Question 1
(a) Topic: 6.4 Sampling and estimation
(b) Topic: 6.4 Sampling and estimation
The heights of a certain species of deer are known to have standard deviation 0.35 m. A zoologist takes a random sample of 150 of these deer and finds that the mean height of the deer in the sample is 1.42 m.
(a) Calculate a 96% confidence interval for the population mean height.
(b) Bubay says that 96% of deer of this species are likely to have heights that are within this confidence interval. Explain briefly whether Bubay is correct.
▶️Answer/Explanation
Solution :-
Part (a): Calculating the 96% Confidence Interval
Given:
- Sample mean (\(\bar{x}\)) = 1.42 m
- Population standard deviation (\(\sigma\)) = 0.35 m
- Sample size (n) = 150
- Confidence level = 96%
For a 96% confidence interval, the z-score (z) is found using a standard normal distribution table or calculator. The z-score corresponding to 96% confidence is approximately 2.054 or 2.055.
The formula for the confidence interval is:
\(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\)
Plugging in the values:
\(1.42 \pm z \frac{0.35}{\sqrt{150}}\)
Lower bound: 1.36 m
Upper bound: 1.48 m
Therefore, the 96% confidence interval is approximately (1.36 m, 1.48 m) (3 sf).
Part (b): Explanation of Bubay’s statement
Bubay is incorrect.
The confidence interval (CI) is about the population mean, not individual values.
Question 2
Topic: 6.2 Linear combinations of random variables
The masses, in kilograms, of small and large bags of wheat have the independent distributions N(16.0, 0.4) and N(51.0, 0.9) respectively.
Find the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag.
▶️Answer/Explanation
Solution :-
Let \(S_1, S_2, S_3\) be the masses of the three small bags, and \(L\) be the mass of the large bag.
1. Mean of \(S_1 + S_2 + S_3 – L\):
\(E(S_1 + S_2 + S_3 – L) = 16 \times 3 – 51 = 48 – 51 = -3\)
(B1: Oe, using \(L – (S_1 + S_2 + S_3)\))
2. Variance of \(S_1 + S_2 + S_3 – L\):
\(Var(S_1 + S_2 + S_3 – L) = 3 \times 0.4 + 0.9 = 1.2 + 0.9 = 2.1\)
(M1)
3. Standardization:
\(\frac{0 – (-3)}{\sqrt{2.1}} = \frac{3}{\sqrt{2.1}} \approx 2.070\)
(M1: For standardizing with their values)
4. Probability Calculation:
\(1 – \Phi(2.070)\)
(M1: For area consistent with their values)
5. Final Answer:
\(0.0192\) (3 sf)
(A1)
Therefore, the probability that the total mass of 3 randomly chosen small bags is greater than the mass of one randomly chosen large bag is approximately 0.0192.
Question 3
(a) Topic: 6.4 Sampling and estimation
(b) Topic: 6.4 Sampling and estimation
The times, \(T\) minutes, taken by a random sample of 75 students to complete a test were noted. The results were summarised by \(\Sigma t = 230\) and \(\Sigma t^2 = 930\).
(a) Calculate unbiased estimates of the population mean and variance of \(T\).
You should now assume that your estimates from part (a) are the true values of the population mean and variance of \(T\).
(b) The times taken by another random sample of 75 students were noted, and the sample mean, \(\overline{T}\), was found.
Find the value of \(a\) such that \(P(\overline{T} > a) = 0.234\).
▶️Answer/Explanation
Solution :-
Part (a): Calculating the 96% Confidence Interval
Given:
- Sample mean (\(\bar{t}\)) = \(\frac{230}{75} = 3.066… \text{ or } 3.07 \text{ (3 sf) or } \frac{46}{15}\) (B1)
- Population standard deviation (\(S^2\)) = \(\frac{75}{74} \left( \frac{930}{75} – \left( \frac{230}{75} \right)^2 \right) \text{ or } \frac{1}{74} \left( 930 – \frac{230^2}{75} \right)\) (M1: Use of correct formula)
- Sample size (n) = 75
- Confidence level = 96%
For a 96% confidence interval, the z-score (z) is found using a standard normal distribution table or calculator. The z-score corresponding to 96% confidence is approximately 2.054 or 2.055.
The formula for the confidence interval is:
\(\bar{x} \pm z \frac{\sigma}{\sqrt{n}}\)
Plugging in the values:
\(1.42 \pm z \frac{0.35}{\sqrt{150}}\)
Part (b): Finding the value of \(a\)
\([\Phi^{-1} (1 – 0.234)] = 0.726\) (B1)
\(\frac{a – ‘3.0667’}{\sqrt{\frac{‘3.04’}{75}}} = \pm ‘0.726’\) (M1: Ft their 0.726 but must be a z value. Note using 0.766 is M0. Must have sqrt 75.)
