Question 7 - A-Level Maths

CAIE FP2 2015 June — Question 7 7 marks

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2015
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Confidence intervals
Type	Recover sample stats from CI
Difficulty	Challenging +1.2 This question requires working backwards from a confidence interval to find sample statistics, involving manipulation of the t-distribution formula and solving simultaneous equations. While it tests understanding of confidence intervals beyond routine application, the algebraic steps are straightforward once the correct formulas are identified. It's moderately harder than average due to the reverse-engineering aspect, but remains a standard Further Maths statistics exercise.
Spec	5.05d Confidence intervals: using normal distribution

7 A random sample of 8 sunflower plants is taken from the large number grown by a gardener, and the heights of the plants are measured. A 95\% confidence interval for the population mean, \(\mu\) metres, is calculated from the sample data as \(1.17 < \mu < 2.03\). Given that the height of a sunflower plant is denoted by \(x\) metres, find the values of \(\Sigma x\) and \(\Sigma x ^ { 2 }\) for this sample of 8 plants.

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Question 7:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
\(\Sigma x = 8\bar{x} = 8 \times \frac{1}{2}(1.17 + 2.03) = 8 \times 1.6 = 12.8\)	M1 A1	Find \(\Sigma x\) via sample mean \(\bar{x}\)
\(t\sqrt{(s^2/8)} = \frac{1}{2}(2.03 - 1.17)\ [= 0.43]\)	M1	Find estimate of population variance \(s^2\)
\(t_{7,\ 0.975} = 2.36[5]\)	A1	Use correct tabular value (1.96 leads to 23.2) (to 3 d.p.)
\(s^2 = 0.2645\) or \(32/121\) or \(0.5143^2\)	A1
\(s^2 = \{\Sigma x^2 - (\Sigma x)^2/8\}/7\)		Find \(\Sigma x^2\) from \(s^2\) (M0 for \(s^2 = \{\ldots\}/8\))
\(\Sigma x^2 = 7 \times 0.2645 + 12.8^2/8 = 22.3\)	M1 A1
Total: 7 marks

## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Sigma x = 8\bar{x} = 8 \times \frac{1}{2}(1.17 + 2.03) = 8 \times 1.6 = 12.8$ | M1 A1 | Find $\Sigma x$ via sample mean $\bar{x}$ |
| $t\sqrt{(s^2/8)} = \frac{1}{2}(2.03 - 1.17)\ [= 0.43]$ | M1 | Find estimate of population variance $s^2$ |
| $t_{7,\ 0.975} = 2.36[5]$ | A1 | Use correct tabular value (1.96 leads to 23.2) (to 3 d.p.) |
| $s^2 = 0.2645$ *or* $32/121$ *or* $0.5143^2$ | A1 | |
| $s^2 = \{\Sigma x^2 - (\Sigma x)^2/8\}/7$ | | Find $\Sigma x^2$ from $s^2$ (M0 for $s^2 = \{\ldots\}/8$) |
| $\Sigma x^2 = 7 \times 0.2645 + 12.8^2/8 = 22.3$ | M1 A1 | |
| **Total: 7 marks** | | |

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7 A random sample of 8 sunflower plants is taken from the large number grown by a gardener, and the heights of the plants are measured. A 95\% confidence interval for the population mean, $\mu$ metres, is calculated from the sample data as $1.17 < \mu < 2.03$. Given that the height of a sunflower plant is denoted by $x$ metres, find the values of $\Sigma x$ and $\Sigma x ^ { 2 }$ for this sample of 8 plants.

\hfill \mbox{\textit{CAIE FP2 2015 Q7 [7]}}

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