| Exam Board | CAIE |
|---|---|
| Module | Further Paper 4 (Further Paper 4) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Confidence intervals |
| Type | Recover sample stats from CI |
| Difficulty | Challenging +1.2 This is a reverse-engineering problem requiring students to work backwards from a confidence interval to find sample statistics. While it involves multiple steps (finding sample mean from interval midpoint, using t-distribution critical value for n=12 at 98% confidence, then solving for standard deviation and finally computing sum of x and sum of x²), each step follows a standard algorithmic procedure. The conceptual demand is moderate—understanding the structure of a confidence interval—but the execution is mechanical calculation rather than problem-solving or proof. Slightly above average difficulty due to the reverse direction and multi-step nature, but still a standard Further Statistics exercise. |
| Spec | 5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{x} = \frac{1}{2}(68.66 + 64.22) = 66.44\) | M1 | |
| \(\sum x = 12 \times 66.44 = 797.28\) | A1 | May come from use of incorrect \(t\) value. Accept AWRT 797. |
| \(\frac{ts}{\sqrt{12}} = \frac{1}{2}(68.66 - 64.22)\) | M1 | Must be with a \(t\) value (not \(z\) value). |
| With \(t = 2.718\) gives \(s^2 = 8.0055\) | A1 | Accept AWRT 8.01; may be recovered by final answer exactly 53059(.34). \(s = 2.8294\). |
| Use \(s^2 = \frac{1}{11}\left(\sum x^2 - \frac{(\sum x)^2}{12}\right)\) to find an expression for \(\sum x^2\) | M1 | Must see an expression for \(\sum x^2\). |
| \(\sum x^2 = 53059\) | A1 | AWRT 53100 but not from wrong working. |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Population is normally distributed | B1 | Underlying distribution is normal B1. Pulse rates are normal B1. Population mean is normally distributed B0. (Underlying) data is normal B0. |
| Total: 1 |
# Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = \frac{1}{2}(68.66 + 64.22) = 66.44$ | M1 | |
| $\sum x = 12 \times 66.44 = 797.28$ | A1 | May come from use of incorrect $t$ value. Accept AWRT 797. |
| $\frac{ts}{\sqrt{12}} = \frac{1}{2}(68.66 - 64.22)$ | M1 | Must be with a $t$ value (not $z$ value). |
| With $t = 2.718$ gives $s^2 = 8.0055$ | A1 | Accept AWRT 8.01; may be recovered by final answer exactly 53059(.34). $s = 2.8294$. |
| Use $s^2 = \frac{1}{11}\left(\sum x^2 - \frac{(\sum x)^2}{12}\right)$ to find an expression for $\sum x^2$ | M1 | Must see an expression for $\sum x^2$. |
| $\sum x^2 = 53059$ | A1 | AWRT 53100 but not from wrong working. |
| **Total: 6** | | |
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# Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Population is normally distributed | B1 | Underlying distribution is normal B1. Pulse rates are normal B1. Population **mean** is normally distributed B0. (Underlying) **data** is normal B0. |
| **Total: 1** | | |
---2 A rowing club has a large number of members.A random sample of 12 of these members is taken and the pulse rate,$x$ beats per minute(bpm),of each is measured after a 30 -minute training session.A $98 \%$ confidence interval for the population mean pulse rate,$\mu \mathrm { bpm }$ ,is calculated from the sample as $64.22 < \mu < 68.66$ .
\begin{enumerate}[label=(\alph*)]
\item Find the values of $\sum x$ and $\sum x ^ { 2 }$ .
\item State an assumption that is necessary for the confidence interval to be valid.\\
\includegraphics[max width=\textwidth, alt={}, center]{b5ff998a-fcb6-4a1b-ae86-ec66b0dccc3c-04_2718_38_141_2009}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q2 [7]}}