Pre-U Pre-U 9795/2 2015 June — Question 1 4 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2015
SessionJune
Marks4
TopicLinear combinations of normal random variables
TypeTwo or more different variables
DifficultyStandard +0.8 This question requires understanding of linear combinations of independent normal variables, including mean and variance properties. Part (i) involves setting up and solving simultaneous equations from the given distribution of 2X-5Y. Part (ii) requires finding the distribution of X-Y and computing a probability. While the concepts are standard for Further Maths statistics, the algebraic manipulation and multi-step reasoning elevate it slightly above average difficulty.
Spec5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions

1 The independent random variables \(X\) and \(Y\) are such that $$X \sim \mathrm {~N} ( \mu , 11 ) , \quad Y \sim \mathrm {~N} \left( 10 , \sigma ^ { 2 } \right) \quad \text { and } \quad 2 X - 5 Y \sim \mathrm {~N} ( 0,144 ) .$$ Find
  1. the values of \(\mu\) and \(\sigma ^ { 2 }\),
  2. \(\mathrm { P } ( X - Y > 10 )\).

Question 1(i) and 1(ii)
(i)
\(2\mu - 50 = 0 \Rightarrow \mu = \mathbf{25}\) — M1A1
\(44 + 25\sigma^2 = 144 \Rightarrow \sigma^2 = \mathbf{4}\) — M1A1
Notes: \(44 + 5\sigma^2\) or \(22 + 5\sigma^2 \rightarrow 20\) or \(24.4\): M1A0
[4]
(ii)
\(X - Y \sim \mathrm{N}(15, 15)\) — M1
\(z = \dfrac{10 - 15}{\sqrt{15}} = -1.291\) — M1, A1
\(\mathrm{P}(X - Y > 10) = \mathbf{0.902}\) — A1
Notes: \(\mathrm{N}(\mu - 10,\ 11 + \sigma^2)\); Standardise, including \(\sqrt{\sigma^2}\); Allow \(+1.29(1)\); \([\mathrm{N}(15, 31) \rightarrow 0.898 \rightarrow 0.815\): M1M1A0A0\(]\)
[4]
**Question 1(i) and 1(ii)**

**(i)**
$2\mu - 50 = 0 \Rightarrow \mu = \mathbf{25}$ — M1A1

$44 + 25\sigma^2 = 144 \Rightarrow \sigma^2 = \mathbf{4}$ — M1A1

Notes: $44 + 5\sigma^2$ or $22 + 5\sigma^2 \rightarrow 20$ or $24.4$: M1A0

**[4]**

**(ii)**
$X - Y \sim \mathrm{N}(15, 15)$ — M1

$z = \dfrac{10 - 15}{\sqrt{15}} = -1.291$ — M1, A1

$\mathrm{P}(X - Y > 10) = \mathbf{0.902}$ — A1

Notes: $\mathrm{N}(\mu - 10,\ 11 + \sigma^2)$; Standardise, including $\sqrt{\sigma^2}$; Allow $+1.29(1)$; $[\mathrm{N}(15, 31) \rightarrow 0.898 \rightarrow 0.815$: M1M1A0A0$]$

**[4]**
1 The independent random variables $X$ and $Y$ are such that

$$X \sim \mathrm {~N} ( \mu , 11 ) , \quad Y \sim \mathrm {~N} \left( 10 , \sigma ^ { 2 } \right) \quad \text { and } \quad 2 X - 5 Y \sim \mathrm {~N} ( 0,144 ) .$$

Find\\
(i) the values of $\mu$ and $\sigma ^ { 2 }$,\\
(ii) $\mathrm { P } ( X - Y > 10 )$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q1 [4]}}