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CIE 9709 2025 March Paper 62 Q1

A Level / CIE / S2

CIE 9709 2025 March Paper 62 Paper · Question 1

题目

Problem

The random variables XX and YY have the independent distributions N(44,16)\mathrm{N}(44,16) and N(30,9)\mathrm{N}(30,9) respectively.

Find P(X−Y<15)\mathrm{P}(X-Y<15).

(5)
题目中文翻译

随机变量 XX 和 YY 相互独立,分别服从 N(44,16)\mathrm{N}(44,16) 和 N(30,9)\mathrm{N}(30,9) 分布。

求 P(X−Y<15)\mathrm{P}(X-Y<15)。

(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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独立正态变量之差仍为正态分布。均值相减,方差相加,再标准化。

答题过程

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Since XX and YY are independent,

X−Y∼ N(44−30,16+9)= N(14,25).\begin{align*} X-Y\sim&\,\mathrm{N}(44-30,16+9)\\[4mm] =&\,\mathrm{N}(14,25). \end{align*}

Hence

P(X−Y<15)= Φ(15−1425)= Φ(0.2)= 0.579(3 s.f.).\begin{align*} \mathrm{P}(X-Y<15)=&\,\Phi\bigg(\frac{15-14}{\sqrt{25}}\bigg)\\[4mm] =&\,\Phi(0.2)\\[4mm] =&\,\boxed{0.579}\quad\text{(3 s.f.).} \end{align*}