\(a = 3.21 \text{ (3 sf) }\)
Question 4
(a) Topic: 6.3 Continuous random variables
(b) Topic: 6.3 Continuous random variables
A random variable X has probability density function f defined by:
\( f(x) = \begin{cases} \frac{a}{x^2} – \frac{18}{x^3} & 2 \le x \le 3 \\ 0 & \text{otherwise} \end{cases} \)
where \( a \) is a constant.
(a) Show that \( a = \frac{27}{2} \).
(b) Show that \( E(X) = \frac{27}{2} \ln\frac{3}{2} – 3 \).
▶️Answer/Explanation
Solution :-
Part (a): Showing \( a = \frac{27}{2} \)
\( \int_{2}^{3} \left( \frac{a}{x^2} – \frac{18}{x^3} \right) dx = 1 \)
\( \left[ -\frac{a}{x} + \frac{9}{x^2} \right]_{2}^{3} = 1 \)
\( \left[ -\frac{a}{3} + 1 + \frac{a}{2} – \frac{9}{4} \right] = 1 \)
\( a = \frac{27}{2} \) (As Given)
Part (b): Showing \( E(X) = \frac{27}{2} \ln\frac{3}{2} – 3 \)
\( \int_{2}^{3} \left( \frac{27}{2x} – \frac{18}{x^2} \right) dx \)
\( \left[ \frac{27}{2} \ln x + \frac{18}{x} \right]_{2}^{3} \) or \( \left[ \frac{27}{2} \ln 2x + \frac{18}{x} \right]_{2}^{3} \)
\( = \frac{27}{2} \ln 3 + 6 – \frac{27}{2} \ln 2 – 9 = \frac{27}{2} \ln \frac{3}{2} – 3 \) (As Given)
Question 5
(a) Topic: 6.5 Hypothesis tests
(b) Topic: 6.5 Hypothesis tests
The lengths, in centimetres, of worms of a certain kind are normally distributed with mean \( \mu \) and standard deviation 2.3. An article in a magazine states that the value of \( \mu \) is 12.7. A scientist wishes to test whether this value is correct. He measures the lengths, \( x \) cm, of a random sample of 50 worms of this kind and finds that \( \sum x = 597.1 \). He plans to carry out a test, at the 1% significance level, of whether the true value of \( \mu \) is different from 12.7.
(a) State, with a reason, whether he should use a one-tailed or a two-tailed test.
(b) Carry out the test.
▶️Answer/Explanation
Solution :-
Part (a): One-tailed or Two-tailed Test
Two-tailed test because looking for difference.
Part (b): Carrying out the test
1. State the null and alternative hypotheses:
\( H_0: \mu = 12.7 \)
\( H_1: \mu \neq 12.7 \)
2. Calculate the test statistic (z-score):
\( z = \frac{\frac{597.1}{50} – 12.7}{\frac{2.3}{\sqrt{50}}} \)
\( z = \frac{11.942 – 12.7}{\frac{2.3}{\sqrt{50}}} \)
\( z \approx -2.330 \)
(Accept -2.336 or -2.337)
3. Compare the test statistic with the critical values:
\( |-2.330| < 2.576 \) or \( 2.330 < 2.576 \)
or \( 0.00989 > 0.005 \)
or \( 0.0097 > 0.005 \)
(Accept 2.574 to 2.579)
Or use of CV:
\( 12.7 – 2.576 \times \frac{2.3}{\sqrt{50}} = 11.862 \)
\( 11.942 > 11.862 \)
4. Conclusion:
[Not reject \( H_0 \)] There is insufficient evidence to suggest that \( \mu \) is not 12.7.
(In context, not definite, e.g. not \( \mu = 12.7 \)).
(No contradictions)
(SC use of 1 tailed test can score B0M1A1M1 for comparison with 0.01 A0 max 3/5)
Question 6
(a) Topic: 6.1 The Poisson distribution
(b) Topic: 6.1 The Poisson distribution
(c) Topic: 6.1 The Poisson distribution
The numbers of customers arriving at service desks A and B during a 10-minute period have the independent distributions \( Po(1.8) \) and \( Po(2.1) \) respectively.
(a) Find the probability that during a randomly chosen 15-minute period more than 2 customers will arrive at desk A.
(b) Find the probability that during a randomly chosen 5-minute period the total number of customers arriving at both desks is less than 4.
(c) An inspector waits at desk B. She wants to wait long enough to be 90% certain of seeing at least one customer arrive at the desk. Find the minimum time for which she should wait, giving your answer correct to the nearest minute.
