CAIE S2 2002 June — Question 1 4 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2002
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeFind CI width or confidence level
DifficultyModerate -0.8 This is a straightforward confidence interval width calculation requiring only the formula width = 2z*σ/√n with given values. It tests basic recall of the confidence interval formula and simple arithmetic, with no conceptual complexity or problem-solving required beyond direct substitution.
Spec5.05d Confidence intervals: using normal distribution

1 The result of a fitness trial is a random variable \(X\) which is normally distributed with mean \(\mu\) and standard deviation 2.4. A researcher uses the results from a random sample of 90 trials to calculate a \(98 \%\) confidence interval for \(\mu\). What is the width of this interval?

\(\bar{x} \pm 2.326 \times \frac{2.4}{\sqrt{90}}\)
\(2.326 \times \frac{2.4}{\sqrt{90}} \times 2\)
AnswerMarks Guidance
Width \(= 1.18\)B1, M1, M1, A1 (4 marks) For \(z\) value of 2.33; For expression of correct form involving \(\sqrt{90}\) in denominator; For subtracting lower from upper, or multiplying half-width by 2; For correct answer
$\bar{x} \pm 2.326 \times \frac{2.4}{\sqrt{90}}$

$2.326 \times \frac{2.4}{\sqrt{90}} \times 2$
Width $= 1.18$ | B1, M1, M1, A1 (4 marks) | For $z$ value of 2.33; For expression of correct form involving $\sqrt{90}$ in denominator; For subtracting lower from upper, or multiplying half-width by 2; For correct answer

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1 The result of a fitness trial is a random variable $X$ which is normally distributed with mean $\mu$ and standard deviation 2.4. A researcher uses the results from a random sample of 90 trials to calculate a $98 \%$ confidence interval for $\mu$. What is the width of this interval?

\hfill \mbox{\textit{CAIE S2 2002 Q1 [4]}}