CAIE S2 2006 June — Question 2 4 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2006
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeExpectation and variance with context application
DifficultyModerate -0.5 This is a straightforward application of standard results for linear combinations of independent random variables (E(aX+bY) and Var(aX+bY) formulas). The context is clear, all values are given, and it requires only direct substitution into well-known formulas with no problem-solving or conceptual insight needed. Slightly easier than average due to its routine nature, though the weighted combination adds minor computational care.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)

2 A mathematics module is assessed by an examination and by coursework. The examination makes up \(75 \%\) of the total assessment and the coursework makes up \(25 \%\). Examination marks, \(X\), are distributed with mean 53.2 and standard deviation 9.3. Coursework marks, \(Y\), are distributed with mean 78.0 and standard deviation 5.1. Examination marks and coursework marks are independent. Find the mean and standard deviation of the combined mark \(0.75 X + 0.25 Y\).

Part (i):
AnswerMarks Guidance
mean \(= 0.75 \times 53.2 + 0.25 \times 78 = 59.4\)B1, M1, M1, A1 (4 marks) Correct answer. \(0.75\) and \(0.25\) multiplied by var or sd. \(0.75\) or \(0.75^2\) mult by \(9.3^2\) and \(0.25\) or \(0.25^2\) mult by \(5.1^2 = 59.4\) or sd \(= 7.09\), correct answer
var \(= 0.75^2 \times 9.3^2 + 0.25^2 \times 5.1^2 = 50.27\)
sd \(= 7.09\)
**Part (i):**
mean $= 0.75 \times 53.2 + 0.25 \times 78 = 59.4$ | B1, M1, M1, A1 (4 marks) | Correct answer. $0.75$ and $0.25$ multiplied by var or sd. $0.75$ or $0.75^2$ mult by $9.3^2$ and $0.25$ or $0.25^2$ mult by $5.1^2 = 59.4$ or sd $= 7.09$, correct answer

var $= 0.75^2 \times 9.3^2 + 0.25^2 \times 5.1^2 = 50.27$

sd $= 7.09$
2 A mathematics module is assessed by an examination and by coursework. The examination makes up $75 \%$ of the total assessment and the coursework makes up $25 \%$. Examination marks, $X$, are distributed with mean 53.2 and standard deviation 9.3. Coursework marks, $Y$, are distributed with mean 78.0 and standard deviation 5.1. Examination marks and coursework marks are independent. Find the mean and standard deviation of the combined mark $0.75 X + 0.25 Y$.

\hfill \mbox{\textit{CAIE S2 2006 Q2 [4]}}