▶️Answer/Explanation
Solution :-
Part (a):
\( \lambda = 1.8 \times \frac{15}{10} = 2.7 \)
\( P(X > 2) = 1 – P(X \leq 2) = 1 – [P(X=0) + P(X=1) + P(X=2)] \)
\( = 1 – \left[ e^{-2.7} \times \left(1 + 2.7 + \frac{2.7^2}{2} \right) \right] \)
\( = 1 – \left[ e^{-2.7} \times (1 + 2.7 + 3.645) \right] \)
\( = 1 – (0.06721 + 0.1815 + 0.2450) \)
\( = 0.506 \) (3 sf)
Part (b):
\( \lambda_A (5 \text{ min}) = 1.8 \times \frac{5}{10} = 0.9 \)
\( \lambda_B (5 \text{ min}) = 2.1 \times \frac{5}{10} = 1.05 \)
\( \lambda_{total} = 0.9 + 1.05 = 1.95 \)
\( P(X < 4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) \)
\( = e^{-1.95} \times \left[ 1 + 1.95 + \frac{1.95^2}{2} + \frac{1.95^3}{6} \right] \)
\( = e^{-1.95} \times [1 + 1.95 + 1.90125 + 1.2358] \)
\( = 0.1423 + 0.2774 + 0.2705 + 0.1758 \)
\( = 0.866 \)
Part (c):
\( P(X \geq 1) \geq 0.90 \)
\( 1 – P(X=0) \geq 0.90 \)
\( 1 – e^{-2.1x} \geq 0.90 \)
\( e^{-2.1x} \leq 0.1 \)
\( -2.1x \leq \ln(0.1) \)
\( x \geq \frac{-\ln(0.1)}{2.1} \)
\( x \geq \frac{2.3026}{2.1} \)
\( x \geq 1.096 \)
She must wait for at least 11 minutes.
Question 7
(a) Topic: 6.5 Hypothesis tests
(b) Topic: 6.5 Hypothesis tests
(c) Topic: 6.5 Hypothesis tests
(d) Topic: 6.5 Hypothesis tests
The number of accidents per year on a certain road has the distribution \(\text{Po}(\lambda)\). In the past, the value of \(\lambda\) was 1.1. Recently, a new speed limit was imposed, and the council wishes to test whether the value of \(\lambda\) has decreased. The council notes the total number, \(X\), of accidents during two randomly chosen years after the speed limit was introduced and carries out a test at the 5% significance level.
(a) Calculate the probability of a Type I error.
(b) Given that \(X = 2\), carry out the test.
(c) The council decides to carry out another similar test at the 5% significance level using the same hypotheses and two different randomly chosen years. Given that the true value of \(\lambda\) is 0.6, calculate the probability of a Type II error.
(d) Using \(\lambda = 0.6\) and a suitable approximating distribution, find the probability that there will be more than 10 accidents in 30 years.
▶️ Answer/Explanation
(a) Calculate the probability of a Type I error.
\[ P(X \leq 2) = e^{-6.6} \left( 1 + 6.6 + \frac{6.6^2}{2} \right) = 0.0400 \ (< 0.05) \]
or \[ e^{-6.6} \left( 1 + 6.6 + 21.78 \right) \]
or \[ 0.001360 + 0.008978 + 0.02963 \]
\[ P(X \leq 3) = e^{-6.6} \left( 1 + 6.6 + \frac{6.6^2}{2} \right) + e^{-6.6} \cdot \frac{6.6^3}{6} = 0.105 \ (> 0.05) \]
\[ P(\text{Type I error}) = 0.0400 \] (3 sf)
(b) Given that \(X = 2\), carry out the test.
\[ H_0: \lambda = 6.6, \quad H_1: \lambda < 6.6 \]
\[ P(X \leq 2) = 0.0400 \ (< 0.05) \]
Since \(0.0400 < 0.05\), we reject \(H_0\).
There is evidence to suggest that the mean number of accidents has decreased.
(c) Calculate the probability of a Type II error.
\[ P(X > 2) = 1 – e^{-1.2} \left( 1 + 1.2 + \frac{1.2^2}{2} \right) \]
or \[ = 1 – e^{-1.2} \left( 1 + 1.2 + 0.72 \right) \]
or \[ = 1 – (0.3012 + 0.3614 + 0.2169) \]
\[ = 0.121 \] (3 sf) or \(0.120\).
(d) Find the probability of more than 10 accidents in 30 years.
Approximating with \(N(18, 18)\), we standardize:
\[ Z = \frac{10.5 – 18}{\sqrt{18}} = -1.768 \]
\[ P(X > 10) = P(Z > -1.768) = \Phi(1.768) \]
Using standard normal tables:
\[ P(X > 10) = 0.961 \text{ or } 0.962 \] (3 sf)