| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| 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 normal approximation. Part (i) requires basic calculation of sampling distribution parameters and a z-score lookup, part (ii) tests understanding of when CLT applies (large n=200 makes normality assumption unnecessary), and part (iii) is a routine hypothesis test. All steps are mechanical with no novel insight required, making it slightly easier than average. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation2.05a Hypothesis testing language: null, alternative, p-value, significance2.05c Significance levels: one-tail and two-tail5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Var}(\bar{X}) = \frac{121}{200}\) or \(\text{SD of }\bar{X} = \frac{11}{\sqrt{200}}\) | ||
| \((\pm)\frac{354-352}{\frac{11}{\sqrt{200}}} (= \pm2.571)\) | M1, A1 | Or with cc attempted. Allow no \(\sqrt{}\). Must include 200 or \(\sqrt{200}\); \(2.57(1)\) or correct expression |
| \(1 - \Phi(\text{``}2.571\text{''}) = 1 - 0.9949 = 0.0051\) | M1, A1[4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (No) | "No" must be seen or implied, but gains no marks by itself | |
| \(n\) is large, \(\bar{X}\) (approx) normal distribution or CLT applies | B1, B1[2] | \(n \geq 30\); (SR: Both statements correct, but wrong or no conclusion scores B1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| H₀: Pop mean \(= 352\); H₁: Pop mean \(\neq 352\) | B1 | Allow '\(\mu\)' but not just 'mean' |
| \(\pm\frac{356-352}{\frac{11}{\sqrt{50}}} \quad (\pm= 2.57(1))\) | M1, A1 | Must have \(\sqrt{50}\); Correct statement or \(2.57(1)\) |
| Comp with \(z = \pm1.96\) (signs consistent); Evidence that pop mean has changed | B1\(\sqrt{}\)[4] | Correct comparison and correct conclusion, follow through one tail test |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(\bar{X}) = \frac{121}{200}$ or $\text{SD of }\bar{X} = \frac{11}{\sqrt{200}}$ | | |
| $(\pm)\frac{354-352}{\frac{11}{\sqrt{200}}} (= \pm2.571)$ | M1, A1 | Or with cc attempted. Allow no $\sqrt{}$. Must include 200 or $\sqrt{200}$; $2.57(1)$ or correct expression |
| $1 - \Phi(\text{``}2.571\text{''}) = 1 - 0.9949 = 0.0051$ | M1, A1[4] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (No) | | "No" must be seen or implied, but gains no marks by itself |
| $n$ is large, $\bar{X}$ (approx) normal distribution or CLT applies | B1, B1[2] | $n \geq 30$; (SR: Both statements correct, but wrong or no conclusion scores B1) |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| H₀: Pop mean $= 352$; H₁: Pop mean $\neq 352$ | B1 | Allow '$\mu$' but not just 'mean' |
| $\pm\frac{356-352}{\frac{11}{\sqrt{50}}} \quad (\pm= 2.57(1))$ | M1, A1 | Must have $\sqrt{50}$; Correct statement or $2.57(1)$ |
| Comp with $z = \pm1.96$ (signs consistent); Evidence that pop mean has changed | B1$\sqrt{}$[4] | Correct comparison and correct conclusion, follow through one tail test |7 Previous records have shown that the number of cars entering Bampor on any day has mean 352 and variance 121.\\
(i) Find the probability that the mean number of cars entering Bampor during a random sample of 200 days is more than 354 .\\
(ii) State, with a reason, whether it was necessary to assume that the number of cars entering Bampor on any day has a normal distribution in order to find the probability in part (i).\\
(iii) It is thought that the population mean may recently have changed. The number of cars entering Bampor during the day was recorded for each of a random sample of 50 days and the sample mean was found to be 356 . Assuming that the variance is unchanged, test at the $5 \%$ significance level whether the population mean is still 352 .
\hfill \mbox{\textit{CAIE S2 2011 Q7 [10]}}