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CIE 9709 2022 June Paper 63 Q5

A Level / CIE / S2

CIE 9709 2022 June Paper 63 Paper · Question 5

题目

Problem

The number of clients who arrive at an information desk has a Poisson distribution with mean 2.22.2 per 55-minute period.

(a) Find the probability that, in a randomly chosen 1515-minute period, exactly 66 clients arrive at the desk.

(3)

(b) If more than 44 clients arrive during a 55-minute period, they cannot all be served.

Find the probability that, during a randomly chosen 55-minute period, not all the clients who arrive at the desk can be served.

(2)

(c) Use a suitable approximating distribution to find the probability that, during a randomly chosen 11-hour period, fewer than 2020 clients arrive at the desk.

(4)
题目中文翻译

到达某咨询台的顾客人数服从泊松分布，平均每 55 分钟有 2.22.2 人到达。

(a) 随机选取一段 1515 分钟的时间，求恰好有 66 名顾客到达咨询台的概率。

(3)

(b) 如果一段 55 分钟的时间内有超过 44 名顾客到达，咨询台就无法为所有人提供服务。

随机选取一段 55 分钟的时间，求到达咨询台的顾客中有人无法得到服务的概率。

(2)

(c) 随机选取一段 11 小时的时间，用适当的近似分布求到达咨询台的顾客少于 2020 人的概率。

(4)
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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15分钟的泊松参数为6.6，求恰好6人的概率。

答题过程

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The 15-minute count N15∼Po(3(2.2))=Po(6.6)N_{15}\sim\mathrm{Po}(3(2.2))=\mathrm{Po}(6.6). Thus

P(N15=6)=e−6.66.666!=0.156(3 s.f.).\mathrm{P}(N_{15}=6)=e^{-6.6}\frac{6.6^6}{6!} =\boxed{0.156}\quad\text{(3 s.f.).}

(b)

解法一

思路

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不能全部服务对应5分钟到达人数超过4，用不超过4的补集求概率。

答题过程

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For N5∼Po(2.2)N_5\sim\mathrm{Po}(2.2),

P(N5>4)= 1−e−2.2(1+2.2+2.222!) −e−2.2(2.233!+2.244!)= 0.0725(3 s.f.).\begin{align*} \mathrm{P}(N_5>4) =&\,1-e^{-2.2}\bigg(1+2.2+\frac{2.2^2}{2!}\bigg)\\[4mm] &\,-e^{-2.2}\bigg(\frac{2.2^3}{3!}+\frac{2.2^4}{4!}\bigg)\\[4mm] =&\,\boxed{0.0725}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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一小时参数为26.4，用同均值方差正态近似，少于20修正为19.5。

答题过程

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Over one hour,

N60∼ Po(12(2.2))=Po(26.4)≈ N(26.4,26.4).\begin{align*} N_{60}\sim&\,\mathrm{Po}(12(2.2))=\mathrm{Po}(26.4)\\[4mm] \approx&\,\mathrm{N}(26.4,26.4). \end{align*}

Using a continuity correction,

P(N60<20)≈ Φ(19.5−26.426.4)= 0.0897(3 s.f.).\begin{align*} \mathrm{P}(N_{60}<20)\approx&\,\Phi\bigg(\frac{19.5-26.4}{\sqrt{26.4}}\bigg)\\[4mm] =&\,\boxed{0.0897}\quad\text{(3 s.f.).} \end{align*}