CAIE S2 2009 November — Question 2 4 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2009
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeSample size determination
DifficultyStandard +0.8 This question requires understanding of confidence intervals, the relationship between sample size and interval width, and solving an inequality involving the normal distribution. While the setup is standard S2 material, students must correctly formulate that width = 2×1.96×(1.5/√n) ≤ 1 and solve for n, requiring algebraic manipulation and rounding up appropriately. It's more challenging than routine confidence interval calculations but doesn't require novel insight.
Spec5.05d Confidence intervals: using normal distribution

2 The lengths of sewing needles in travel sewing kits are distributed normally with mean \(\mu \mathrm { mm }\) and standard deviation 1.5 mm . A random sample of \(n\) needles is taken. Find the smallest value of \(n\) such that the width of a \(95 \%\) confidence interval for the population mean is at most 1 mm .

AnswerMarks Guidance
\(1.96 \times \frac{1.5}{\sqrt{n}} < \frac{1}{2}\)B1 \(1.96 \times \frac{1.5}{\sqrt{n}}\) seen
B1Confidence interval halved
M1Solving an equation in their \(z\), 1.5, \(n\) (2 and sq rt not needed)
\(n = 35\)A1 [4] Correct answer (condone \(n \geq 35\))
$1.96 \times \frac{1.5}{\sqrt{n}} < \frac{1}{2}$ | B1 | $1.96 \times \frac{1.5}{\sqrt{n}}$ seen
 | B1 | Confidence interval halved
 | M1 | Solving an equation in their $z$, 1.5, $n$ (2 and sq rt not needed)
$n = 35$ | A1 [4] | Correct answer (condone $n \geq 35$)
2 The lengths of sewing needles in travel sewing kits are distributed normally with mean $\mu \mathrm { mm }$ and standard deviation 1.5 mm . A random sample of $n$ needles is taken. Find the smallest value of $n$ such that the width of a $95 \%$ confidence interval for the population mean is at most 1 mm .

\hfill \mbox{\textit{CAIE S2 2009 Q2 [4]}}