| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Justifying CLT for confidence intervals |
| Difficulty | Standard +0.3 This is a straightforward application of standard S2 techniques: calculating sample statistics, constructing a confidence interval using the CLT, and explaining when CLT applies. Part (d) requires understanding that confidence level relates to the probability the interval contains the mean, making it slightly above routine recall but still a standard textbook exercise with no novel problem-solving required. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05b Unbiased estimates: of population mean and variance5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{4509}{90}\ [= 50.1]\) | B1 | |
| \(\dfrac{90}{89}\!\left(\dfrac{225950}{90} - 50.1^2\right)\) or \(\dfrac{1}{89}\!\left(225950 - \dfrac{4509^2}{90}\right)\) | M1 | Attempted. Use of biased = 0.5455 scores M0A0 |
| \(\dfrac{491}{890}\) or \(0.552\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(50.1 \pm z\sqrt{\dfrac{491}{890 \div 90}}\) | M1 | Expression of the correct form, allow any \(z\)-value but must be a \(z\)-value |
| \(z = 2.326\) | B1 | Accept 2.326 to 2.329 |
| \(49.9\) to \(50.3\) (3 sf) | A1 | FT from biased variance. Must be an interval. |
| Answer | Marks | Guidance |
|---|---|---|
| Population of masses is unknown | B1 | Accept population of masses is not normal |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 - 0.98\) | M1 | 0.02 seen |
| \(0.02 \div 2 = 0.01\) | A1 | As final answer |
## Question 4:
### Part 4(a):
$\dfrac{4509}{90}\ [= 50.1]$ | **B1** |
$\dfrac{90}{89}\!\left(\dfrac{225950}{90} - 50.1^2\right)$ or $\dfrac{1}{89}\!\left(225950 - \dfrac{4509^2}{90}\right)$ | **M1** | Attempted. Use of biased = 0.5455 scores M0A0
$\dfrac{491}{890}$ or $0.552$ (3 sf) | **A1** |
### Part 4(b):
$50.1 \pm z\sqrt{\dfrac{491}{890 \div 90}}$ | **M1** | Expression of the correct form, allow any $z$-value but must be a $z$-value
$z = 2.326$ | **B1** | Accept 2.326 to 2.329
$49.9$ to $50.3$ (3 sf) | **A1** | FT from biased variance. Must be an interval.
### Part 4(c):
Population of masses is unknown | **B1** | Accept population of masses is not normal
### Part 4(d):
$1 - 0.98$ | **M1** | 0.02 seen
$0.02 \div 2 = 0.01$ | **A1** | As final answer
---4 The masses, $m$ kilograms, of flour in a random sample of 90 sacks of flour are summarised as follows.
$$n = 90 \quad \Sigma m = 4509 \quad \Sigma m ^ { 2 } = 225950$$
\begin{enumerate}[label=(\alph*)]
\item Find unbiased estimates of the population mean and variance.
\item Calculate a $98 \%$ confidence interval for the population mean.
\item Explain why it was necessary to use the Central Limit theorem in answering part (b).
\item Find the probability that the confidence interval found in part (b) is wholly above the true value of the population mean.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2021 Q4 [9]}}