AQA S3 2007 June — Question 4 6 marks

Exam BoardAQA
ModuleS3 (Statistics 3)
Year2007
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentral limit theorem
TypeSample size determination
DifficultyStandard +0.8 This is a moderately challenging S3 question requiring students to work backwards from confidence interval width to determine sample size. It involves manipulating the formula for CI width (2z*σ/√n), solving for n, and understanding the relationship between precision and sample size. While the calculation itself is straightforward once set up, the conceptual step of relating width to the margin of error and correctly applying the 99% confidence level makes this harder than routine CI construction problems.
Spec5.05d Confidence intervals: using normal distribution

4 A machine is used to fill 5-litre plastic containers with vinegar. The volume, in litres, of vinegar in a container filled by the machine may be assumed to be normally distributed with mean \(\mu\) and standard deviation 0.08 . A quality control inspector requires a \(99 \%\) confidence interval for \(\mu\) to be constructed such that it has a width of at most 0.05 litres. Calculate, to the nearest 5, the sample size necessary in order to achieve the inspector's requirement.

Question 4:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(98\% \Rightarrow z = 2.5758\)B1 AWFW 2.57 to 2.58
CI width is \(2 \times \frac{z\sigma}{\sqrt{n}}\)M1 Used; allow \(\frac{z\sigma}{\sqrt{n}}\)
\(2 \times \frac{2.5758 \times 0.08}{\sqrt{n}} = 0.05\)A1\(\checkmark\) OE; \(\checkmark\) on \(z\); allow no '\(2\times\)'
\(\sqrt{n} = 8.24256\)m1 Solving for \(\sqrt{n}\) or \(n\)
\(n = 67.9 \Rightarrow 68\)A1\(\checkmark\) AWRT; \(\checkmark\) on \(z\)
To nearest 5, \(n = 70\)A1 CAO
Total6
# Question 4:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $98\% \Rightarrow z = 2.5758$ | B1 | AWFW 2.57 to 2.58 |
| CI width is $2 \times \frac{z\sigma}{\sqrt{n}}$ | M1 | Used; allow $\frac{z\sigma}{\sqrt{n}}$ |
| $2 \times \frac{2.5758 \times 0.08}{\sqrt{n}} = 0.05$ | A1$\checkmark$ | OE; $\checkmark$ on $z$; allow no '$2\times$' |
| $\sqrt{n} = 8.24256$ | m1 | Solving for $\sqrt{n}$ or $n$ |
| $n = 67.9 \Rightarrow 68$ | A1$\checkmark$ | AWRT; $\checkmark$ on $z$ |
| To nearest 5, $n = 70$ | A1 | CAO |
| **Total** | **6** | |

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4 A machine is used to fill 5-litre plastic containers with vinegar. The volume, in litres, of vinegar in a container filled by the machine may be assumed to be normally distributed with mean $\mu$ and standard deviation 0.08 .

A quality control inspector requires a $99 \%$ confidence interval for $\mu$ to be constructed such that it has a width of at most 0.05 litres.

Calculate, to the nearest 5, the sample size necessary in order to achieve the inspector's requirement.

\hfill \mbox{\textit{AQA S3 2007 Q4 [6]}}