CAIE S2 2024 June — Question 4 6 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeDirect comparison with scalar multiple (different variables)
DifficultyStandard +0.3 Part (a) is a standard symmetry result requiring minimal calculation. Part (b) requires forming a linear combination (X₁ - 2X₂), finding its distribution using variance rules, then a routine normal probability calculation. This is a textbook application of linear combinations with no novel insight required, making it slightly easier than average.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

4 A random variable \(X\) has the distribution \(\mathrm { N } ( 10,12 )\). Two independent values of \(X\), denoted by \(X _ { 1 }\) and \(X _ { 2 }\), are chosen at random.
  1. Write down the value of \(\mathrm { P } \left( X _ { 1 } > X _ { 2 } \right)\).
  2. Find \(\mathrm { P } \left( X _ { 1 } > 2 X _ { 2 } - 3 \right)\).

Question 4:
Part (a):
AnswerMarks Guidance
AnswerMark Guidance
\(0.5\)B1
Total: 1
Part (b):
AnswerMarks Guidance
AnswerMark Guidance
\(E(X_1 - 2X_2 + 3) = 10 - 20 + 3\ [= -7]\) or \(E(2X_2 - X_1 - 3) = 20 - 10 - 3\ [= 7]\)B1 Or equivalent using \(X_1 - 2X_2 = 10 - 20\ [= -10]\) or \(2X_2 - X_1 = 20 - 10\ [= +10]\)
\(\text{Var}(X_1 - 2X_2 + 3) = 12 + 2^2 \times 12 + 0\) \([= 60]\)B1
\(\frac{0-(-7)}{\sqrt{60}}\) \([= 0.904]\)M1 Or numerator \(3-{'}10{'}\) or \(-3-({'-10'})\), but not \('-3-10'\) (i.e. numerator must be \('7'\) or \('-7'\))
\(1 - \Phi(0.904)\)M1 For area consistent with their working
\(= 0.183\)A1
Total: 5
## Question 4:

**Part (a):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $0.5$ | B1 | |
| **Total: 1** | | |

**Part (b):**

| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X_1 - 2X_2 + 3) = 10 - 20 + 3\ [= -7]$ or $E(2X_2 - X_1 - 3) = 20 - 10 - 3\ [= 7]$ | B1 | Or equivalent using $X_1 - 2X_2 = 10 - 20\ [= -10]$ or $2X_2 - X_1 = 20 - 10\ [= +10]$ |
| $\text{Var}(X_1 - 2X_2 + 3) = 12 + 2^2 \times 12 + 0$ $[= 60]$ | B1 | |
| $\frac{0-(-7)}{\sqrt{60}}$ $[= 0.904]$ | M1 | Or numerator $3-{'}10{'}$ or $-3-({'-10'})$, but not $'-3-10'$ (i.e. numerator must be $'7'$ or $'-7'$) |
| $1 - \Phi(0.904)$ | M1 | For area consistent with their working |
| $= 0.183$ | A1 | |
| **Total: 5** | | |

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4 A random variable $X$ has the distribution $\mathrm { N } ( 10,12 )$. Two independent values of $X$, denoted by $X _ { 1 }$ and $X _ { 2 }$, are chosen at random.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $\mathrm { P } \left( X _ { 1 } > X _ { 2 } \right)$.
\item Find $\mathrm { P } \left( X _ { 1 } > 2 X _ { 2 } - 3 \right)$.
\end{enumerate}

\hfill \mbox{\textit{CAIE S2 2024 Q4 [6]}}