| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.8 Part (a) is routine confidence interval calculation. Part (b) tests conceptual understanding of confidence intervals vs prediction intervals—a common misconception. Part (c) requires conditional probability reasoning with independent events, which is non-standard and requires careful thought about what 'at least one contains the mean' implies for the probability calculation. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(81.5 \pm z \times \frac{5.8}{\sqrt{20}}\) | M1 | For a correct expression (accept if only one side of the interval calculated). Any \(z\) (must be a \(z\)). |
| \(z = 2.326\) | B1 | |
| \(78.5\) to \(84.5\) (3sf) | A1 | Must be an interval. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Not true. C.I. is for mean time, not individual times. | B1 | OE. Both comments needed. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{0.98^2}{1 - 0.02^2}\) | M1 | Attempt \(\frac{P(\text{both contain } \mu)}{P(\text{at least one contains } \mu)}\) with numerator attempt \(0.98^2\) and denominator attempt involving \(0.02\). Must see their quotient. |
| \(= 0.961\) (3sf) | A1 | NB: \([0.98^{2=}]\ 0.9604\) scores M0 A0. |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $81.5 \pm z \times \frac{5.8}{\sqrt{20}}$ | M1 | For a correct expression (accept if only one side of the interval calculated). Any $z$ (must be a $z$). |
| $z = 2.326$ | B1 | |
| $78.5$ to $84.5$ (3sf) | A1 | Must be an interval. |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Not true. C.I. is for **mean** time, not individual times. | B1 | OE. Both comments needed. |
## Question 3(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{0.98^2}{1 - 0.02^2}$ | M1 | Attempt $\frac{P(\text{both contain } \mu)}{P(\text{at least one contains } \mu)}$ with numerator attempt $0.98^2$ and denominator attempt involving $0.02$. Must see their quotient. |
| $= 0.961$ (3sf) | A1 | NB: $[0.98^{2=}]\ 0.9604$ scores M0 A0. |3 The time taken in minutes for a certain daily train journey has a normal distribution with standard deviation 5.8. For a random sample of 20 days the journey times were noted and the mean journey time was found to be 81.5 minutes.
\begin{enumerate}[label=(\alph*)]
\item Calculate a $98 \%$ confidence interval for the population mean journey time.\\
A student was asked for the meaning of this confidence interval. The student replied as follows.\\
'The times for $98 \%$ of these journeys are likely to be within the confidence interval.'
\item Explain briefly whether this statement is true or not.\\
Two independent 98\% confidence intervals are found.
\item Given that at least one of these intervals contains the population mean, find the probability that both intervals contain the population mean.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q3 [6]}}