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CIE 9709 2025 November Paper 62 Q1

A Level / CIE / S2

CIE 9709 2025 Nov Paper 62 Paper · Question 1

题目

Problem

The number, XX, of used computers donated to a charity has a constant average rate of 2.42.4 computers per week.

(a) State a necessary condition for XX to have a Poisson distribution.

(1)

Now assume that XX has a Poisson distribution.

(b) Calculate the probability that the number of computers donated during a 4-week period is more than 66 and less than 99.

(3)

(c) Use a suitable approximating distribution to calculate the probability that more than 5050 computers are donated during a 20-week period.

(4)
题目中文翻译

某慈善机构收到捐赠的二手电脑数量为 XX,平均捐赠速率恒为每周 2.42.4 台。

(a) 写出 XX 服从泊松分布的一个必要条件。

(1)

现在假设 XX 服从泊松分布。

(b) 计算在一段 44 周的时间内,捐赠电脑的数量大于 66 且小于 99 的概率。

(3)

(c) 用一个合适的近似分布,计算在一段 2020 周的时间内捐赠超过 5050 台电脑的概率。

(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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平均速率恒定已经给出,补充计算机捐赠之间的独立性;回答必须联系具体情境。

答题过程

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A necessary condition is that computers are donated independently of one another.

(b)

解法一

思路

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四周内的平均捐赠数乘以四,严格介于六与九之间只包含七台和八台。

答题过程

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Let NN be the number donated in four weeks. Then N∼Po(4(2.4))=Po(9.6)N\sim\mathrm{Po}(4(2.4))=\mathrm{Po}(9.6). Hence

P(6<N<9)= P(N=7)+P(N=8)= e−9.6(9.677!+9.688!)≈ 0.222.\begin{align*} \mathrm P(6<N<9) =&\,\mathrm P(N=7)+\mathrm P(N=8)\\[4mm] =&\,\mathrm e^{-9.6}\bigg(\frac{9.6^7}{7!}+\frac{9.6^8}{8!}\bigg)\\[4mm] \approx&\,\boxed{0.222}. \end{align*}

(c)

解法一

思路

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二十周的参数为 4848,用同均值同方差的正态近似。超过五十台对应至少五十一台,边界修正为 50.550.5。

答题过程

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Over twenty weeks the total TT has distribution Po(20(2.4))=Po(48)\mathrm{Po}(20(2.4))=\mathrm{Po}(48). Since 48>1548>15, approximate it by Y∼N(48,48)Y\sim\mathrm N(48,48).

With a continuity correction,

P(T>50)= P(T≥51)≈ P(Y>50.5)= 1−Φ(50.5−4848)= 1−Φ(0.360844…)≈ 0.359.\begin{align*} \mathrm P(T>50) =&\,\mathrm P(T\ge51)\\[4mm] \approx&\,\mathrm P(Y>50.5)\\[4mm] =&\,1-\Phi\bigg(\frac{50.5-48}{\sqrt{48}}\bigg)\\[4mm] =&\,1-\Phi(0.360844\ldots)\\[4mm] \approx&\,\boxed{0.359}. \end{align*}