CAIE FP2 2013 November — Question 7 7 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypePooled variance estimation
DifficultyStandard +0.8 This is a Further Maths statistics question requiring knowledge of pooled variance estimation formula and algebraic manipulation. While the concept is standard for FM students, it requires careful setup of the pooled variance formula with different sample sizes (n and 2n), substitution of given summations, and solving a resulting equation. The multi-step algebraic work and the specific FM context place it moderately above average difficulty.
Spec5.05b Unbiased estimates: of population mean and variance

7 Two independent random variables \(X\) and \(Y\) have distributions with the same variance \(\sigma ^ { 2 }\). Random samples of \(n\) observations of \(X\) and \(2 n\) observations of \(Y\) are taken and the results are summarised by $$\Sigma x = 10.0 , \quad \Sigma x ^ { 2 } = 25.0 , \quad \Sigma y = 15.0 , \quad \Sigma y ^ { 2 } = 43.5 .$$ Given that the pooled estimate of \(\sigma ^ { 2 }\) is 2 , find the value of \(n\).

Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\Sigma(x - \bar{x})^2 = 25 - \frac{10^2}{n}\)B1 Find \(\Sigma(x-\bar{x})^2\)
\(\Sigma(y - \bar{y})^2 = 43.5 - \frac{15^2}{2n}\)B1 Find \(\Sigma(y-\bar{y})^2\)
\(\frac{(25 - \frac{10^2}{n} + 43.5 - \frac{15^2}{2n})}{(3n-2)} = 2\)M1 A1 Equate pooled estimate of \(\sigma^2\) to 2
\(12n^2 - 145n + 425 = 0\)M1 Rearrange to give quadratic in \(n\)
\(n = \frac{(145 \pm 25)}{24} = 5\) (reject \(\neq 7.08\))M1 A1 Find \(n\) (integer value)
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Sigma(x - \bar{x})^2 = 25 - \frac{10^2}{n}$ | B1 | Find $\Sigma(x-\bar{x})^2$ |
| $\Sigma(y - \bar{y})^2 = 43.5 - \frac{15^2}{2n}$ | B1 | Find $\Sigma(y-\bar{y})^2$ |
| $\frac{(25 - \frac{10^2}{n} + 43.5 - \frac{15^2}{2n})}{(3n-2)} = 2$ | M1 A1 | Equate pooled estimate of $\sigma^2$ to 2 |
| $12n^2 - 145n + 425 = 0$ | M1 | Rearrange to give quadratic in $n$ |
| $n = \frac{(145 \pm 25)}{24} = 5$ (reject $\neq 7.08$) | M1 A1 | Find $n$ (integer value) |

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

$$\Sigma x = 10.0 , \quad \Sigma x ^ { 2 } = 25.0 , \quad \Sigma y = 15.0 , \quad \Sigma y ^ { 2 } = 43.5 .$$

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

\hfill \mbox{\textit{CAIE FP2 2013 Q7 [7]}}