Standard +0.3 This is a straightforward application of the confidence interval width formula for a normal distribution with known standard deviation. Students need to recall that width = 2 × z × σ/√n, rearrange to solve for n, and round up appropriately. While it requires understanding of confidence intervals, it's a standard textbook exercise with no conceptual surprises or multi-step reasoning.
1 The result of a memory test is known to be normally distributed with mean \(\mu\) and standard deviation 1.9. It is required to have a \(95 \%\) confidence interval for \(\mu\) with a total width of less than 2.0 . Find the least possible number of tests needed to achieve this.
\(1.9 \times 1.96 < 1\) and \(n > 13.9 (13.87)\) and \(n = 14\)
M1, A1, M1, A1
For equality or inequality involving width or equivalent and term in \(\frac{1}{\sqrt{n}}\) and a z-value; For correct inequality; For solving a relevant equation; For correct answer cwo
$1.9 \times 1.96 < 1$ and $n > 13.9 (13.87)$ and $n = 14$ | M1, A1, M1, A1 | For equality or inequality involving width or equivalent and term in $\frac{1}{\sqrt{n}}$ and a z-value; For correct inequality; For solving a relevant equation; For correct answer cwo
1 The result of a memory test is known to be normally distributed with mean $\mu$ and standard deviation 1.9. It is required to have a $95 \%$ confidence interval for $\mu$ with a total width of less than 2.0 . Find the least possible number of tests needed to achieve this.
\hfill \mbox{\textit{CAIE S2 2003 Q1 [4]}}