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)
228 exam-style questions on CCEA A Level Maths 4.5.1 Statistical distributions (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.