for the population mean diameter. How many of these 40 confidence intervals would be expected to include the true value of the population mean diameter?
▶️Answer/Explanation
(a) \[ \text{Est}(\mu) = \frac{2520}{200} = 12.6 \] \[ \text{Est}(\sigma^2) = \frac{200}{199} \left( \frac{31852}{200} – 12.6^2 \right) = 0.5025 \] \[ z = 1.96 \] \[ 12.6 \pm 1.96 \times \sqrt{\frac{0.5025}{200}} \] Confidence interval: \( 12.5 \) to \( 12.7 \) (3 s.f.)
(b) \[ 0.95 \times 40 = 38 \]
2. [Maximum mark: 5]
Arvind uses an ordinary fair 6-sided die to play a game. He believes he has a system to predict the score when the die is thrown. Before each throw of the die, he writes down what he thinks the score will be. He claims that he can write the correct score more often than he would if he were just guessing. His friend Laxmi tests his claim by asking him to write down the score before each of 15 throws of the die. Arvind writes the correct score on exactly 5 out of 15 throws. Test Arvind’s claim at the 10% significance level.
▶️Answer/Explanation
Hypotheses:
\( H_0: p = \frac{1}{6} \) (Arvind is guessing)
\( H_1: p > \frac{1}{6} \) (Arvind can predict scores better than guessing)
Test:
Using binomial distribution \( X \sim B(15, \frac{1}{6}) \):
Calculate \( P(X \geq 5) = 1 – P(X \leq 4) \approx 0.0898 \) (3 s.f.)
Conclusion:
Since \( 0.0898 < 0.10 \), reject \( H_0 \). There is evidence at the 10% level that Arvind can predict scores better than guessing.
3. [Maximum mark: 5]
The lengths, in centimetres, of two types of insect, A and B, are modelled by the random variables \( X \sim N(6.2, 0.36) \) and \( Y \sim N(2.4, 0.25) \) respectively. Find the probability that the length of a randomly chosen type A insect is greater than the sum of the lengths of 3 randomly chosen type B insects.
▶️Answer/Explanation
Let \( D = X – (Y_1 + Y_2 + Y_3) \):
\[ E(D) = 6.2 – 3 \times 2.4 = -1 \]
\[ \text{Var}(D) = 0.36 + 3 \times 0.25 = 1.11 \]
Standardize \( D \):
\[ P(D > 0) = P\left(Z > \frac{0 – (-1)}{\sqrt{1.11}}\right) = P(Z > 0.949) \approx 0.171 \text{ (3 s.f.)} \]
4. [Maximum mark: 8]
The independent random variables \( X \) and \( Y \) have distributions \( \text{Po}(2) \) and \( \text{B}(20, \frac{1}{4}) \) respectively.
(a) Find the mean and standard deviation of \( X – 3Y \.
(b) Find \( P(Y = 15X) \).
▶️Answer/Explanation
(a)
\[ E(X) = 2, \text{Var}(X) = 2 \]
\[ E(Y) = 20 \times \frac{1}{4} = 5, \text{Var}(Y) = 20 \times \frac{1}{4} \times \frac{3}{4} = \frac{15}{4} \]
For \( X – 3Y \):
\[ E(X – 3Y) = 2 – 3 \times 5 = -13 \]
\[ \text{Var}(X – 3Y) = 2 + 9 \times \frac{15}{4} = 35.75 \]
Standard deviation: \( \sqrt{35.75} \approx 5.98 \) (3 s.f.)
(b)
Possible pairs: \( (0, 0) \) and \( (1, 15) \).
\[ P(Y = 15X) = P(X=0, Y=0) + P(X=1, Y=15) \]
\[ = e^{-2} \times \left(\frac{3}{4}\right)^{20} + e^{-2} \times 2 \times \binom{20}{15} \left(\frac{1}{4}\right)^{15} \left(\frac{3}{4}\right)^5 \]
\[ \approx 0.000430 \text{ (3 s.f.)} \]
5. [Maximum mark: 10]
Cars arrive at a fuel station at random and at a constant average rate of 13.5 per hour.
(a) Find the probability that more than 4 cars arrive during a 20-minute period.
