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CIE 9709 2024 June Paper 61 Q2

A Level / CIE / S2

CIE 9709 2024 June Paper 61 Paper · Question 2

题目

Problem

The random variable XX has the distribution N(31.2,10.42)\mathrm{N}(31.2,10.4^2). Two independent random values of XX, denoted by X1X_1 and X2X_2, are chosen.

Find P(X1>3X2)\mathrm{P}(X_1>3X_2).

(5)
题目中文翻译

随机变量 XX 服从 N(31.2,10.42)\mathrm{N}(31.2,10.4^2) 分布。从 XX 中独立地随机选取两个值，分别记为 X1X_1 和 X2X_2。

求 P(X1>3X2)\mathrm{P}(X_1>3X_2)。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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建立X1减3X2的正态分布；乘3会使方差乘9，而两项独立使方差相加。

答题过程

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Since X1X_1 and X2X_2 are independent,

E(X1−3X2)= 31.2−3(31.2)= −62.4,Var(X1−3X2)= 10.42+32(10.42)= 1081.6.\begin{align*} \mathrm{E}(X_1-3X_2)=&\,31.2-3(31.2)\\[4mm] =&\,-62.4,\\[4mm] \mathrm{Var}(X_1-3X_2)=&\,10.4^2+3^2(10.4^2)\\[4mm] =&\,1081.6. \end{align*}

Thus D=X1−3X2∼N(−62.4,1081.6)D=X_1-3X_2\sim\mathrm{N}(-62.4,1081.6), and

P(X1>3X2)= P(D>0)= 1−Φ(62.41081.6)= 0.0288898…= 0.0289(3 s.f.).\begin{align*} \mathrm{P}(X_1>3X_2)=&\,\mathrm{P}(D>0)\\[4mm] =&\,1-\Phi\bigg(\frac{62.4}{\sqrt{1081.6}}\bigg)\\[4mm] =&\,0.0288898\ldots\\[4mm] =&\,\boxed{0.0289}\quad\text{(3 s.f.).} \end{align*}