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CIE 9709 2024 March Paper 62 Q4

A Level / CIE / S2

CIE 9709 2024 March Paper 62 Paper · Question 4

题目

Problem

Each year a transport firm uses XX litres of gasoline and YY litres of diesel fuel, where XX and YY have the independent distributions X∼N(10 700,9502)X\sim\mathrm{N}(10\,700,950^2) and Y∼N(13 400,12102)Y\sim\mathrm{N}(13\,400,1210^2).

(a) Find the probability that in a randomly chosen year the firm uses more gasoline than diesel fuel.

(5)

The costs per litre of gasoline and diesel fuel are $0.80 and $0.85 respectively.

(b) Find the probability that the total cost of gasoline and diesel fuel in a randomly chosen year is between $20 000 and $22 000.

(5)
题目中文翻译

一家运输公司每年使用 XX 升汽油和 YY 升柴油，其中 XX 和 YY 相互独立，分别服从 X∼N(10 700,9502)X\sim\mathrm{N}(10\,700,950^2) 和 Y∼N(13 400,12102)Y\sim\mathrm{N}(13\,400,1210^2)。

(a) 随机选取一年，求该公司汽油用量超过柴油用量的概率。

(5)

汽油和柴油的每升价格分别为 0.800.80 美元和 0.850.85 美元。

(b) 随机选取一年，求汽油和柴油的总费用介于 20 00020\,000 美元与 22 00022\,000 美元之间的概率。

(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

(a)

解法一

思路

展开

独立正态用量之差为正态，均值相减，方差相加。

答题过程

展开

By independence,

D=X−Y.D=X-Y.

Then

D∼ N(10700−13400, 9502+12102)= N(−2700,2366600).\begin{align*} D\sim&\,\mathrm{N}(10700-13400,\ 950^2+1210^2)\\[4mm] =&\,\mathrm{N}(-2700,2366600). \end{align*}

Hence

P(X>Y)= 1−Φ(27002366600)= 0.0396213…= 0.0396(3 s.f.).\begin{align*} \mathrm{P}(X>Y)=&\,1-\Phi\bigg(\frac{2700}{\sqrt{2366600}}\bigg)\\[4mm] =&\,0.0396213\ldots\\[4mm] =&\,\boxed{0.0396}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

展开

成本是0.80X加0.85Y，均值按系数线性组合，方差按系数平方组合，再求两个边界累计概率之差。

答题过程

展开

Let C=0.80X+0.85YC=0.80X+0.85Y be the total annual fuel cost, in dollars. Then

E(C)= 0.80(10700)+0.85(13400)= 19950,\begin{align*} \mathrm{E}(C)=&\,0.80(10700)+0.85(13400)\\[4mm] =&\,19950, \end{align*}

and, by independence,

Var(C)= (0.80)2(950)2+(0.85)2(1210)2= 1635412.25.\begin{align*} \mathrm{Var}(C)=&\,(0.80)^2(950)^2+(0.85)^2(1210)^2\\[4mm] =&\,1635412.25. \end{align*}

Thus C∼N(19950,1635412.25)C\sim\mathrm{N}(19950,1635412.25), giving

 P(20000<C<22000)= Φ(22000−199501635412.25) −Φ(20000−199501635412.25)= 0.429941…= 0.430(3 s.f.).\begin{align*} &\,\mathrm{P}(20000<C<22000)\\[4mm] =&\,\Phi\bigg(\frac{22000-19950}{\sqrt{1635412.25}}\bigg)\\[4mm] &\,-\Phi\bigg(\frac{20000-19950}{\sqrt{1635412.25}}\bigg)\\[4mm] =&\,0.429941\ldots\\[4mm] =&\,\boxed{0.430}\quad\text{(3 s.f.).} \end{align*}