| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2010 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Expectation and variance with context application |
| Difficulty | Moderate -0.3 This is a straightforward application of standard results for linear combinations of independent normal variables (part i) followed by a routine one-sample z-test (part ii). Both parts require only direct recall and application of formulas with no problem-solving insight or novel reasoning, making it slightly easier than average. |
| Spec | 5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(E(F) = 28 + \frac{1}{2} \times 52 = 54\) | B1 | |
| \(\text{Var}(F) = 5.6^2 + \frac{1}{4} \times 12.4^2\) | M1 | |
| \(= 69.8\) | A1 [3] | \(\sqrt{69.8}\) or \(8.35\): M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(H_0\): Grinford mean \(= 54\); \(H_1\): Grinford mean \(< 54\) | B1ft | Allow "\(\mu\)", otherwise undefined mean: B0; ft their 54 |
| \(\frac{49 - 54}{\sqrt{\frac{69.8}{10}}}\) | M1 | Standardising must have \(\sqrt{10}\) |
| \(= -1.89(3)\) or \(-1.89(2)\), allow \(+\) | A1 | |
| Comp with \(-1.645\) (or \(1.893\) with \(1.645\)) | M1 | Comp \(P(z < -1.893)\) with \(0.05\); Allow comparison with \(1.96\) for consistent 2-tail test |
| Evidence that Grinford mean lower | A1ft [5] | Allow "Accept Grinford mean lower". No contradictions. OR Alt methods: \((x-54)/(\sqrt{69.8/10}) = 1.645\) giving \(x = 49.65\), compare with 49 scores M1A1M1A1ft. oe. No mixed methods. |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(F) = 28 + \frac{1}{2} \times 52 = 54$ | B1 | |
| $\text{Var}(F) = 5.6^2 + \frac{1}{4} \times 12.4^2$ | M1 | |
| $= 69.8$ | A1 [3] | $\sqrt{69.8}$ or $8.35$: M1A0 |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $H_0$: Grinford mean $= 54$; $H_1$: Grinford mean $< 54$ | B1ft | Allow "$\mu$", otherwise undefined mean: B0; ft their 54 |
| $\frac{49 - 54}{\sqrt{\frac{69.8}{10}}}$ | M1 | Standardising must have $\sqrt{10}$ |
| $= -1.89(3)$ or $-1.89(2)$, allow $+$ | A1 | |
| Comp with $-1.645$ (or $1.893$ with $1.645$) | M1 | Comp $P(z < -1.893)$ with $0.05$; Allow comparison with $1.96$ for consistent 2-tail test |
| Evidence that Grinford mean lower | A1ft [5] | Allow "Accept Grinford mean lower". No contradictions. OR Alt methods: $(x-54)/(\sqrt{69.8/10}) = 1.645$ giving $x = 49.65$, compare with 49 scores M1A1M1A1ft. oe. No mixed methods. |
---5 The marks of candidates in Mathematics and English in 2009 were represented by the independent random variables $X$ and $Y$ with distributions $\mathrm { N } \left( 28,5.6 ^ { 2 } \right)$ and $\mathrm { N } \left( 52,12.4 ^ { 2 } \right)$ respectively. Each candidate's marks were combined to give a final mark $F$, where $F = X + \frac { 1 } { 2 } Y$.\\
(i) Find $\mathrm { E } ( F )$ and $\operatorname { Var } ( F )$.\\
(ii) The final marks of a random sample of 10 candidates from Grinford in 2009 had a mean of 49. Test at the 5\% significance level whether this result suggests that the mean final mark of all candidates from Grinford in 2009 was lower than elsewhere.
\hfill \mbox{\textit{CAIE S2 2010 Q5 [8]}}