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CIE 9709 2026 June Paper 63 Q4

A Level / CIE / S2

CIE 9709 2026 June Paper 63 Paper · Question 4

题目

Problem

Cars arrive at a junction at a constant average rate of 3.23.2 per minute.

(a) State a condition for the number of cars arriving per minute to have a Poisson distribution.

(1)

It is now given that the number of cars arriving per minute has the distribution Po(3.2)\mathrm{Po}(3.2).

(b) Find the probability that at least 44 cars arrive during a randomly chosen 3-minute period.

(3)

(c) Given that at least 44 cars arrive during a particular randomly chosen 3-minute period, find the probability that exactly 55 cars arrive during this period.

(2)

(d) Use a suitable approximating distribution to find the probability that more than 6060 cars arrive during a randomly chosen 20-minute period.

(5)
题目中文翻译

汽车到达某路口的平均速率恒为每分钟 3.23.2 辆。

(a) 写出每分钟到达的汽车数服从泊松分布的一个条件。

(1)

现已知每分钟到达的汽车数服从分布 Po(3.2)\mathrm{Po}(3.2)。

(b) 求在随机选取的一段 33 分钟内至少有 44 辆汽车到达的概率。

(3)

(c) 已知在某段随机选取的 33 分钟内至少有 44 辆汽车到达，求这段时间内恰有 55 辆汽车到达的概率。

(2)

(d) 用一个合适的近似分布，求在随机选取的一段 2020 分钟内到达的汽车数超过 6060 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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平均速率恒定还不够；补充汽车到达的独立性即可满足题目所求的一个条件。

答题过程

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One suitable condition is that cars arrive independently of one another\boxed{\text{cars arrive independently of one another}}.

(b)

解法一

思路

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时间长度变成三分钟，Poisson 参数也乘以 33。至少四辆用“总概率减至多三辆”计算。

答题过程

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Let NN be the number of cars arriving in 33 minutes. Then N∼Po(3.2×3)=Po(9.6)N\sim\mathrm{Po}(3.2\times3)=\mathrm{Po}(9.6). Thus

P(N≥4)= 1−P(N≤3)= 1−e−9.6(1+9.6) −e−9.6(9.622!+9.633!)≈ 0.9861741≈ 0.986.\begin{align*} \mathrm P(N\ge4) =&\,1-\mathrm P(N\le3)\\[4mm] =&\,1-\mathrm e^{-9.6}(1+9.6)\\[4mm] &\,-\mathrm e^{-9.6}\bigg(\frac{9.6^2}{2!}+\frac{9.6^3}{3!}\bigg)\\[4mm] \approx&\,0.9861741\\[4mm] \approx&\,\boxed{0.986}. \end{align*}

(c)

解法一

思路

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恰好五辆已经包含在至少四辆这个事件内，因此条件概率的分子就是五辆的单点概率。

答题过程

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Since {N=5}⊆{N≥4}\{N=5\}\subseteq\{N\ge4\},

P(N=5∣N≥4)= P(N=5)P(N≥4)= e−9.69.65/5!0.9861741…≈ 0.0467.\begin{align*} \mathrm P(N=5\mid N\ge4) =&\,\frac{\mathrm P(N=5)}{\mathrm P(N\ge4)}\\[4mm] =&\,\frac{\mathrm e^{-9.6}9.6^5/5!}{0.9861741\ldots}\\[4mm] \approx&\,\boxed{0.0467}. \end{align*}

(d)

解法一

思路

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二十分钟内的平均车数为 6464，适用正态近似。超过六十辆表示至少六十一辆，连续性修正边界为 60.560.5。

答题过程

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Let MM be the number of cars arriving in 2020 minutes. Then

M∼Po(3.2×20)=Po(64).M\sim\mathrm{Po}(3.2\times20)=\mathrm{Po}(64).

Since 64>1564>15, use Y∼N(64,64)Y\sim\mathrm N(64,64) as a suitable approximation. With a continuity correction,

P(M>60)= P(M≥61)≈ P(Y>60.5)= 1−Φ(60.5−648)= 1−Φ(−0.4375)= Φ(0.4375)≈0.669.\begin{align*} \mathrm P(M>60) =&\,\mathrm P(M\ge61)\\[4mm] \approx&\,\mathrm P(Y>60.5)\\[4mm] =&\,1-\Phi\bigg(\frac{60.5-64}{8}\bigg)\\[4mm] =&\,1-\Phi(-0.4375)\\[4mm] =&\,\Phi(0.4375)\approx\boxed{0.669}. \end{align*}