| Exam Board | AQA |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Sample size determination |
| Difficulty | Standard +0.8 This S3 question requires understanding confidence intervals, sample size determination, and working with parameters expressed in terms of μ. Students must set up the inequality 2×1.96×(μ/2)/√n ≤ 0.2μ, simplify algebraically to isolate n, and solve to get n ≥ 384.16. While the individual steps are standard A-level techniques, the abstract parameterization and multi-step algebraic manipulation make this moderately challenging for typical S3 students. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Width of CI \(= 2 \times 1.96 \times \frac{\sigma}{\sqrt{n}}\) | M1 | Setting up CI width expression |
| \(\sigma = \frac{\mu}{2}\), so width \(= 2 \times 1.96 \times \frac{\mu}{2\sqrt{n}} = \frac{1.96\mu}{\sqrt{n}}\) | A1 | Correct substitution of \(\sigma = \frac{\mu}{2}\) |
| Require \(\frac{1.96\mu}{\sqrt{n}} \leq 0.2\mu\) | M1 | Setting up inequality with \(0.2\mu\) |
| \(\sqrt{n} \geq \frac{1.96}{0.2} = 9.8\) | A1 | Correct rearrangement |
| \(n \geq 96.04\) | DM1 | Squaring to find \(n\) |
| Minimum \(n = 100\) | A1 | Rounding up to nearest 10 |
## Question 4:
**Minimum sample size for 95% CI with width at most $0.2\mu$**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Width of CI $= 2 \times 1.96 \times \frac{\sigma}{\sqrt{n}}$ | M1 | Setting up CI width expression |
| $\sigma = \frac{\mu}{2}$, so width $= 2 \times 1.96 \times \frac{\mu}{2\sqrt{n}} = \frac{1.96\mu}{\sqrt{n}}$ | A1 | Correct substitution of $\sigma = \frac{\mu}{2}$ |
| Require $\frac{1.96\mu}{\sqrt{n}} \leq 0.2\mu$ | M1 | Setting up inequality with $0.2\mu$ |
| $\sqrt{n} \geq \frac{1.96}{0.2} = 9.8$ | A1 | Correct rearrangement |
| $n \geq 96.04$ | DM1 | Squaring to find $n$ |
| Minimum $n = 100$ | A1 | Rounding up to nearest 10 |4 \\
The waiting time at a hospital's A\&E department may be modelled by a normal distribution with mean $\mu$ and standard deviation $\frac { \mu } { 2 }$. \\
The department's manager wishes a $95 \%$ confidence interval for $\mu$ to be constructed such that it has a width of at most $0.2 \mu$. \\
Calculate, to the nearest 10, an estimate of the minimum sample size necessary in order to achieve the manager's wish. \\
(5 marks) \\
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\hfill \mbox{\textit{AQA S3 2011 Q4 [5]}}