CAIE S2 2019 March — Question 2 5 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2019
SessionMarch
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeTwo or more different variables
DifficultyStandard +0.8 This requires understanding that X > 3Y can be rewritten as X - 3Y > 0, then applying the linear combination property that X - 3Y ~ N(9.2 - 3(3.0), 12.1 + 9(8.6)), and finally standardizing to find the probability. While the technique is standard for Further Maths Statistics, it requires multiple conceptual steps and careful handling of variance scaling, making it moderately challenging but within typical S2 scope.
Spec5.04b Linear combinations: of normal distributions
2 The independent random variables \(X\) and \(Y\) have the distributions \(\mathrm { N } ( 9.2,12.1 )\) and \(\mathrm { N } ( 3.0,8.6 )\) respectively. Find \(\mathrm { P } ( X > 3 Y )\).
Question 2:
AnswerMarks Guidance
\(E(X - 3Y) = 0.2\)B1 oe
\(\text{Var}(X - 3Y) = 12.1 + 9 \times 8.6\ (= 89.5)\)B1
\(\dfrac{0 - 0.2}{\sqrt{"89.5"}}\ (= -0.021)\)M1 For area consistent with their working
\(\Phi(`0.021`)\)M1
\(= 0.508\) (3 sfs)A1
Total: 5 marks
**Question 2:**

$E(X - 3Y) = 0.2$ | B1 | oe

$\text{Var}(X - 3Y) = 12.1 + 9 \times 8.6\ (= 89.5)$ | B1 |

$\dfrac{0 - 0.2}{\sqrt{"89.5"}}\ (= -0.021)$ | M1 | For area consistent with their working

$\Phi(`0.021`)$ | M1 |

$= 0.508$ (3 sfs) | A1 |

**Total: 5 marks**
2 The independent random variables $X$ and $Y$ have the distributions $\mathrm { N } ( 9.2,12.1 )$ and $\mathrm { N } ( 3.0,8.6 )$ respectively. Find $\mathrm { P } ( X > 3 Y )$.\\

\hfill \mbox{\textit{CAIE S2 2019 Q2 [5]}}
This paper (7 questions)
View full paper
Q1 4 Q2 5 Q3 6 Q4 7 Q5 8 Q6 10 Q7 10