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CIE 9709 2022 June Paper 63 Q4

A Level / CIE / S2

CIE 9709 2022 June Paper 63 Paper · Question 4

题目

Problem

Each box of Seeds & Raisins contains SS grams of seeds and RR grams of raisins. The weight of a box, when empty, is BB grams. SS, RR and BB are independent random variables, where S∼N(300,45)S\sim\mathrm{N}(300,45), R∼N(200,25)R\sim\mathrm{N}(200,25) and B∼N(15,4)\mathrm{B}\sim\mathrm{N}(15,4). A full box of Seeds & Raisins is chosen at random.

(a) Find the probability that the total weight of the box and its contents is more than 500500 grams.

(5)

(b) Find the probability that the weight of seeds in the box is less than 1.41.4 times the weight of raisins in the box.

(5)
题目中文翻译

每盒 Seeds & Raisins 含有 SS 克种子和 RR 克葡萄干。空盒重 BB 克。SS、RR 和 BB 是相互独立的随机变量，其中 S∼N(300,45)S\sim\mathrm{N}(300,45)，R∼N(200,25)R\sim\mathrm{N}(200,25)，B∼N(15,4)\mathrm{B}\sim\mathrm{N}(15,4)。随机选取一整盒 Seeds & Raisins。

(a) 求盒子与盒内食品的总重量超过 500500 克的概率。

(5)

(b) 求盒中种子的重量小于葡萄干重量的 1.41.4 倍的概率。

(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

(a)

解法一

思路

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满盒总质量为三项独立正态变量之和，均值方差分别相加。

答题过程

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Let T=S+R+BT=S+R+B. By independence,

T∼ N(300+200+15, 45+25+4)= N(515,74).\begin{align*} T\sim&\,\mathrm{N}(300+200+15,\ 45+25+4)\\[4mm] =&\,\mathrm{N}(515,74). \end{align*}

Hence

P(T>500)= 1−Φ(500−51574)= 0.959(3 s.f.).\begin{align*} \mathrm{P}(T>500)=&\,1-\Phi\bigg(\frac{500-515}{\sqrt{74}}\bigg)\\[4mm] =&\,\boxed{0.959}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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建立S减1.4R的正态分布，求差小于0的概率，方差中R项乘1.4平方。

答题过程

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By independence,

D=S−1.4R.D=S-1.4R.

Then

D∼ N(300−1.4(200), 45+(1.4)2(25))= N(20,94).\begin{align*} D\sim&\,\mathrm{N}(300-1.4(200),\ 45+(1.4)^2(25))\\[4mm] =&\,\mathrm{N}(20,94). \end{align*}

Therefore

P(S<1.4R)= Φ(0−2094)= 0.0196(3 s.f.).\begin{align*} \mathrm{P}(S<1.4R)=&\,\Phi\bigg(\frac{0-20}{\sqrt{94}}\bigg)\\[4mm] =&\,\boxed{0.0196}\quad\text{(3 s.f.).} \end{align*}