CAIE FP2 2011 June — Question 6 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2011
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypePooled variance estimate calculation
DifficultyStandard +0.8 This question requires understanding of pooled variance estimation formula and algebraic manipulation to solve for n. While the concept is A-level appropriate, it involves working backwards from the pooled estimate through a multi-step calculation with summations, requiring careful algebraic handling of the formula with two samples of different sizes—more demanding than routine variance calculations.
Spec5.05b Unbiased estimates: of population mean and variance

6 The independent random variables \(X\) and \(Y\) have distributions with the same variance \(\sigma ^ { 2 }\). Random samples of 5 observations of \(X\) and \(n\) observations of \(Y\) are made and the results are summarised by $$\Sigma x = 5.5 , \quad \Sigma x ^ { 2 } = 15.05 , \quad \Sigma y = 8.0 , \quad \Sigma y ^ { 2 } = 36.4$$ Given that the pooled estimate of \(\sigma ^ { 2 }\) is 3 , find the value of \(n\).

Question 6:
AnswerMarks Guidance
Working/AnswerMark Guidance
Find pooled estimate: \(\frac{(15{\cdot}05 - 5{\cdot}5^2/5 + 36{\cdot}4 - 8^2/n)}{(3+n)}\)M1 A1
Equate to 3 and rearrange: \(45{\cdot}4 - 64/n = 9 + 3n\)M1 A1
\(3n^2 - 36{\cdot}4n + 64 = 0\)A1
Solve for \(n\): \(n = (36{\cdot}4 \pm 23{\cdot}6)/6 = 10\)M1 A1 [Subtotal: 7]
Total: 7
## Question 6:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Find pooled estimate: $\frac{(15{\cdot}05 - 5{\cdot}5^2/5 + 36{\cdot}4 - 8^2/n)}{(3+n)}$ | M1 A1 | |
| Equate to 3 and rearrange: $45{\cdot}4 - 64/n = 9 + 3n$ | M1 A1 | |
| $3n^2 - 36{\cdot}4n + 64 = 0$ | A1 | |
| Solve for $n$: $n = (36{\cdot}4 \pm 23{\cdot}6)/6 = 10$ | M1 A1 | [Subtotal: 7] |

**Total: 7**

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6 The independent random variables $X$ and $Y$ have distributions with the same variance $\sigma ^ { 2 }$. Random samples of 5 observations of $X$ and $n$ observations of $Y$ are made and the results are summarised by

$$\Sigma x = 5.5 , \quad \Sigma x ^ { 2 } = 15.05 , \quad \Sigma y = 8.0 , \quad \Sigma y ^ { 2 } = 36.4$$

Given that the pooled estimate of $\sigma ^ { 2 }$ is 3 , find the value of $n$.

\hfill \mbox{\textit{CAIE FP2 2011 Q6 [7]}}