1. [Maximum mark: 3]
The number of characters in emails sent by a particular company is modelled by the distribution \( N(1250, 480^2) \). Find the probability that the mean number of characters in a random sample of 100 emails sent by the company is more than 1300.
▶️Answer/Explanation
Solution:
Standardize the sample mean:
\( z = \frac{1300 – 1250}{\frac{480}{\sqrt{100}}} = \frac{50}{48} \approx 1.042 \)
Find the probability:
\( P(Z > 1.042) = 1 – \Phi(1.042) \approx 0.149 \) (3 s.f.)
2. [Maximum mark: 6]
Anton believes that 10% of students at his college are left-handed. Aliya believes that this is an underestimate. She plans to carry out a hypothesis test of the null hypothesis \( p = 0.1 \) against the alternative hypothesis \( p > 0.1 \), where \( p \) is the actual proportion of students at the college that are left-handed. She chooses a random sample of 20 students from the college. She will reject the null hypothesis if at least 5 of these students are left-handed.
(a) Explain what is meant by a Type I error in this context.
(b) Find the probability of a Type I error in the test.
(c) Given that the true value of \( p \) is 0.3, find the probability of a Type II error in the test.
▶️Answer/Explanation
(a) A Type I error occurs when Aliya concludes that more than 10% of the students are left-handed when, in reality, the true proportion is 10%.
(b) Calculate \( P(X \geq 5) \) under \( H_0 \) (\( p = 0.1 \)):
Using binomial distribution \( B(20, 0.1) \):
\( P(X \geq 5) = 1 – P(X \leq 4) \approx 0.0432 \) (3 s.f.).
(c) Calculate \( P(X \leq 4) \) under \( p = 0.3 \):
Using binomial distribution \( B(20, 0.3) \):
\( P(X \leq 4) \approx 0.238 \) (3 s.f.).
3. [Maximum mark: 9]
Batteries of type A are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type B, has a shorter mean life than type A batteries.
(a) Give a reason for using a sample rather than the whole population in carrying out this test.
(b) Test, at the 2% significance level, whether type B batteries have a shorter mean life than type A batteries.
(c) Calculate a 94% confidence interval for the population mean life of type B batteries.
▶️Answer/Explanation
(a) Testing the entire population would be impractical due to time, cost, or the destructive nature of the test.
(b) Hypotheses:
\( H_0: \mu = 150 \)
\( H_1: \mu < 150 \)
Test statistic:
\( z = \frac{147 – 150}{\frac{15}{\sqrt{120}}} \approx -2.191 \)
Critical value at 2% significance level: \( -2.054 \)
Since \( -2.191 < -2.054 \), reject \( H_0 \). There is evidence that type B batteries have a shorter mean life.
(c) 94% confidence interval:
\( 147 \pm 1.881 \times \frac{15}{\sqrt{120}} \)
Interval: \( (144.4, 149.6) \) hours (3 s.f.).
4. [Maximum mark: 10]
Each box of Seeds & Raisins contains \( S \) grams of seeds and \( R \) grams of raisins. The weight of a box, when empty, is \( B \) grams. \( S \), \( R \), and \( B \) are independent random variables, where \( S \sim N(300, 45) \), \( R \sim N(200, 25) \), and \( B \sim N(15, 4) \). A full box of Seeds & Raisins is chosen at random.
(a) Find the probability that the total weight of the box and its contents is more than 500 grams.
(b) Find the probability that the weight of seeds in the box is less than 1.4 times the weight of raisins in the box.
▶️Answer/Explanation
(a) Total weight \( T = S + R + B \sim N(515, 74) \):
\( P(T > 500) = P\left(Z > \frac{500 – 515}{\sqrt{74}}}\right) \approx P(Z > -1.744) \approx 0.959 \) (3 s.f.).
(b) Define \( D = S – 1.4R \sim N(20, 94) \):
\( P(D < 0) = P\left(Z < \frac{0 – 20}{\sqrt{94}}}\right) \approx P(Z < -2.063) \approx 0.0196 \) (3 s.f.).
5. [Maximum mark: 9]
The number of clients who arrive at an information desk has a Poisson distribution with mean 2.2 per 5-minute period.
(a) Find the probability that, in a randomly chosen 15-minute period, exactly 6 clients arrive at the desk.
(b) If more than 4 clients arrive during a 5-minute period, they cannot all be served. Find the probability that, during a randomly chosen 5-minute period, not all the clients who arrive at the desk can be served.
(c) Use a suitable approximating distribution to find the probability that, during a randomly chosen 1-hour period, fewer than 20 clients arrive at the desk.
▶️Answer/Explanation
(a) For 15 minutes, \( \lambda = 6.6 \):
\( P(X = 6) = e^{-6.6} \times \frac{6.6^6}{6!} \approx 0.156 \) (3 s.f.).
(b) \( P(X > 4) = 1 – P(X \leq 4) \approx 0.0725 \) (3 s.f.).
(c) Approximate with \( N(26.4, 26.4) \):
\( P(X < 20) \approx P\left(Z < \frac{19.5 – 26.4}{\sqrt{26.4}}}\right) \approx P(Z < -1.343) \approx 0.0897 \) (3 s.f.).
6. [Maximum mark: 4]
A random sample of 5 values of a variable \( X \) is given below: 2, 3, 3, 5, \( a \).
(a) Find an expression, in terms of \( a \), for the mean of these values.
(b) Find and simplify a quadratic equation in terms of \( a \) and hence find the value of \( a \).
▶️Answer/Explanation
(a) Mean \( = \frac{2 + 3 + 3 + 5 + a}{5} = \frac{13 + a}{5} \).
(b) Using the unbiased variance formula:
\( \frac{5}{4}\left(\frac{47 + a^2}{5} – \left(\frac{13 + a}{5}\right)^2\right) = 4 \)
Simplifies to \( 2a^2 – 13a – 7 = 0 \).
Solution: \( a = 7 \) (since \( a > 0 \)).
7. [Maximum mark: 10]
The random variables \( X \) and \( W \) have probability density functions \( f \) and \( g \) defined as follows:
\[ f(x) = \begin{cases} p(a^2 – x^2) & 0 \leq x \leq a, \\ 0 & \text{otherwise}, \end{cases} \]
\[ g(w) = \begin{cases} q(a^2 – w^2) & -a \leq w \leq a, \\ 0 & \text{otherwise}, \end{cases} \]
where \( a \), \( p \), and \( q \) are constants.
(a) (i) Write down the value of \( P(X \geq 0) \).
(ii) Write down the value of \( P(W \geq 0) \).
(iii) Write down an expression for \( q \) in terms of \( p \) only.
(b) Given that \( E(X) = 3 \), find the value of \( a \).
▶️Answer/Explanation
(a)(i) \( P(X \geq 0) = 1 \) (since \( X \) is defined for \( 0 \leq x \leq a \)).
(a)(ii) \( P(W \geq 0) = 0.5 \) (symmetry about 0).
(a)(iii) \( q = \frac{1}{2}p \).
(b) Integrate \( f(x) \) to find \( p \):
\( \int_0^a p(a^2 – x^2) dx = 1 \Rightarrow p = \frac{3}{2a^3} \).
Calculate \( E(X) \):
\( \int_0^a x \cdot p(a^2 – x^2) dx = 3 \Rightarrow a = 8 \).