| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for sampling distribution |
| Difficulty | Standard +0.3 This is a straightforward application of the Central Limit Theorem with standard procedures: (i) requires calculating a probability using CLT with n=50, (ii) tests understanding that CLT applies for large samples regardless of population distribution, and (iii) is a routine one-tailed hypothesis test. All parts follow textbook templates with no novel insight required, 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{14 - 14.2}{\frac{3.1}{\sqrt{50}}} \;(= -0.456)\) | M1 | For stand'n; must have \(\sqrt{50}\) |
| \(1 - \Phi("0.456")\) | M1 | For area consistent with their working |
| \(= 0.324\) (3 sfs) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| No because \(n\) large | B1 | Accept \(n > 30\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 14.2\); \(H_1: \mu < 14.2\) | B1 | or 'pop mean', but not just 'mean' |
| \(\frac{13.5 - 14.2}{\frac{3.1}{\sqrt{100}}}\) | M1 | For stand'n; must have \(\sqrt{100}\) |
| \(= -2.258\) | A1 | |
| comp \(-2.054\) (or \(-2.055\)) | M1 | Valid comparison of \(z\) values or areas \((0.0119 < 0.02)\) |
| There is evidence (at 2% level) that mean mass in this area \(< 14.2\) | A1ft | Ft their \(z\). Correct conclusion no contradictions |
**Question 5(i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{14 - 14.2}{\frac{3.1}{\sqrt{50}}} \;(= -0.456)$ | M1 | For stand'n; must have $\sqrt{50}$ |
| $1 - \Phi("0.456")$ | M1 | For area consistent with their working |
| $= 0.324$ (3 sfs) | A1 | |
**Question 5(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| No because $n$ large | B1 | Accept $n > 30$ |
**Question 5(iii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 14.2$; $H_1: \mu < 14.2$ | B1 | or 'pop mean', but not just 'mean' |
| $\frac{13.5 - 14.2}{\frac{3.1}{\sqrt{100}}}$ | M1 | For stand'n; must have $\sqrt{100}$ |
| $= -2.258$ | A1 | |
| comp $-2.054$ (or $-2.055$) | M1 | Valid comparison of $z$ values or areas $(0.0119 < 0.02)$ |
| There is evidence (at 2% level) that mean mass in this area $< 14.2$ | A1ft | Ft their $z$. Correct conclusion no contradictions |5 The mass, in kilograms, of rocks in a certain area has mean 14.2 and standard deviation 3.1.\\
(i) Find the probability that the mean mass of a random sample of 50 of these rocks is less than 14.0 kg .\\
(ii) Explain whether it was necessary to assume that the population of the masses of these rocks is normally distributed.\\
(iii) A geologist suspects that rocks in another area have a mean mass which is less than 14.2 kg . A random sample of 100 rocks in this area has sample mean 13.5 kg . Assuming that the standard deviation for rocks in this area is also 3.1 kg , test at the $2 \%$ significance level whether the geologist is correct.\\
\hfill \mbox{\textit{CAIE S2 2018 Q5 [9]}}