(b) Use an approximating distribution to find the probability that the number of cars that arrive during a 12-hour period is between 150 and 160 inclusive.
(c) Independently of cars, trucks arrive at the fuel station at random and at a constant average rate of 3.6 per 15-minute period. Find the probability that the total number of cars and trucks arriving at the fuel station during a 10-minute period is more than 3 and less than 7.
▶️Answer/Explanation
(a)
For 20 minutes, \( \lambda = \frac{13.5}{3} = 4.5 \):
\[ P(X > 4) = 1 – P(X \leq 4) \approx 0.468 \text{ (3 s.f.)} \]
(b)
For 12 hours, \( \lambda = 13.5 \times 12 = 162 \). Approximate with \( N(162, 162) \):
\[ P(150 \leq X \leq 160) \approx P(149.5 \leq X \leq 160.5) \]
\[ \approx \Phi\left(\frac{160.5 – 162}{\sqrt{162}}\right) – \Phi\left(\frac{149.5 – 162}{\sqrt{162}}\right) \approx 0.290 \text{ (3 s.f.)} \]
(c)
For 10 minutes:
Cars: \( \lambda = \frac{13.5}{6} = 2.25 \)
Trucks: \( \lambda = 3.6 \times \frac{2}{3} = 2.4 \)
Total \( \lambda = 4.65 \):
\[ P(4 \leq X \leq 6) \approx 0.494 \text{ (3 s.f.)} \]
6. [Maximum mark: 9]
A random variable \( X \) has probability density function \( f \). The graph of \( f(x) \) is a straight line segment parallel to the \( x \)-axis from \( x = 0 \) to \( x = a \), where \( a \) is a positive constant.
(a) State, in terms of \( a \), the median of \( X \).
(b) Find \( P(X > \frac{3}{4}a) \).
(c) Show that \( \text{Var}(X) = \frac{1}{12}a^2 \).
(d) Given that \( P(X < b) = p \), where \( 0 < b < \frac{1}{2}a \), find \( P\left(\frac{1}{3}b < X < a – \frac{1}{3}b\right) \) in terms of \( p \).
▶️Answer/Explanation
(a) Median = \( \frac{a}{2} \)
(b) \( P(X > \frac{3}{4}a) = \frac{1}{4} \)
(c)
Since \( f(x) = \frac{1}{a} \) for \( 0 \leq x \leq a \):
\[ E(X) = \frac{a}{2} \]
\[ E(X^2) = \int_0^a x^2 \cdot \frac{1}{a} \, dx = \frac{a^2}{3} \]
\[ \text{Var}(X) = \frac{a^2}{3} – \left(\frac{a}{2}\right)^2 = \frac{a^2}{12} \]
(d)
Given \( P(X < b) = p \), then \( P\left(\frac{1}{3}b < X < a – \frac{1}{3}b\right) = 1 – \frac{2p}{3} \).
7. [Maximum mark: 6]
In the past, the mean time for Jenny’s morning run was 28.2 minutes. She does some extra training and she wishes to test whether her mean time has been reduced. After the training Jenny takes a random sample of 40 morning runs. She decides that if the sample mean run time is less than 27 minutes she will conclude that the training has been effective. You may assume that, after the training, Jenny’s run time has a standard deviation of 4.0 minutes.
(a) State suitable null and alternative hypotheses for Jenny’s test.
(b) Find the probability that Jenny will make a Type I error.
(c) Jenny found that the sample mean run time was 27.2 minutes. Explain briefly whether it is possible for her to make a Type I error or a Type II error or both.
▶️Answer/Explanation
(a)
\( H_0: \mu = 28.2 \) (no change)
\( H_1: \mu < 28.2 \) (mean time reduced)
(b)
Type I error occurs if \( H_0 \) is rejected when true:
\[ P(\bar{X} < 27 \mid \mu = 28.2) = P\left(Z < \frac{27 – 28.2}{4/\sqrt{40}}\right) \approx 0.0289 \text{ (3 s.f.)} \]
(c)
Since \( \bar{X} = 27.2 > 27 \), Jenny does not reject \( H_0 \). Thus:
– Type I error is impossible (since \( H_0 \) is not rejected).
– Type II error is possible (if \( \mu < 28.2 \) but \( H_0 \) is not rejected).