| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Moderate -0.5 This question tests standard confidence interval calculations with known population standard deviation. Part (a) is a direct application of the formula using z-values, and part (b) requires working backwards from width to find the confidence level. Both parts are routine calculations with no conceptual challenges beyond remembering the formula—easier than average but not trivial due to the reverse calculation in part (b). |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(45 \pm z \times \dfrac{6}{\sqrt{200}}\) | M1 | For expression of correct form, any \(z\). Accept one side of interval for M1. |
| \(z = 1.96\) | B1 | Must be seen. |
| \(44.2\) to \(45.8\) (3 sf) | A1 | Must be an interval. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(z \times \dfrac{11}{\sqrt{200}} = 2\) | M1 | Or \(\ldots = 4\) for M1 |
| \(z = 2.571\) | A1 | Accept 3 sf if nothing better seen. |
| \(\phi(\textit{their } `2.571`) = 0.9949\) and \(\textit{their } `0.9949` - (1 - \textit{their } `0.9949`) [= 0.9898]\) | M1 | OE. For area consistent with their values. Must be seen. |
| \(\alpha = 99.0\) (3 sf) | A1 | Allow 99. cwo. Final answer of 0.99 scores A0. |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $45 \pm z \times \dfrac{6}{\sqrt{200}}$ | M1 | For expression of correct form, any $z$. Accept one side of interval for M1. |
| $z = 1.96$ | B1 | Must be seen. |
| $44.2$ to $45.8$ (3 sf) | A1 | Must be an interval. |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $z \times \dfrac{11}{\sqrt{200}} = 2$ | M1 | Or $\ldots = 4$ for M1 |
| $z = 2.571$ | A1 | Accept 3 sf if nothing better seen. |
| $\phi(\textit{their } `2.571`) = 0.9949$ **and** $\textit{their } `0.9949` - (1 - \textit{their } `0.9949`) [= 0.9898]$ | M1 | OE. For area consistent with their values. Must be seen. |
| $\alpha = 99.0$ (3 sf) | A1 | Allow 99. cwo. Final answer of 0.99 scores A0. |
---5 A builders' merchant sells stones of different sizes.
\begin{enumerate}[label=(\alph*)]
\item The masses of size $A$ stones have standard deviation 6 grams. The mean mass of a random sample of 200 size $A$ stones is 45 grams.
Find a 95\% confidence interval for the population mean mass of size $A$ stones.
\item The masses of size $B$ stones have standard deviation 11 grams. Using a random sample of size 200, an $\alpha \%$ confidence interval for the population mean mass is found to have width 4 grams.
Find $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q5 [7]}}