CAIE FP2 2009 June — Question 6 6 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2009
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeCI with two different confidence levels same sample
DifficultyStandard +0.8 This question requires understanding the relationship between confidence intervals at different levels, working backwards from a given interval to find the sample mean and margin of error, then applying the ratio of critical t-values. While the calculation itself is straightforward once the method is identified, recognizing that the ratio t₀.₀₅/t₀.₀₂₅ for 19 degrees of freedom connects the two intervals requires solid conceptual understanding beyond routine application of formulas.
Spec5.05d Confidence intervals: using normal distribution
6 The times taken by employees in a factory to complete a certain task have a normal distribution with mean \(\mu\) seconds and standard deviation \(\sigma\) seconds, both of which are unknown. Based on a random sample of 20 employees, the symmetric \(95 \%\) confidence interval for \(\mu\) is \(( 481,509 )\). Calculate a symmetric \(90 \%\) confidence interval for \(\mu\).
[0pt] [6]
Question 6:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(\bar{x} = \tfrac{1}{2}(481 + 509) = 495\)M1 A1 Find sample mean
\(\bar{x} \pm ts/\sqrt{n}\), any \(t\) or \(z\ [s = 29.9]\)M1 Use or imply confidence interval formula
\((t_{19,\,0.95}\ /\ t_{19,\,0.975})\ 14\)M1 Find 90% interval semi-width
\((1.725/2.086\) *or* \(1.645/1.96\) lose A1 only)
\(= (1.729/2.093)\cdot 14 = 11.6\)A1
\([483.4,\ 506.6]\) *or* \([483, 507]\)A1 Hence 90% confidence interval. Part mark: 6. Total: [6] 
## Question 6:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\bar{x} = \tfrac{1}{2}(481 + 509) = 495$ | M1 A1 | Find sample mean |
| $\bar{x} \pm ts/\sqrt{n}$, any $t$ or $z\ [s = 29.9]$ | M1 | Use or imply confidence interval formula |
| $(t_{19,\,0.95}\ /\ t_{19,\,0.975})\ 14$ | M1 | Find 90% interval semi-width |
| $(1.725/2.086$ *or* $1.645/1.96$ lose A1 only) | | |
| $= (1.729/2.093)\cdot 14 = 11.6$ | A1 | |
| $[483.4,\ 506.6]$ *or* $[483, 507]$ | A1 | Hence 90% confidence interval. Part mark: 6. Total: **[6]** |

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6 The times taken by employees in a factory to complete a certain task have a normal distribution with mean $\mu$ seconds and standard deviation $\sigma$ seconds, both of which are unknown. Based on a random sample of 20 employees, the symmetric $95 \%$ confidence interval for $\mu$ is $( 481,509 )$. Calculate a symmetric $90 \%$ confidence interval for $\mu$.\\[0pt]
[6]

\hfill \mbox{\textit{CAIE FP2 2009 Q6 [6]}}
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Q1 5 Q2 7 Q3 8 Q4 11 Q5 12 Q6 6 Q7 8 Q8 8 Q9 9 Q10 12 Q11 EITHER Q11 OR