| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Sampling method explanation |
| Difficulty | Moderate -0.8 This question tests basic understanding of sampling rationale and routine application of CLT with given summary statistics. Part (a) requires simple explanations, part (b)(i) is standard calculation of sample mean and variance, part (b)(ii) is straightforward CLT application with normal tables, and part (b)(iii) requires brief conceptual explanation. All components are textbook exercises with no problem-solving or novel insight required. |
| Spec | 2.01a Population and sample: terminology5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Pop too large / Not all pop accessible | B1 [1] | Time consuming or similar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Testing involves destruction | B1 [1] | Or similar |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{9850}{500} = (19.7)\) | B1 | |
| \(\frac{500}{499}\left(\frac{194125}{500} - \left(\frac{9850}{500}\right)^2\right)\) | M1 | Allow with \(\sqrt{\phantom{x}}\). Method must be seen or clearly implied |
| \(= 0.160(32)\) (3 sfs) or \(80/499\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{19.73 - 19.7}{\sqrt{\frac{"0.160"}{60}}}\) | M1 | For standardising |
| \(= 0.580\) or \(0.581\) | A1ft | ft their mean and var in (b)(i) |
| \(1 - \Phi("0.580")\) \((= 1 - 0.7191)\) | M1 | Correct tail |
| \(= 0.281\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| "Yes" must be seen or implied to gain marks | B1 | |
| \(X\) not nec'y normal | B1 | or \(\bar{X}\) is approx N |
| Sample large | [2] | (SR Both reasons correct, but wrong or no conclusion scores SR B1) |
## Question 7(a):
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Pop too large / Not all pop accessible | B1 [1] | Time consuming or similar |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Testing involves destruction | B1 [1] | Or similar |
---
## Question 7(b):
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{9850}{500} = (19.7)$ | B1 | |
| $\frac{500}{499}\left(\frac{194125}{500} - \left(\frac{9850}{500}\right)^2\right)$ | M1 | Allow with $\sqrt{\phantom{x}}$. Method must be seen or clearly implied |
| $= 0.160(32)$ (3 sfs) or $80/499$ | A1 [3] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{19.73 - 19.7}{\sqrt{\frac{"0.160"}{60}}}$ | M1 | For standardising |
| $= 0.580$ or $0.581$ | A1ft | ft their mean and var in **(b)(i)** |
| $1 - \Phi("0.580")$ $(= 1 - 0.7191)$ | M1 | Correct tail |
| $= 0.281$ | A1 [4] | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| "Yes" must be seen or implied to gain marks | B1 | |
| $X$ not nec'y normal | B1 | or $\bar{X}$ is approx N |
| Sample large | [2] | (SR Both reasons correct, but wrong or no conclusion scores SR B1) |7
\begin{enumerate}[label=(\alph*)]
\item Give a reason why sampling would be required in order to reach a conclusion about
\begin{enumerate}[label=(\roman*)]
\item the mean height of adult males in England,
\item the mean weight that can be supported by a single cable of a certain type without the cable breaking.
\end{enumerate}\item The weights, in kg , of sacks of potatoes are represented by the random variable $X$ with mean $\mu$ and standard deviation $\sigma$. The weights of a random sample of 500 sacks of potatoes are found and the results are summarised below.
$$n = 500 , \quad \Sigma x = 9850 , \quad \Sigma x ^ { 2 } = 194125 .$$
\begin{enumerate}[label=(\roman*)]
\item Calculate unbiased estimates of $\mu$ and $\sigma ^ { 2 }$.
\item A further random sample of 60 sacks of potatoes is taken. Using your values from part (b) (i), find the probability that the mean weight of this sample exceeds 19.73 kg .
\item Explain whether it was necessary to use the Central Limit Theorem in your calculation in part (b) (ii).
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2010 Q7 [11]}}