Standard +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.
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 .
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]}}