| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Approximating Binomial to Normal Distribution |
| Type | Confidence interval for proportion |
| Difficulty | Standard +0.8 This question requires working backwards from confidence interval width to find sample proportions, involving algebraic manipulation of the normal approximation formula and solving a quadratic equation. Part (b) tests understanding of confidence interval interpretation with independent samples, which is conceptually subtle. The reverse-engineering aspect and the non-trivial quadratic make this harder than standard CI construction questions. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(z = 1.645\) | B1 | |
| \(z \times \frac{\sqrt{\frac{x}{100} \times (1 - \frac{x}{100})}}{100} = 0.07896\) | M1 | OE. Equation of correct form. Accept \(p = x/100\). Any \(z\). Allow missing factor of 2. |
| \([x(100-x) = 100^3 \times 0.07896^2 \div 1.645^2]\) \(x^2 - 100x + 2304 = 0\) | A1 | Any correct (likely scalar multiple) three-term quadratic equation in \(x\) or \(p\) with simplified coefficients. Accept \(p^2 - p + 0.2304 = 0\) or \(p(1-p) = 0.2304\) |
| \(x = 36\) or \(64\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(0.1^2 = 0.01\) | B1 | Accept either. |
| 1 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $z = 1.645$ | B1 | |
| $z \times \frac{\sqrt{\frac{x}{100} \times (1 - \frac{x}{100})}}{100} = 0.07896$ | M1 | OE. Equation of correct form. Accept $p = x/100$. Any $z$. Allow missing factor of 2. |
| $[x(100-x) = 100^3 \times 0.07896^2 \div 1.645^2]$ $x^2 - 100x + 2304 = 0$ | A1 | Any correct (likely scalar multiple) three-term quadratic equation in $x$ or $p$ with simplified coefficients. Accept $p^2 - p + 0.2304 = 0$ or $p(1-p) = 0.2304$ |
| $x = 36$ or $64$ | A1 | |
| | **4** | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $0.1^2 = 0.01$ | B1 | Accept either. |
| | **1** | |3 In a random sample of 100 students at Luciana's college, $x$ students said that they liked exams. Luciana used this result to find an approximate $90 \%$ confidence interval for the proportion, $p$, of all students at her college who liked exams. Her confidence interval had width 0.15792 .
\begin{enumerate}[label=(\alph*)]
\item Find the two possible values of $x$.\\
Suzma independently took another random sample and found another approximate $90 \%$ confidence interval for $p$.
\item Find the probability that neither of the two confidence intervals contains the true value of $p$. [1]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2023 Q3 [5]}}