| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Calculate CI from summary stats |
| Difficulty | Standard +0.3 This is a straightforward confidence interval question requiring standard formula application (part a), recall of confidence interval interpretation (part b), and understanding a common misconception (part c). The calculation is routine with given variance, and the conceptual parts test basic understanding rather than deep insight. Slightly easier than average due to minimal problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Part (a) | M1, M1 A1, A2 | \(\bar{x} = \frac{1419}{30} = 47.3\); C.I. is \(x \pm 1.96\frac{s}{\sqrt{n}} = 47.3 \pm 1.96 \cdot \frac{5}{\sqrt{30}}\); giving (45.51, 49.09) |
| Part (b) | B1 | \(\frac{19}{20}\) |
| Part (c) | B1, (7) | it either does or doesn't include true mean \(\therefore\) probability is 0 or 1 |
**Part (a)** | M1, M1 A1, A2 | $\bar{x} = \frac{1419}{30} = 47.3$; C.I. is $x \pm 1.96\frac{s}{\sqrt{n}} = 47.3 \pm 1.96 \cdot \frac{5}{\sqrt{30}}$; giving (45.51, 49.09)
**Part (b)** | B1 | $\frac{19}{20}$
**Part (c)** | B1, (7) | it either does or doesn't include true mean $\therefore$ probability is 0 or 1A teacher gives each student in his class a list of 30 numbers. All the numbers have been generated at random by a computer from a normal distribution with a fixed mean and variance. The teacher tells the class that the variance of the distribution is 25 and asks each of them to calculate a 95\% confidence interval based on their list of numbers.
The sum of the numbers given to one student is 1419.
\begin{enumerate}[label=(\alph*)]
\item Find the confidence interval that should be obtained by this student. [5]
\end{enumerate}
Assuming that all the students calculate their confidence intervals correctly,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item state the proportion of the students you would expect to have a confidence interval that includes the true mean of the distribution, [1]
\item explain why the probability of any one student's confidence interval including the true mean is not 0.95 [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q2 [7]}}