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CIE 9709 2025 November Paper 61 Q6

A Level / CIE / S2

CIE 9709 2025 Nov Paper 61 Paper · Question 6

题目

Problem

The masses, in kilograms, of large and small bags of potatoes have the independent distribution N(2.5,0.05)\mathrm{N}(2.5,0.05) and N(0.8,0.02)\mathrm{N}(0.8,0.02) respectively.

(a) Find the probability that the total mass of a randomly chosen large bag of potatoes and a randomly chosen small bag of potatoes is more than 3.55 kg3.55\,\mathrm{kg}.

(5)

(b) Find the probability that the mass of a randomly chosen large bag of potatoes is less than 33 times the mass of a randomly chosen small bag of potatoes.

(5)
题目中文翻译

大袋和小袋土豆的质量（以千克为单位）相互独立，分别服从分布 N(2.5,0.05)\mathrm{N}(2.5,0.05) 和 N(0.8,0.02)\mathrm{N}(0.8,0.02)。

(a) 求随机选取的一大袋土豆和随机选取的一小袋土豆的总质量超过 3.55 kg3.55\,\mathrm{kg} 的概率。

(5)

(b) 求随机选取的一大袋土豆的质量小于随机选取的一小袋土豆的质量的 33 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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独立正态质量相加，均值和方差分别相加；连续质量不作连续性修正。

答题过程

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Let LL and SS be the masses of a large and a small bag. By independence,

L+S∼ N(2.5+0.8, 0.05+0.02)= N(3.3,0.07).\begin{align*} L+S\sim&\,\mathrm N(2.5+0.8,\,0.05+0.02)\\[4mm] =&\,\mathrm N(3.3,0.07). \end{align*}

Hence

P(L+S>3.55)= 1−Φ(3.55−3.30.07)= 1−Φ(0.944911…)≈ 0.172.\begin{align*} \mathrm P(L+S>3.55) =&\,1-\Phi\bigg(\frac{3.55-3.3}{\sqrt{0.07}}\bigg)\\[4mm] =&\,1-\Phi(0.944911\ldots)\\[4mm] \approx&\,\boxed{0.172}. \end{align*}

(b)

解法一

思路

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本问是同一个小袋质量的三倍，不是三个独立小袋的总和，因此其方差要乘以 323^2。

答题过程

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Put D=L−3SD=L-3S. Then

E(D)= 2.5−3(0.8)=0.1,Var⁡(D)= 0.05+(−3)2(0.02)= 0.23.\begin{align*} \mathrm E(D)=&\,2.5-3(0.8)=0.1,\\[4mm] \operatorname{Var}(D)=&\,0.05+(-3)^2(0.02)\\[4mm] =&\,0.23. \end{align*}

Thus D∼N(0.1,0.23)D\sim\mathrm N(0.1,0.23), and

P(L<3S)= P(D<0)= Φ(0−0.10.23)= Φ(−0.208514…)≈ 0.417.\begin{align*} \mathrm P(L<3S) =&\,\mathrm P(D<0)\\[4mm] =&\,\Phi\bigg(\frac{0-0.1}{\sqrt{0.23}}\bigg)\\[4mm] =&\,\Phi(-0.208514\ldots)\\[4mm] \approx&\,\boxed{0.417}. \end{align*}