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CIE 9709 2024 June Paper 63 Q3

A Level / CIE / S2

CIE 9709 2024 June Paper 63 Paper · Question 3

题目

Problem

The masses in kilograms of large and small bags of cement have the independent distributions N(50,2.4)\mathrm{N}(50,2.4) and N(26,1.8)\mathrm{N}(26,1.8) respectively.

Find the probability that the total mass of 55 randomly chosen large bags of cement is greater than the total mass of 1010 randomly chosen small bags of cement.

(5)
题目中文翻译

大袋和小袋水泥的质量（以千克为单位）相互独立，分别服从 N(50,2.4)\mathrm{N}(50,2.4) 和 N(26,1.8)\mathrm{N}(26,1.8) 分布。

随机选取 55 大袋和 1010 小袋水泥，求 55 大袋的总质量大于 1010 小袋的总质量的概率。

(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

解法一

思路

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分别相加5袋与10袋的独立质量，再取差。每袋都是新的随机变量，所以总方差按袋数相加。

答题过程

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Let LL be the total mass of the five large bags and SS the total mass of the ten small bags. By independence,

L∼ N(250,12),S∼ N(260,18),\begin{align*} L\sim&\,\mathrm{N}(250,12),\\[4mm] S\sim&\,\mathrm{N}(260,18), \end{align*}

so

D=L−S.D=L-S.

Then

D∼ N(250−260, 12+18)= N(−10,30).\begin{align*} D\sim&\,\mathrm{N}(250-260,\ 12+18)\\[4mm] =&\,\mathrm{N}(-10,30). \end{align*}

Therefore

P(L>S)= 1−Φ(0−(−10)30)= 0.0339446…= 0.0339(3 s.f.).\begin{align*} \mathrm{P}(L>S)=&\,1-\Phi\bigg(\frac{0-(-10)}{\sqrt{30}}\bigg)\\[4mm] =&\,0.0339446\ldots\\[4mm] =&\,\boxed{0.0339}\quad\text{(3 s.f.).} \end{align*}