CAIE FP2 2011 June — Question 8 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
Marks9
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Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeCI from raw data list
DifficultyStandard +0.3 This is a two-part question combining standard confidence interval calculation (using t-distribution with small sample) and a routine sample size determination formula. The second part requires rearranging n ≥ (2×1.96×σ/width)², which is a textbook application. While it involves two different scenarios (unknown vs known variance), both are straightforward applications of standard formulas with no conceptual challenges or novel problem-solving required.
Spec2.05e Hypothesis test for normal mean: known variance
8 In a crossword competition the times, \(x\) minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows. $$\Sigma x = 210.9 \quad \Sigma ( x - \bar { x } ) ^ { 2 } = 151.2$$ The time to complete a crossword has a normal distribution with mean \(\mu\) minutes. Calculate a \(95 \%\) confidence interval for \(\mu\). Assume now that the standard deviation of the population is known to be 5.6 minutes. Find the smallest sample size that would lead to a \(95 \%\) confidence interval for \(\mu\) of width at most 5 minutes.
Question 8:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\bar{d} = 35.15\)B1 Calculate sample mean
\(s^2 = 30.24\ [5.52]\) (or \(25.2\ [5.02^2]\))B1 Estimate population variance (allow biased)
\(35.15 \pm t\sqrt{(30.24/6)}\)M1 Find confidence interval (allow \(z\) in place of \(t\)); inconsistent use of 5 or 6 loses M1
\(t_{5,\,0.975} = 2.571\) (2 d.p.)A1 Use correct tabular value
\(35.15 \pm 5.77\) or \([29.4,\ 40.9]\)A1 Evaluate C.I. correct to 3 s.f. (needs correct \(s\), \(t\))
\(1.96 \times 5.6/\sqrt{n} \leq\) (or \(<\)) \(2.5\)M1 A1 State inequality involving sample size \(n\); equality or wrong critical value loses A1
\(4.39^2 = 19.3\)A1 Solve for limiting value of \(n\)
\(n_{\min} = 20\)A1 State smallest sample size
Total: [9]
## Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{d} = 35.15$ | B1 | Calculate sample mean |
| $s^2 = 30.24\ [5.52]$ (or $25.2\ [5.02^2]$) | B1 | Estimate population variance (allow biased) |
| $35.15 \pm t\sqrt{(30.24/6)}$ | M1 | Find confidence interval (allow $z$ in place of $t$); inconsistent use of 5 or 6 loses M1 |
| $t_{5,\,0.975} = 2.571$ (2 d.p.) | A1 | Use correct tabular value |
| $35.15 \pm 5.77$ or $[29.4,\ 40.9]$ | A1 | Evaluate C.I. correct to 3 s.f. (needs correct $s$, $t$) |
| $1.96 \times 5.6/\sqrt{n} \leq$ (or $<$) $2.5$ | M1 A1 | State inequality involving sample size $n$; equality or wrong critical value loses A1 |
| $4.39^2 = 19.3$ | A1 | Solve for limiting value of $n$ |
| $n_{\min} = 20$ | A1 | State smallest sample size |

**Total: [9]**

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8 In a crossword competition the times, $x$ minutes, taken by a random sample of 6 entrants to complete a crossword are summarised as follows.

$$\Sigma x = 210.9 \quad \Sigma ( x - \bar { x } ) ^ { 2 } = 151.2$$

The time to complete a crossword has a normal distribution with mean $\mu$ minutes. Calculate a $95 \%$ confidence interval for $\mu$.

Assume now that the standard deviation of the population is known to be 5.6 minutes. Find the smallest sample size that would lead to a $95 \%$ confidence interval for $\mu$ of width at most 5 minutes.

\hfill \mbox{\textit{CAIE FP2 2011 Q8 [9]}}
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