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CIE 9709 2023 November Paper 62 Q6

A Level / CIE / S2

CIE 9709 2023 Nov Paper 62 Paper · Question 6

题目

Problem

A factory makes loaves of bread in batches. One batch of loaves contains XX kilograms of dried yeast and YY kilograms of flour, where XX and YY have the independent distributions N(0.7,0.022)\mathrm{N}(0.7,0.02^2) and N(100.0,3.02)\mathrm{N}(100.0,3.0^2) respectively.

Dried yeast costs $13.50 per kilogram and flour costs $0.90 per kilogram. For making one batch of bread the total of all other costs is $55. The factory sells each batch of bread for $200.

Find the probability that the profit made on one randomly chosen batch of bread is greater than $40.

(7)
题目中文翻译

一家工厂分批制作面包。每批面包含 XX 千克干酵母和 YY 千克面粉,其中 XX 和 YY 相互独立,分别服从 N(0.7,0.022)\mathrm{N}(0.7,0.02^2) 和 N(100.0,3.02)\mathrm{N}(100.0,3.0^2) 分布。

干酵母每千克售价 13.5013.50 美元,面粉每千克售价 0.900.90 美元。每批面包的其他制作成本合计为 5555 美元。工厂每批面包卖 200200 美元。

随机选取一批面包,求该批面包的利润超过 4040 美元的概率。

(7)
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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先求随机原料成本的正态分布。利润大于40等价于原料成本小于105,固定收入和其他成本只影响均值,不增加方差。

答题过程

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Let C=13.50X+0.90YC=13.50X+0.90Y be the cost of dried yeast and flour, in dollars. By independence,

E(C)= 13.50(0.7)+0.90(100)= 99.45,\begin{align*} \mathrm{E}(C)=&\,13.50(0.7)+0.90(100)\\[4mm] =&\,99.45, \end{align*}

and

Var(C)= (13.50)2(0.02)2+(0.90)2(3.0)2= 0.0729+7.29=7.3629.\begin{align*} \mathrm{Var}(C)=&\,(13.50)^2(0.02)^2+(0.90)^2(3.0)^2\\[4mm] =&\,0.0729+7.29=7.3629. \end{align*}

Thus C∼N(99.45,7.3629)C\sim\mathrm{N}(99.45,7.3629).

The profit is P=200−55−C=145−CP=200-55-C=145-C, so

P>40⟺C<105.P>40\quad\Longleftrightarrow\quad C<105.

Therefore

P(P>40)= Φ(105−99.457.3629)= 0.979590…= 0.980(3 s.f.).\begin{align*} \mathrm{P}(P>40)=&\,\Phi\bigg(\frac{105-99.45}{\sqrt{7.3629}}\bigg)\\[4mm] =&\,0.979590\ldots\\[4mm] =&\,\boxed{0.980}\quad\text{(3 s.f.).} \end{align*}