June 2025 Paper 2 Q14
14 The masses of stones on a certain beach are normally distributed with mean 0.36 kg and standard deviation 0.12 kg.
Some geologists suspect that the mean mass, \(\mu\) kg, of stones on a second beach is different from the mean for the first beach. They carry out a hypothesis test, at the 5% significance level, of the null hypothesis \(\mu = 0.36\) against the alternative hypothesis \(\mu \neq 0.36\).
They weigh each of a random sample of 40 stones on the second beach and find the sample mean, \(\bar{X}\) kg, of their masses. You may assume that the standard deviation of the masses of stones on the second beach is 0.12 kg.
| Scheme | Marks | AO |
|---|---|---|
| \(\left[X \sim \mathrm{N}\left(0.36, 0.12^2\right) \Rightarrow\right]\) \(\mathrm{P}(0.3 \lt X \lt 0.4) = 0.322\) | M1 | 3.3 |
| \(\left[0.322^2\right] = 0.104\) (3sf) | A1 | 3.4 |
| [2] |
Notes
M1: May be implied by A1. Allow this mark for awrt 0.32
A1: awrt 0.104
| Scheme | Marks | AO |
|---|---|---|
| \(\bar{X} \sim \mathrm{N}\left(0.36, \dfrac{0.12^2}{40}\right)\) | M1 | 3.3 |
| \(\mathrm{P}(\bar{X} \lt 0.32) = 0.0175\) (3sf) | A1 | 3.4 |
| [2] |
Notes
M1: Must be seen. May see \(\sigma^2 = 0.00036\).
Condone \(\bar{X} \sim \mathrm{N}\left(0.36, \frac{0.12}{\sqrt{40}}\right)\) for this mark only (i.e. M1A0).
Accept \(\mu = 0.36\) and \(\sigma^2 = \frac{0.12^2}{40}\) or \(\sigma = \frac{0.12}{\sqrt{40}}\).
A1: awrt 0.0175 Dep on correct distribution seen.
Answer only without working (correct distribution) scores 0/2.
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\bar{X} \sim \mathrm{N}\left(0.36, \dfrac{0.12^2}{40}\right)\right]\) \(\mathrm{P}(\bar{X} \lt a) = 0.025\) or \(a = 0.36 - 1.96 \times \dfrac{0.12}{\sqrt{40}}\) | M1 | 3.1a |
| \(a = 0.323\) | A1 | 1.1 |
| \(\mathrm{P}(\bar{X} \gt b) = 0.025\) or \(b = 0.36 + 1.96 \times \dfrac{0.12}{\sqrt{40}}\) | M1 | 3.4 |
| \(b = 0.397\) | A1 | 1.1 |
| \(0.323 \lt \bar{X} \lt 0.397\) (3sf) | A1 | 3.2a |
| [5] |
Notes
Condone use of \(X\) for first four marks (but not final A1). Condone misuse of \(\Phi^{-1}\) but only if correct distribution and correct values seen.
M1: Must see either of these steps (could be shown on a graph if accompanied by correct \(a\) to 1sf), or use of symmetry soi if \(b\) found first (i.e. \(a = 0.36 - |\text{‘}b\text{’} - 0.36|\)).
M1: Or \(\mathrm{P}(\bar{X} \lt b)\,[= 1 - 0.025] = 0.975\)
Must see either of these steps (could be shown on a graph if accompanied by correct \(b\) to 1sf), or use of symmetry soi if \(a\) found first (i.e. \(b = 0.36 + |\text{‘}a\text{’} - 0.36|\)).
A1: Must be given correctly using \(\bar{X}\) for this mark.
Accept \(\bar{X} \gt 0.323\) and \(\bar{X} \lt 0.397\)
Allow \(\leqslant\) for \(\lt\)
SCB2 for 0.323 and 0.397 without working (B1 for each max. 2/5)