题目

Problem

Cars arrive at a fuel station at random and at a constant average rate of $13.5$ per hour.

(a) Find the probability that more than $4$ cars arrive during a $20$ -minute period.

(3)

(b) Use an approximating distribution to find the probability that the number of cars that arrive during a $12$ -hour period is between $150$ and $160$ inclusive.

(4)

Independently of cars, trucks arrive at the fuel station at random and at a constant average rate of $3.6$ per $15$ -minute period.

(c) Find the probability that the total number of cars and trucks arriving at the fuel station during a $10$ -minute period is more than $3$ and less than $7$ .

(3)

题目中文翻译

小汽车随机到达某加油站，平均到达速率恒为每小时 $13.5$ 辆。

(a) 求一段 $20$ 分钟的时间内到达的小汽车超过 $4$ 辆的概率。

(3)

(b) 用近似分布求一段 $12$ 小时的时间内到达的小汽车数介于 $150$ 与 $160$ 之间（含端点）的概率。

(4)

卡车也随机到达该加油站，其到达与小汽车相互独立，平均到达速率恒为每 $15$ 分钟 $3.6$ 辆。

(c) 求一段 $10$ 分钟的时间内到达该加油站的小汽车和卡车的总数多于 $3$ 辆且少于 $7$ 辆的概率。

(3)

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

20分钟的泊松参数为4.5，超过4辆取不超过4辆的补集。

答题过程

展开

The car count $C_{20}$ in 20 minutes has distribution $\mathrm{Po}(13.5/3)=\mathrm{Po}(4.5)$ . Thus

\begin{align*} \mathrm{P}(C_{20}>4)=&\,1-e^{-4.5}\bigg(1+4.5+\frac{4.5^2}{2!}\bigg)\\[4mm] &\,-e^{-4.5}\bigg(\frac{4.5^3}{3!}+\frac{4.5^4}{4!}\bigg)\\[4mm] =&\,\boxed{0.468}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

展开

12小时参数162足够大，采用正态近似。包含150与160两个端点，连续性修正为149.5至160.5。

答题过程

展开

Over 12 hours,

\begin{align*} C_{12}\sim&\,\mathrm{Po}(12(13.5))=\mathrm{Po}(162)\\[4mm] \approx&\,\mathrm{N}(162,162). \end{align*}

With continuity corrections at both boundaries,

\begin{align*} &\,\mathrm{P}(150\leq C_{12}\leq160)\\[4mm] \approx&\,\Phi\bigg(\frac{160.5-162}{\sqrt{162}}\bigg)\\[4mm] &\,-\Phi\bigg(\frac{149.5-162}{\sqrt{162}}\bigg)\\[4mm] =&\,0.290066\ldots\\[4mm] =&\,\boxed{0.290}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

展开

把两种车的参数都换为10分钟，再利用独立泊松和，严格3至7之间只含4、5、6。

答题过程

展开

The 10-minute means are

\begin{align*} \lambda_{\mathrm{cars}}=&\,13.5\frac{10}{60}=2.25,\\[4mm] \lambda_{\mathrm{trucks}}=&\,3.6\frac{10}{15}=2.4. \end{align*}

By independence, the total $N\sim\mathrm{Po}(4.65)$ . Hence

\begin{align*} \mathrm{P}(3<N<7)=&\,e^{-4.65}\bigg(\frac{4.65^4}{4!}+\frac{4.65^5}{5!}\bigg)\\[4mm] &\,+e^{-4.65}\frac{4.65^6}{6!}\\[4mm] =&\,0.493742\ldots\\[4mm] =&\,\boxed{0.494}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2022 June Paper 61 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程