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CIE 9709 2023 March Paper 62 Q2

A Level / CIE / S2

CIE 9709 2023 March Paper 62 Paper · Question 2

题目

Problem

The number of orders arriving at a shop during an 88-hour working day is modelled by the random variable XX with distribution Po(25.2)\mathrm{Po}(25.2).

(a) State two assumptions that are required for the Poisson model to be valid in this context.

(2)

(b)

(i) Find the probability that the number of orders that arrive in a randomly chosen 33-hour period is between 33 and 55 inclusive.

(3)

(ii) Find the probability that, in two randomly chosen 11-hour periods, exactly 11 order will arrive in one of the 11-hour periods, and at least 22 orders will arrive in the other 11-hour period.

(4)

(c) The shop can only deal with a maximum of 120120 orders during any 3636-hour period.

Use a suitable approximating distribution to find the probability that, in a randomly chosen 3636-hour period, there will be too many orders for the shop to deal with.

(4)
题目中文翻译

一家商店在一个 88 小时的工作日内收到的订单数，用服从 Po(25.2)\mathrm{Po}(25.2) 分布的随机变量 XX 描述。

(a) 写出在此情境中泊松模型成立所需的两个假设。

(2)

(b)

(i) 随机选取一段 33 小时的时间，求这段时间内收到的订单数介于 33 与 55 之间（含端点）的概率。

(3)

(ii) 随机选取两段各长 11 小时的时间，求其中一段恰好收到 11 份订单、另一段至少收到 22 份订单的概率。

(4)

(c) 该商店在任意一段 3636 小时的时间内最多只能处理 120120 份订单。

随机选取一段 3636 小时的时间，用适当的近似分布求这段时间内的订单数超过商店处理能力的概率。

(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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给出两条情境化的泊松模型假设：平均到达速率恒定及各订单独立到达。

答题过程

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Orders must arrive at a constant average rate, and the arrival of one order must be independent of the arrivals of other orders.

(b)(i)

解法一

思路

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3小时的参数为25.2乘3/8等于9.45，直接相加3、4、5次概率。

答题过程

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Let N3N_3 be the number of orders in three hours. Then

N3∼Po(25.238)=Po(9.45).N_3\sim\mathrm{Po}\bigg(25.2\frac38\bigg)=\mathrm{Po}(9.45).

Therefore

 P(3≤N3≤5)= e−9.45(9.4533!+9.4544!) +e−9.459.4555!= 0.0866346…= 0.0866(3 s.f.).\begin{align*} &\,\mathrm{P}(3\leq N_3\leq5)\\[4mm] =&\,e^{-9.45}\bigg(\frac{9.45^3}{3!}+\frac{9.45^4}{4!}\bigg)\\[4mm] &\,+e^{-9.45}\frac{9.45^5}{5!}\\[4mm] =&\,0.0866346\ldots\\[4mm] =&\,\boxed{0.0866}\quad\text{(3 s.f.).} \end{align*}

(b)(ii)

解法一

思路

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每小时参数为3.15，两个时段中恰好1单与至少2单可交换，故两种不相交情况的概率相加，即乘2。

答题过程

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For one hour, N1∼Po(25.2/8)=Po(3.15)N_1\sim\mathrm{Po}(25.2/8)=\mathrm{Po}(3.15). Hence

P(N1=1)=e−3.15(3.15),\mathrm{P}(N_1=1)=e^{-3.15}(3.15),

and

P(N1≥2)=1−e−3.15(1+3.15).\mathrm{P}(N_1\geq2)=1-e^{-3.15}(1+3.15).

Either period can be the one with exactly one order. Thus the required probability is

 2e−3.15(3.15)[1−e−3.15(1+3.15)]≈ 0.222(3 s.f.).\begin{align*} &\,2e^{-3.15}(3.15)\bigg[1-e^{-3.15}(1+3.15)\bigg]\\[4mm] \approx&\,\boxed{0.222}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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36小时总数参数为113.4，超过120单对应正态连续性修正边界120.5。

答题过程

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For the 36-hour total,

N36∼ Po(25.2368)=Po(113.4)≈ N(113.4,113.4).\begin{align*} N_{36}\sim&\,\mathrm{Po}\bigg(25.2\frac{36}{8}\bigg)=\mathrm{Po}(113.4)\\[4mm] \approx&\,\mathrm{N}(113.4,113.4). \end{align*}

Using a continuity correction,

P(N36>120)≈ 1−Φ(120.5−113.4113.4)= 0.252471…= 0.252(3 s.f.).\begin{align*} \mathrm{P}(N_{36}>120)\approx&\,1-\Phi\bigg(\frac{120.5-113.4}{\sqrt{113.4}}\bigg)\\[4mm] =&\,0.252471\ldots\\[4mm] =&\,\boxed{0.252}\quad\text{(3 s.f.).} \end{align*}