题目

Problem

The volumes of regular and large coffees at a certain café have independent normal distributions. The mean and standard deviation of the volumes of regular coffees are $215\,\mathrm{cm}^3$ and $2.8\,\mathrm{cm}^3$ respectively. The mean and standard deviation of the volumes of large coffees are $330\,\mathrm{cm}^3$ and $3.3\,\mathrm{cm}^3$ respectively. Find the probability that the total volume of three randomly chosen regular coffees is greater than the total volume of two randomly chosen large coffees.

(5)

题目中文翻译

某咖啡馆的普通杯和大杯咖啡的体积相互独立，且都服从正态分布。普通杯咖啡体积的均值和标准差分别为 $215\,\mathrm{cm}^3$ 和 $2.8\,\mathrm{cm}^3$ 。大杯咖啡体积的均值和标准差分别为 $330\,\mathrm{cm}^3$ 和 $3.3\,\mathrm{cm}^3$ 。求随机选取的三杯普通杯咖啡的总体积大于随机选取的两杯大杯咖啡的总体积的概率。

(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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三杯普通咖啡是三个独立样本之和，不是同一随机变量乘以三；因此其方差是 $3\sigma^2$ ，不是 $9\sigma^2$ 。

答题过程

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Let $R$ be the total volume of three regular coffees and $L$ the total volume of two large coffees. Independence gives

\begin{align*} R\sim&\,\mathrm N(3(215),\,3(2.8^2)),\\[4mm] L\sim&\,\mathrm N(2(330),\,2(3.3^2)). \end{align*}

Hence $D=R-L$ is normal with

\begin{align*} \mathrm E(D)=&\,645-660=-15,\\[4mm] \operatorname{Var}(D)=&\,3(2.8^2)+2(3.3^2)\\[4mm] =&\,45.3. \end{align*}

Therefore

\begin{align*} \mathrm P(R>L) =&\,\mathrm P(D>0)\\[4mm] =&\,1-\Phi\bigg(\frac{0-(-15)}{\sqrt{45.3}}\bigg)\\[4mm] =&\,1-\Phi(2.22865\ldots)\\[4mm] \approx&\,\boxed{0.0129}. \end{align*}

CIE 9709 2026 June Paper 65 Q2

题目

解答

解法一

思路

答题过程