CAIE Further Paper 4 2022 November — Question 1 7 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2022
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeStandard CI with summary statistics
DifficultyStandard +0.3 This is a straightforward two-sample confidence interval calculation with summary statistics provided. Students must compute sample means and variances from the summations, then apply the standard formula for difference of means. While it requires careful arithmetic and knowledge of the CI formula, it involves no conceptual challenges or novel problem-solving—just routine application of a standard procedure from the Further Statistics syllabus.
Spec5.05d Confidence intervals: using normal distribution

1 Jasmine is researching the heights of pine trees in forests in two regions \(A\) and \(B\). She chooses a random sample of 50 pine trees in region \(A\) and records their heights, \(x \mathrm {~m}\). She also chooses a random sample of 60 pine trees in region \(B\) and records their heights, \(y \mathrm {~m}\). Her results are summarised as follows. $$\sum x = 1625 \quad \sum x ^ { 2 } = 53200 \quad \sum y = 1854 \quad \sum y ^ { 2 } = 57900$$ Find a \(95 \%\) confidence interval for the difference between the population mean heights of pine trees in regions \(A\) and \(B\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\bar{x} = \frac{1625}{50} = 32.5\), \(s_x^2 = \frac{1}{49}\left(53200 - \frac{1625^2}{50}\right) = 7.908\)M1 \(\frac{775}{98}\)
\(\bar{y} = \frac{1854}{60} = 30.9\), \(s_y^2 = \frac{1}{59}\left(57900 - \frac{1854^2}{60}\right) = 10.3627\)A1 \(\frac{3057}{295}\); Both correct
\(s^2 = \frac{7.908}{50} + \frac{10.3627}{60}\ [= 0.33087]\)M1 A1
CI: \((32.5 - 30.9) \pm 1.96s\)M1 A1 Correct form with a \(z\)-value. Correct with 1.96.
\(1.6 \pm 1.1274 = [0.473, 2.73]\)A1 At least 3sf.
7Pooled variance used: M1A1 M0A0 M1A1A0 max 4/7
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\bar{x} = \frac{1625}{50} = 32.5$, $s_x^2 = \frac{1}{49}\left(53200 - \frac{1625^2}{50}\right) = 7.908$ | M1 | $\frac{775}{98}$ |
| $\bar{y} = \frac{1854}{60} = 30.9$, $s_y^2 = \frac{1}{59}\left(57900 - \frac{1854^2}{60}\right) = 10.3627$ | A1 | $\frac{3057}{295}$; Both correct |
| $s^2 = \frac{7.908}{50} + \frac{10.3627}{60}\ [= 0.33087]$ | M1 A1 | |
| CI: $(32.5 - 30.9) \pm 1.96s$ | M1 A1 | Correct form with a $z$-value. Correct with 1.96. |
| $1.6 \pm 1.1274 = [0.473, 2.73]$ | A1 | At least 3sf. |
| | **7** | Pooled variance used: M1A1 M0A0 M1A1A0 max 4/7 |
1 Jasmine is researching the heights of pine trees in forests in two regions $A$ and $B$. She chooses a random sample of 50 pine trees in region $A$ and records their heights, $x \mathrm {~m}$. She also chooses a random sample of 60 pine trees in region $B$ and records their heights, $y \mathrm {~m}$. Her results are summarised as follows.

$$\sum x = 1625 \quad \sum x ^ { 2 } = 53200 \quad \sum y = 1854 \quad \sum y ^ { 2 } = 57900$$

Find a $95 \%$ confidence interval for the difference between the population mean heights of pine trees in regions $A$ and $B$.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q1 [7]}}