| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2003 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single normal population sample mean |
| Difficulty | Moderate -0.3 This is a straightforward application of sampling distribution of the mean and a basic hypothesis test. Part (i) requires standardizing X̄ using σ/√n (routine CLT application), and part (ii) is a standard two-tailed z-test with clearly stated hypotheses and significance level. Both parts are textbook exercises requiring only direct application of formulas with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(P(\bar{X} > 1800) = 1 - \Phi\left(\frac{1800 - 1850}{117/\sqrt{26}}\right) = \Phi(2.179) = 0.985\) | B1, M1, A1 | For \(\frac{117}{\sqrt{26}}\) (or equiv); For standardising and use of tables; For correct answer cwo |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Accept \(H_0\), no significant change | B1, M1, A1, M1, A1ft | Both hypotheses correct; Standardising attempt including standard error; Correct test statistic (\(+/-\)); Comparing with \(z = \pm 1.645\) with \(+\) or \(−\) with \(−\) (or equiv area comparison). ft 1 tail test z=1.282; For correct conclusion on their test statistic and their z. No contradictions. |
| [5] |
**(i)** $P(\bar{X} > 1800) = 1 - \Phi\left(\frac{1800 - 1850}{117/\sqrt{26}}\right) = \Phi(2.179) = 0.985$ | B1, M1, A1 | For $\frac{117}{\sqrt{26}}$ (or equiv); For standardising and use of tables; For correct answer cwo
| [3]
**(ii)** $H_0: \mu = 1850$; $H_1: \mu \neq 1850$
Test statistic $= \frac{1833 - 1850}{117/\sqrt{26}} = -0.7409$
Critical value $z = \pm 1.645$
Accept $H_0$, no significant change | B1, M1, A1, M1, A1ft | Both hypotheses correct; Standardising attempt including standard error; Correct test statistic ($+/-$); Comparing with $z = \pm 1.645$ with $+$ or $−$ with $−$ (or equiv area comparison). ft 1 tail test z=1.282; For correct conclusion on their test statistic and their z. No contradictions.
| [5]5 The distance driven in a week by a long-distance lorry driver is a normally distributed random variable with mean 1850 km and standard deviation 117 km .\\
(i) Find the probability that in a random sample of 26 weeks his average distance driven per week is more than 1800 km .\\
(ii) New driving regulations are introduced and in a random sample of 26 weeks after their introduction the lorry driver drives a total of 47658 km . Assuming the standard deviation remains unchanged, test at the $10 \%$ level whether his mean weekly driving distance has changed.
\hfill \mbox{\textit{CAIE S2 2003 Q5 [8]}}