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CIE 9709 2023 November Paper 63 Q4

A Level / CIE / S2

CIE 9709 2023 Nov Paper 63 Paper · Question 4

题目

Problem

The masses, in kilograms, of chemicals AA and BB produced per day by a factory are modelled by the independent random variables XX and YY respectively, where X∼N(10.3,5.76)X\sim\mathrm{N}(10.3,5.76) and Y∼N(11.4,9.61)Y\sim\mathrm{N}(11.4,9.61).

The income generated by the chemicals is $2.50 per kilogram for AA and $3.25 per kilogram for BB.

(a) Find the mean and variance of the daily income generated by chemical AA.

(2)

(b) Find the probability that, on a randomly chosen day, the income generated by chemical AA is greater than the income generated by chemical BB.

(6)
题目中文翻译

某工厂每天生产的化学品 AA 和 BB 的质量（以千克为单位）分别用相互独立的随机变量 XX 和 YY 描述，其中 X∼N(10.3,5.76)X\sim\mathrm{N}(10.3,5.76)，Y∼N(11.4,9.61)Y\sim\mathrm{N}(11.4,9.61)。

化学品 AA 每千克带来 2.502.50 美元的收入，化学品 BB 每千克带来 3.253.25 美元的收入。

(a) 求化学品 AA 每天带来的收入的均值和方差。

(2)

(b) 随机选取一天，求当天化学品 AA 带来的收入高于化学品 BB 带来的收入的概率。

(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

(a)

解法一

思路

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收入为产量乘固定价格，均值乘2.50，方差乘2.50平方。

答题过程

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Let IA=2.50XI_A=2.50X be the daily income from chemical AA, in dollars.

E(IA)=2.50(10.3)=25.75 dollars,\mathrm{E}(I_A)=2.50(10.3)=\boxed{25.75\text{ dollars}},

and

Var(IA)=(2.50)2(5.76)=36 dollars2.\mathrm{Var}(I_A)=(2.50)^2(5.76)=\boxed{36\text{ dollars}^2}.

(b)

解法一

思路

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先求B收入的正态参数，再利用独立性建立A减B收入之差的分布。

答题过程

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For IB=3.25YI_B=3.25Y,

E(IB)= 3.25(11.4)=37.05,Var(IB)= (3.25)2(9.61)= 101.505625.\begin{align*} \mathrm{E}(I_B)=&\,3.25(11.4)=37.05,\\[4mm] \mathrm{Var}(I_B)=&\,(3.25)^2(9.61)\\[4mm] =&\,101.505625. \end{align*}

By independence,

D=IA−IB.D=I_A-I_B.

Then

D∼ N(25.75−37.05, 36+101.505625)= N(−11.3,137.505625).\begin{align*} D\sim&\,\mathrm{N}(25.75-37.05,\ 36+101.505625)\\[4mm] =&\,\mathrm{N}(-11.3,137.505625). \end{align*}

Thus

P(IA>IB)= 1−Φ(11.3137.505625)= 0.167611…= 0.168(3 s.f.).\begin{align*} \mathrm{P}(I_A>I_B)=&\,1-\Phi\bigg(\frac{11.3}{\sqrt{137.505625}}\bigg)\\[4mm] =&\,0.167611\ldots\\[4mm] =&\,\boxed{0.168}\quad\text{(3 s.f.).} \end{align*}