| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Marks | 9 |
| 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 formulas. Part (a) is routine application of CI = x̄ ± z(σ/√n). Part (b) requires rearranging the width formula to solve for n, which is a standard textbook exercise with no conceptual difficulty beyond algebraic manipulation. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks |
|---|---|
| giving \((29.96, 32.84)\) | M1 A1 A2 |
| Answer | Marks | Guidance |
|---|---|---|
| giving \(n > 222.42\) so need 223 observations | M1 A1 A1 M1 A1 | (9) |
(a) $\text{C.I. } \bar{x} \pm 1.6449 \frac{s}{\sqrt{n}} = 31.4 \pm 1.6449 \frac{6.8}{\sqrt{60}}$
giving $(29.96, 32.84)$ | M1 A1 A2 |
(b) width $= 2 \times 1.6449 \times \frac{s.d}{\sqrt{n}} \therefore 2 \times 1.6449 \times \frac{6.8}{\sqrt{n}} < 1.5$
$\therefore \sqrt{n} > 14.91376$
giving $n > 222.42$ so need 223 observations | M1 A1 A1 M1 A1 | (9)
---3. A random variable $X$ is distributed normally with a standard deviation of 6.8
Sixty observations of $X$ are made and found to have a mean of 31.4
\begin{enumerate}[label=(\alph*)]
\item Find a 90\% confidence interval for the mean of $X$.
\item How many observations of $X$ would be needed in order to obtain a $90 \%$ confidence interval for the mean of $X$ with a width of less than 1.5\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 Q3 [9]}}