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CIE 9709 2025 June Paper 65 Q4

A Level / CIE / S2

CIE 9709 2025 June Paper 65 Paper · Question 4

题目

Problem

The numbers of cars and trucks arriving per minute at a fuel station are modelled by independent variables with distributions Po(0.8)\mathrm{Po}(0.8) and Po(0.5)\mathrm{Po}(0.5) respectively.

(a) Find the probability that at least 44 cars and at least 22 trucks arrive at the fuel station during a randomly chosen 5-minute period.

(4)

(b) Use a suitable approximating distribution to find the probability that a total of fewer than 145145 cars and trucks arrive at the fuel station during a randomly chosen 2-hour period.

(5)
题目中文翻译

每分钟到达某加油站的汽车数和卡车数分别用相互独立的变量建模，服从分布 Po(0.8)\mathrm{Po}(0.8) 和 Po(0.5)\mathrm{Po}(0.5)。

(a) 求在随机选取的一段 55 分钟内，至少有 44 辆汽车且至少有 22 辆卡车到达该加油站的概率。

(4)

(b) 用一个合适的近似分布，求在随机选取的一段 22 小时内，到达该加油站的汽车和卡车总数少于 145145 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

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分别转换五分钟内的参数，再将两个独立右尾事件的概率相乘。

答题过程

展开

Let C,TC,T be the numbers of cars and trucks arriving in five minutes. Then C∼Po(4)C\sim\mathrm{Po}(4) and T∼Po(2.5)T\sim\mathrm{Po}(2.5) independently.

P(C≥4)= 1−e−4(1+4) −e−4(422!+433!)≈ 0.566530,\begin{align*} \mathrm P(C\ge4)=&\,1-\mathrm e^{-4}(1+4)\\[4mm] &\,-\mathrm e^{-4}\bigg(\frac{4^2}{2!}+\frac{4^3}{3!}\bigg)\\[4mm] \approx&\,0.566530, \end{align*}

and

P(T≥2)=1−e−2.5(1+2.5)≈0.712703.\mathrm P(T\ge2)=1-\mathrm e^{-2.5}(1+2.5)\approx0.712703.

By independence,

 P(C≥4, T≥2)= P(C≥4)P(T≥2)≈ 0.404.\begin{align*} &\,\mathrm P(C\ge4,\ T\ge2)\\[4mm] =&\,\mathrm P(C\ge4)\mathrm P(T\ge2)\\[4mm] \approx&\,\boxed{0.404}. \end{align*}

(b)

解法一

思路

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两小时是一百二十分钟，合并参数后用正态近似并修正整数边界。

答题过程

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Let NN be the total number of vehicles arriving in two hours. Then

N∼Po(120(0.8+0.5))=Po(156).N\sim\mathrm{Po}(120(0.8+0.5))=\mathrm{Po}(156).

Since 156>15156>15, approximate NN by Y∼N(156,156)Y\sim\mathrm N(156,156). With a continuity correction,

P(N<145)= P(N≤144)≈ P(Y<144.5)= Φ(144.5−156156)≈ 0.179.\begin{align*} \mathrm P(N<145)=&\,\mathrm P(N\le144)\\[4mm] \approx&\,\mathrm P(Y<144.5)\\[4mm] =&\,\Phi\bigg(\frac{144.5-156}{\sqrt{156}}\bigg)\\[4mm] \approx&\,\boxed{0.179}. \end{align*}