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IAL 2025 Oct S2 A Q5

A Level / Edexcel / S2

IAL 2025 Oct A Paper · Question 5

题目

Problem

Cars stop at a service station randomly at a rate of 3 every 5 minutes.

(a) Calculate the probability that in a randomly selected 10 minute period,

(i) exactly 7 cars will stop at the service station,

(ii) more than 7 cars will stop at the service station.

(4)

Using a normal approximation, the probability that more than 40 cars will stop at the service station during a randomly selected nn minute period is 0.2266 correct to 4 significant figures.

(b) Find the value of nn. You must show all stages of your working.

(9)

(Total for Question 5 is 13 marks)

题目中文翻译

汽车以每 5 分钟 3 辆的随机速率停在服务站。

(a) 计算在随机选择的 10 分钟内

(i) 恰好有 7 辆汽车停在服务站的概率，

(ii) 超过 7 辆汽车停在服务站的概率。

使用正态近似，在随机选择的 nn 分钟内超过 40 辆汽车停在服务站的概率为 0.2266（精确到 4 位有效数字）。

(b) 求 nn 的值。你必须展示所有计算步骤。

（第 5 题共 13 分）

解答

(a)(i)

解法一

思路

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10 分钟内的平均车辆数为 2×3=62\times3=6，所以车辆数服从均值为 6 的泊松分布。把 X=7X=7 代入泊松分布的单点概率公式。

答题过程

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Let XX be the number of cars stopping in a 10-minute period. Then

X∼Po⁡(6).X\sim\operatorname{Po}(6).

Therefore,

P(X=7)= e−6677!= 0.137676…= 0.138(3 s.f.).\begin{align*} P(X=7) =&\,\frac{\mathrm{e}^{-6}6^7}{7!}\\ =&\,0.137676\ldots\\ =&\,\boxed{0.138}\quad\text{(3 s.f.)}. \end{align*}

解法二

思路

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官方评分资料也接受使用泊松累积分布表。用 P(X≤7)−P(X≤6)P(X\leq7)-P(X\leq6) 隔离出 X=7X=7 的单点概率。

答题过程

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Using cumulative Poisson probabilities,

P(X=7)= P(X≤7)−P(X≤6)= 0.7440−0.6063= 0.1377.\begin{align*} P(X=7) =&\,P(X\leq7)-P(X\leq6)\\ =&\,0.7440-0.6063\\ =&\,\boxed{0.1377}. \end{align*}

(a)(ii)

解法一

思路

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“超过 7 辆”表示 X>7X>7。使用补事件，把所求概率写成 1−P(X≤7)1-P(X\leq7)，再查泊松累积分布。

答题过程

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Using X∼Po⁡(6)X\sim\operatorname{Po}(6),

P(X>7)= 1−P(X≤7)= 1−0.7440= 0.256.\begin{align*} P(X>7) =&\,1-P(X\leq7)\\ =&\,1-0.7440\\ =&\,\boxed{0.256}. \end{align*}

(b)

解法一

思路

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令 λ\lambda 为 nn 分钟内的平均车辆数，则 λ=(3/5)n=0.6n\lambda=(3/5)n=0.6n。泊松分布用均值和方差均为 λ\lambda 的正态分布近似；“超过 40”经连续性修正成为 Y>40.5Y>40.5。由上尾概率 0.2266 得标准正态值 0.750.75，建立关于 λ\sqrt{\lambda} 的二次方程并求出 nn。

答题过程

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Let YY be the number of cars stopping in an nn-minute period, and let its Poisson mean be λ\lambda. Since the rate is 33 cars per 55 minutes,

λ=35n=0.6n.\lambda=\frac35n=0.6n.

Using a normal approximation,

Y∼˙N(λ,λ).Y\mathrel{\dot\sim}N(\lambda,\lambda).

Applying the continuity correction,

P(Y>40)≈P(Y>40.5)=0.2266.P(Y>40)\approx P(Y>40.5)=0.2266.

Since P(Z>0.75)=0.2266P(Z>0.75)=0.2266,

40.5−λλ=0.75.\frac{40.5-\lambda}{\sqrt{\lambda}}=0.75.

Hence,

λ+0.75λ−40.5=0.\lambda+0.75\sqrt{\lambda}-40.5=0.

Let t=λt=\sqrt{\lambda}, where t>0t>0. Then

t2+0.75t−40.5= 0(t−6)(t+6.75)= 0.\begin{align*} t^2+0.75t-40.5=&\,0\\ (t-6)(t+6.75)=&\,0. \end{align*}

Therefore t=6t=6, so

λ=t2=36.\lambda=t^2=36.

Finally,

0.6n= 36n= 60.\begin{align*} 0.6n=&\,36\\ n=&\,\boxed{60}. \end{align*}