CAIE FP2 2013 June — Question 8 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks9
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 problem with summary statistics provided. Students need to calculate sample means and variances from the given sums, then apply the standard formula for a CI for difference of means. While it requires careful arithmetic and knowledge of the formula, it's a routine application with no conceptual challenges or novel problem-solving required—slightly easier than average due to its mechanical nature.
Spec5.05d Confidence intervals: using normal distribution

8 The number, \(x\), of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country \(A\). The number, \(y\), of the same type of shell was counted at 50 randomly chosen sites, each one metre square, along the coastline in country \(B\). The results are summarised as follows. $$\Sigma x = 1752 \quad \Sigma x ^ { 2 } = 55500 \quad \Sigma y = 1220 \quad \Sigma y ^ { 2 } = 33500$$ Find a 95\% confidence interval for the difference between the mean number of sea shells, per square metre, on the coastlines in country \(A\) and in country \(B\).

Question 8:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\bar{x} = 29.2,\ \bar{y} = 24.4\)B1 Find both sample means
\(s_x^2 = (55500 - 1752^2/60)/59\) and \(s_y^2 = (33500 - 1220^2/50)/49\)M1 A1 Estimate both population variances (to 4 s.f.)
\(s_x^2 = 73.59\) and \(s_y^2 = 76.16\)A1 (allow biased: \(72.36\) and \(74.64\))
\(s^2 = s_x^2/60 + s_y^2/50 = 2.750\) or \(1.658^2\)M1, A1 *EITHER*: Estimate combined variance (3 s.f.)
\((\bar{x} - \bar{y}) \pm 1.96s\)M1 A1 Use this \(s\) to find conf. interval
\(4.8 \pm 3.25\) or \([1.55,\ 8.05]\)A1 Evaluate
\(s^2 = (59s_x^2 + 49s_y^2)/108\) or \((55500 - 1752^2/60 + 33500 - 1220^2/50)/108\)(M1 A1) *OR*: Estimate common variance (to 3 s.f.) (note \(s_x\) and \(s_y\) not needed explicitly)
\(= 74.8\) or \(8.65^2\) or \(3364/45\)
\((\bar{x} - \bar{y}) \pm 1.96s\sqrt{1/60 + 1/50}\)(M1 A1) Use this \(s\) to find conf. interval
\(4.8 \pm 3.24[5]\) or \([1.55[5],\ 8.04[5]]\)(A1) Evaluate
Total: 9 marks
## Question 8:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = 29.2,\ \bar{y} = 24.4$ | B1 | Find both sample means |
| $s_x^2 = (55500 - 1752^2/60)/59$ and $s_y^2 = (33500 - 1220^2/50)/49$ | M1 A1 | Estimate both population variances (to 4 s.f.) |
| $s_x^2 = 73.59$ and $s_y^2 = 76.16$ | A1 | (allow biased: $72.36$ and $74.64$) |
| $s^2 = s_x^2/60 + s_y^2/50 = 2.750$ or $1.658^2$ | M1, A1 | *EITHER*: Estimate combined variance (3 s.f.) |
| $(\bar{x} - \bar{y}) \pm 1.96s$ | M1 A1 | Use this $s$ to find conf. interval |
| $4.8 \pm 3.25$ or $[1.55,\ 8.05]$ | A1 | Evaluate |
| $s^2 = (59s_x^2 + 49s_y^2)/108$ or $(55500 - 1752^2/60 + 33500 - 1220^2/50)/108$ | (M1 A1) | *OR*: Estimate common variance (to 3 s.f.) (note $s_x$ and $s_y$ not needed explicitly) |
| $= 74.8$ or $8.65^2$ or $3364/45$ | | |
| $(\bar{x} - \bar{y}) \pm 1.96s\sqrt{1/60 + 1/50}$ | (M1 A1) | Use this $s$ to find conf. interval |
| $4.8 \pm 3.24[5]$ or $[1.55[5],\ 8.04[5]]$ | (A1) | Evaluate |

**Total: 9 marks**

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8 The number, $x$, of a certain type of sea shell was counted at 60 randomly chosen sites, each one metre square, along the coastline in country $A$. The number, $y$, of the same type of shell was counted at 50 randomly chosen sites, each one metre square, along the coastline in country $B$. The results are summarised as follows.

$$\Sigma x = 1752 \quad \Sigma x ^ { 2 } = 55500 \quad \Sigma y = 1220 \quad \Sigma y ^ { 2 } = 33500$$

Find a 95\% confidence interval for the difference between the mean number of sea shells, per square metre, on the coastlines in country $A$ and in country $B$.

\hfill \mbox{\textit{CAIE FP2 2013 Q8 [9]}}