CAIE Further Paper 4 2024 June — Question 1 4 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeSingle sample confidence interval t-distribution
DifficultyStandard +0.3 This is a straightforward application of the t-distribution confidence interval formula with given summary statistics. Students need to calculate the sample standard deviation, find the critical t-value for 13 degrees of freedom, and apply the standard formula. While it requires knowledge of the t-distribution (a Further Maths topic), the execution is mechanical with no problem-solving or conceptual challenges beyond standard textbook exercises.
Spec5.05d Confidence intervals: using normal distribution

1 The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean \(\mu\) minutes. The times taken, \(x\) minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where \(\bar { x }\) is the sample mean. $$\bar { x } = 42.8 \quad \sum ( x - \bar { x } ) ^ { 2 } = 941.5$$ Find a 95\% confidence interval for \(\mu\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(s^2 = \dfrac{941.5}{13}\)B1 72.423
CI: \(42.8 \pm 2.160\sqrt{\dfrac{s^2}{14}}\)M1 A1 Correct formula with a \(t\) value. 2.160 seen in a CI formula for A1.
\([37.9, 47.7]\)A1 \((37.9, 47.7)\) A1; \(37.9 \leqslant x \leqslant 47.7\) A1 (ignore symbol used inside the inequality); Condone \([47.7, 37.9]\) A1; Final answer \(42.8 \pm 4.91\) A0
4
**Question 1:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $s^2 = \dfrac{941.5}{13}$ | **B1** | 72.423 |
| CI: $42.8 \pm 2.160\sqrt{\dfrac{s^2}{14}}$ | **M1 A1** | Correct formula with a $t$ value. 2.160 seen in a CI formula for A1. |
| $[37.9, 47.7]$ | **A1** | $(37.9, 47.7)$ A1; $37.9 \leqslant x \leqslant 47.7$ A1 (ignore symbol used inside the inequality); Condone $[47.7, 37.9]$ A1; Final answer $42.8 \pm 4.91$ A0 |
| | **4** | |
1 The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean $\mu$ minutes. The times taken, $x$ minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where $\bar { x }$ is the sample mean.

$$\bar { x } = 42.8 \quad \sum ( x - \bar { x } ) ^ { 2 } = 941.5$$

Find a 95\% confidence interval for $\mu$.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2024 Q1 [4]}}