| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Find minimum sample size |
| Difficulty | Moderate -0.3 This is a straightforward confidence interval question requiring standard formulas and calculations. Part (a) involves routine application of the normal distribution CI formula with known σ, part (b) requires rearranging the width formula to solve for n (algebraically simple), and part (c) asks for a basic comment on sampling bias. The arithmetic is slightly fiddly with time conversions, but the statistical concepts are fundamental S3 material with no novel problem-solving required. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a): mean \(= \frac{(46 \times 60) + 15}{20} = 138.75\) | M1 | |
| C.I. \(\bar{x} \pm 1.6449 \frac{s}{\sqrt{n}} = 138.75 \pm 1.6449 \cdot \frac{23}{{\sqrt{20}}}\) | M1 A1 | |
| giving \((130.3, 147.2)\) | A2 | |
| Part (b): width \(= 2 \times 1.6449 \times \frac{23}{\sqrt{n}}\); \(2 \times 1.6449 \times \frac{23}{\sqrt{n}} < 10\) | M1 A1 | |
| \(\therefore \sqrt{n} > 7.56654\) | A1 | |
| giving \(n > 57.25\) so min. value of \(n = 58\) | M1 A1 | |
| Part (c): e.g. she might buy big-budget movies with longer credits | B1 | (11 marks total) |
**Part (a):** mean $= \frac{(46 \times 60) + 15}{20} = 138.75$ | M1 |
C.I. $\bar{x} \pm 1.6449 \frac{s}{\sqrt{n}} = 138.75 \pm 1.6449 \cdot \frac{23}{{\sqrt{20}}}$ | M1 A1 |
giving $(130.3, 147.2)$ | A2 |
**Part (b):** width $= 2 \times 1.6449 \times \frac{23}{\sqrt{n}}$; $2 \times 1.6449 \times \frac{23}{\sqrt{n}} < 10$ | M1 A1 |
$\therefore \sqrt{n} > 7.56654$ | A1 |
giving $n > 57.25$ so min. value of $n = 58$ | M1 A1 |
**Part (c):** e.g. she might buy big-budget movies with longer credits | B1 | (11 marks total)
---A film-buff is interested in how long it takes for the credits to roll at the end of a movie. She takes a random sample of 20 movies from those that she has bought on DVD and finds that the credits on these films last for a total of 46 minutes and 15 seconds
\begin{enumerate}[label=(\alph*)]
\item Assuming that the time for the credits to roll follows a Normal distribution with a standard deviation of 23 seconds, use her data to calculate a 90% confidence interval for the mean time taken for the credits to roll. [5]
\item Find the minimum number of movies she would need to have included in her sample for her confidence interval to have a width of less than 10 seconds. [5]
\item Explain why her sample might not be representative of all movies. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q3 [11]}}