1. [Maximum mark: 7]
Each of a random sample of 80 adults gave an estimate, \( h \) metres, of the height of a particular building. The results were summarised as follows:
\( n = 80 \)
\( \Sigma h = 2048 \)
\( \Sigma h^2 = 52\,760 \)
(a) Calculate unbiased estimates of the population mean and variance.
(b) Using this sample, the upper boundary of an \( \alpha\% \) confidence interval for the population mean is 26.0. Find the value of \( \alpha \).
▶️Answer/Explanation
(a) Unbiased estimate of the population mean:
\( \text{Estimate of } \mu = \frac{2048}{80} = 25.6 \)
Unbiased estimate of the population variance:
\( \text{Estimate of } \sigma^2 = \frac{1}{79} \left( 52\,760 – \frac{2048^2}{80} \right) = 4.19 \) (3 sf)
(b) Using the confidence interval formula:
\( 25.6 + z \sqrt{\frac{4.19}{80}} = 26.0 \)
Solving for \( z \): \( z = 1.748 \)
From standard normal tables, \( \Phi(1.748) = 0.960 \)
Thus, \( \alpha = 2 \times 96.0 – 100 = 92.0 \)
2. [Maximum mark: 5]
In the past, the mean length of a particular variety of worm has been 10.3 cm, with standard deviation 2.6 cm. Following a change in the climate, it is thought that the mean length of this variety of worm has decreased. The lengths of a random sample of 100 worms of this variety are found, and the mean of this sample is 9.8 cm.
Assuming that the standard deviation remains at 2.6 cm, carry out a test at the 2% significance level of whether the mean length has decreased.
▶️Answer/Explanation
Hypotheses:
\( H_0: \mu = 10.3 \)
\( H_1: \mu < 10.3 \)
Test statistic:
\( z = \frac{9.8 – 10.3}{2.6 / \sqrt{100}} = -1.923 \)
Critical value:
At 2% significance level (one-tailed), \( z_{\text{critical}} = -2.054 \).
Conclusion:
Since \( -1.923 > -2.054 \), we do not reject \( H_0 \). There is no evidence at the 2% level that the mean length has decreased.
3. [Maximum mark: 6]
1.6% of adults in a certain town ride a bicycle. A random sample of 200 adults from this town is selected.
(a) Use a suitable approximating distribution to find the probability that more than 3 of these adults ride a bicycle.
(b) Justify your approximating distribution.
▶️Answer/Explanation
(a) Using Poisson approximation:
Mean \( \lambda = 200 \times 0.016 = 3.2 \)
\( P(X > 3) = 1 – P(X \leq 3) = 1 – e^{-3.2} \left(1 + 3.2 + \frac{3.2^2}{2} + \frac{3.2^3}{6}\right) = 0.397 \) (3 sf)
(b) Justification:
– \( n = 200 \) is large.
– \( p = 0.016 \) is small.
– \( np = 3.2 < 5 \), satisfying the conditions for Poisson approximation.
4. [Maximum mark: 8]
The number of faults in cloth made on a certain machine has a Poisson distribution with mean 2.4 per 10 m². An adjustment is made to the machine. It is required to test at the 5% significance level whether the mean number of faults has decreased. A randomly selected 30 m² of cloth is checked and the number of faults is found.
(a) State suitable null and alternative hypotheses for the test.
(b) Find the probability of a Type I error.
(c) Exactly 3 faults are found in the randomly selected 30 m² of cloth. Carry out the test at the 5% significance level.
(d) Given that the number of faults actually has a Poisson distribution with mean 0.5 per 10 m², find the probability of a Type II error.
▶️Answer/Explanation
(a) Hypotheses:
\( H_0: \lambda = 7.2 \) (2.4 × 3)
\( H_1: \lambda < 7.2 \)
(b) Probability of Type I error:
Critical region: \( X \leq 2 \) (since \( P(X \leq 2) \approx 0.0255 < 0.05 \))
Thus, Type I error probability = 0.0255 (3 sf)
(c) Test result:
Since 3 faults is outside the critical region (\( X \leq 2 \)), we do not reject \( H_0 \). No evidence of a decrease.
(d) Probability of Type II error:
Under \( \lambda = 1.5 \) (0.5 × 3), \( P(X > 2) = 1 – P(X \leq 2) = 0.191 \) (3 sf).
5. [Maximum mark: 6]
\( X \) is a random variable with distribution \( B(10, 0.2) \). A random sample of 160 values of \( X \) is taken.
(a) Find the approximate distribution of the sample mean, including the values of the parameters.
(b) Hence find the probability that the sample mean is less than 1.8.
▶️Answer/Explanation
(a) Approximate distribution:
Mean \( \mu = 10 \times 0.2 = 2 \)
Variance \( \sigma^2 = 10 \times 0.2 \times 0.8 = 1.6 \)
Sample mean distribution: \( \text{Normal}\left(2, \frac{1.6}{160}\right) \) or \( \text{Normal}(2, 0.01) \).
(b) Probability calculation:
\( P(\bar{X} < 1.8) = P\left(Z < \frac{1.8 – 2}{\sqrt{0.01}}\right) = P(Z < -2) = 0.0228 \) (3 sf).
6. [Maximum mark: 10]
The masses, in grams, of small and large bags of flour have the distributions \( N(510, 100) \) and \( N(1015, 324) \), respectively. André selects 4 small bags of flour and 2 large bags of flour at random.
(a) Find the probability that the total mass of these 6 bags of flour is less than 4130 g.
(b) Find the probability that the total mass of the 4 small bags is more than the total mass of the 2 large bags.
▶️Answer/Explanation
(a) Total mass distribution:
Mean \( = 4 \times 510 + 2 \times 1015 = 4070 \) g
Variance \( = 4 \times 100 + 2 \times 324 = 1048 \) g²
\( P(T < 4130) = P\left(Z < \frac{4130 – 4070}{\sqrt{1048}}\right) = P(Z < 1.853) = 0.968 \) (3 sf).
(b) Difference in masses:
Mean \( = 4 \times 510 – 2 \times 1015 = 10 \) g
Variance \( = 4 \times 100 + 2 \times 324 = 1048 \) g²
\( P(D > 0) = P\left(Z > \frac{0 – 10}{\sqrt{1048}}\right) = P(Z > -0.309) = 0.621 \) (3 sf).
7. [Maximum mark: 8]
The diagram shows the graph of the probability density function, \( f \), of a random variable \( X \) which takes values between \(-3\) and \( 2 \) only. The graph is symmetrical about the line \( x = -0.5 \), and \( P(X < 0) = p \).
(a) Find \( P(-1 < X < 0) \) in terms of \( p \).
(b) The probability density function is given by:
\( f(x) = \begin{cases} a – b(x^2 + x) & -3 \leq x \leq 2, \\ 0 & \text{otherwise}, \end{cases} \)
where \( a \) and \( b \) are positive constants.
(i) Show that \( 30a – 55b = 6 \).
(ii) Determine the values of \( a \) and \( b \).
▶️Answer/Explanation
(a) By symmetry:
\( P(-1 < X < 0) = 2p – 1 \).
(b)(i) Integrating \( f(x) \):
\( \int_{-3}^2 f(x) \, dx = 1 \) leads to \( 30a – 55b = 6 \).
(b)(ii) Using boundary conditions:
At \( x = -3 \), \( f(-3) = 0 \): \( a – 6b = 0 \).
Solving the system: \( a = \frac{36}{125} \), \( b = \frac{6}{125} \).