Pre-U Pre-U 9795/2 Specimen — Question 7 6 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
SessionSpecimen
Marks6
TopicLinear combinations of normal random variables
TypeDirect comparison with scalar multiple (different variables)
DifficultyChallenging +1.2 This question requires constructing a new random variable (M - 0.75F), finding its distribution using properties of linear combinations of independent normals, then calculating a probability. While it involves multiple steps and understanding that independence allows variance addition, the technique is standard for Further Maths statistics and follows a well-established procedure once the setup is recognized.
Spec2.04f Find normal probabilities: Z transformation5.04b Linear combinations: of normal distributions
The length \(M\) of male snakes of a certain species may be regarded as a normal random variable with mean \(0.45\) metres and standard deviation \(0.06\) metres. The length \(F\) of female snakes of the same species may be regarded as a normal random variable with mean \(0.55\) metres and standard deviation \(0.08\) metres. Assuming that \(M\) and \(F\) are independent, find the probability that a randomly chosen male snake of this species is more than three-quarters of the length of a randomly chosen female snake of this species. [6]
AnswerMarks Guidance
\(M > \frac{3}{4}F \Rightarrow 4M - 3F > 0\) (Mark for identifying \(4M-3F\) OE)M1
\(4M - 3F \sim N(0.15, 0.1152)\)B1, B1
\(z = \frac{0-0.15}{\sqrt{0.1152}} = -0.4419\)M1, A1
\(P(4M-3F > 0) = 0.671\)A1 6
[Or equivalent alternative solution based on \(M - \frac{3}{4}F \sim N(0.0375, 0.0072)\)]
$M > \frac{3}{4}F \Rightarrow 4M - 3F > 0$ (Mark for identifying $4M-3F$ OE) | M1 |

$4M - 3F \sim N(0.15, 0.1152)$ | B1, B1 |

$z = \frac{0-0.15}{\sqrt{0.1152}} = -0.4419$ | M1, A1 |

$P(4M-3F > 0) = 0.671$ | A1 | **6**

[Or equivalent alternative solution based on $M - \frac{3}{4}F \sim N(0.0375, 0.0072)$]
The length $M$ of male snakes of a certain species may be regarded as a normal random variable with mean $0.45$ metres and standard deviation $0.06$ metres. The length $F$ of female snakes of the same species may be regarded as a normal random variable with mean $0.55$ metres and standard deviation $0.08$ metres. Assuming that $M$ and $F$ are independent, find the probability that a randomly chosen male snake of this species is more than three-quarters of the length of a randomly chosen female snake of this species. [6]

\hfill \mbox{\textit{Pre-U Pre-U 9795/2  Q7 [6]}}
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