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CIE 9709 2024 June Paper 61 Q5

A Level / CIE / S2

CIE 9709 2024 June Paper 61 Paper · Question 5

题目

Problem

Sales of cell phones at a certain shop occur singly, randomly and independently.

(a) State one further condition that must be satisfied for the number of sales in a certain time period to be well modelled by a Poisson distribution.

(1)

The average number of sales per hour is 1.21.2.

Assume now that a Poisson distribution is a suitable model.

(b) Find the probability that the number of sales during a randomly chosen 1212-hour period will be more than 1212 and less than 1616.

(3)

(c) Use a suitable approximating distribution to find the probability that the number of sales during a randomly chosen 11-month period (140140 hours) will be less than 150150.

(4)
题目中文翻译

某商店的手机逐部售出，每次售出都是随机且相互独立的。

(a) 要用泊松分布恰当地描述某段时间内售出的手机数量，还必须满足一个条件。写出这个条件。

(1)

每小时平均售出 1.21.2 部手机。

现在假设泊松分布是一个合适的模型。

(b) 随机选取一段 1212 小时的时间，求这段时间内售出的手机多于 1212 部且少于 1616 部的概率。

(3)

(c) 随机选取一段为期 11 个月（140140 小时）的时间，用合适的近似分布求这段时间内售出的手机少于 150150 部的概率。

(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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除已知随机、独立、单次发生外，还需单位时间内的平均销售率保持不变。

答题过程

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Sales must occur at a constant average rate over the time period considered.

(b)

解法一

思路

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12小时参数为14.4，严格介于12与16之间的整数为13、14、15。

答题过程

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Let SS be the number of sales in 12 hours. Then S∼Po(12(1.2))=Po(14.4)S\sim\mathrm{Po}(12(1.2))=\mathrm{Po}(14.4).

P(12<S<16)= e−14.4(14.41313!+14.41414!) +e−14.414.41515!= 0.309046…= 0.309(3 s.f.).\begin{align*} \mathrm{P}(12<S<16)=&\,e^{-14.4}\bigg(\frac{14.4^{13}}{13!}+\frac{14.4^{14}}{14!}\bigg)\\[4mm] &\,+e^{-14.4}\frac{14.4^{15}}{15!}\\[4mm] =&\,0.309046\ldots\\[4mm] =&\,\boxed{0.309}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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140小时泊松参数为168，参数较大可用均值和方差都为168的正态近似。少于150修正为149.5。

答题过程

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The number MM of sales in 140 hours satisfies

M∼ Po(140(1.2))=Po(168)≈ N(168,168).\begin{align*} M\sim&\,\mathrm{Po}(140(1.2))=\mathrm{Po}(168)\\[4mm] \approx&\,\mathrm{N}(168,168). \end{align*}

Using a continuity correction,

P(M<150)≈ Φ(149.5−168168)= 0.0767459…= 0.0767(3 s.f.).\begin{align*} \mathrm{P}(M<150)\approx&\,\Phi\bigg(\frac{149.5-168}{\sqrt{168}}\bigg)\\[4mm] =&\,0.0767459\ldots\\[4mm] =&\,\boxed{0.0767}\quad\text{(3 s.f.).} \end{align*}