An aerospace manufacturer produces high-precision glass panels and rare-earth magnetic components.
The thickness, TTT mm, of a glass panel is known to follow a normal distribution with unknown mean μ\muμ and a fixed standard deviation of 0.080.080.08 mm.
A quality control inspector selects a random sample of 404040 panels. The sum of the thicknesses for these 404040 panels is recorded as 128.8128.8128.8 mm.
Determine a 98%98\%98% confidence interval for the mean thickness of the glass panels, giving your limits to three decimal places. (4)
Explain why it was not necessary to invoke the Central Limit Theorem to find the interval in part (a). (1)
The production manager asserts that the panels are manufactured to a mean thickness of 3.253.253.25 mm.
Evaluate whether the production manager's assertion is consistent with your result from part (a). (2)
The mass, MMM grams, of a magnetic component is modelled by the distribution M∼N(15.5,0.42)M \sim N(15.5, 0.4^2)M∼N(15.5,0.42).
A batch of 101010 such components is selected at random.
Calculate the probability that the mean mass of these 101010 components is less than 15.315.315.3 grams. (3)
390 exam-style questions on OCR (MEI) A Level Maths 2.4 Probability Distributions, covering 2.4.1 Recognise binomial situations, 2.4.2 Probability of success p, 2.4.3 Calculate binomial probabilities, 2.4.4 Mean of the binomial distribution, 2.4.5 Expected frequencies for binomial, 2.4.6 Probability functions and discrete random variables, 2.4.7 Numerical probabilities for a simple distribution, 2.4.8 Normal distribution as a model (A-level only), 2.4.9 Shape of the Normal curve (A-level only), 2.4.10 Linear transformation and standardising (A-level only), 2.4.11 Symmetry and inflection of Normal curve (A-level only), 2.4.12 Calculate probabilities from a Normal distribution (A-level only), 2.4.13 Model with probability distributions, and 2.4 Probability Distributions. Each one has a worked solution and a mark scheme showing where the marks go.