题目

Problem

The mass, in tonnes, of steel produced per day at a factory is normally distributed with mean $65.2$ and standard deviation $3.6$ . It can be assumed that the mass of steel produced each day is independent of other days. The factory makes $50 profit on each tonne of steel produced.

Find the probability that the total profit made in a randomly chosen $7$ -day week is less than $22 000.

(6)

题目中文翻译

某工厂每天生产的钢材质量（以吨为单位）服从均值为 $65.2$ 、标准差为 $3.6$ 的正态分布。可以假设各天的钢材产量相互独立。该工厂每生产一吨钢材可获利 $50$ 美元。

随机选取一个为期 $7$ 天的星期，求这一周的总利润少于 $22\,000$ 美元的概率。

(6)

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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先求七天总产量的正态分布，利润小于22000等价于总产量小于440吨。

答题过程

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Let $S$ be the total steel production over the seven days. By independence,

\begin{align*} S\sim&\,\mathrm{N}(7(65.2),\ 7(3.6)^2)\\[4mm] =&\,\mathrm{N}(456.4,90.72). \end{align*}

The total profit is $P=50S$ , so

P<22000\quad\Longleftrightarrow\quad S<440.

Hence

\begin{align*} \mathrm{P}(P<22000)=&\,\Phi\bigg(\frac{440-456.4}{\sqrt{90.72}}\bigg)\\[4mm] =&\,0.0425494\ldots\\[4mm] =&\,\boxed{0.0425}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2023 June Paper 63 Q4

题目

解答

解法一

思路

答题过程