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CIE 9709 2023 June Paper 62 Q4

A Level / CIE / S2

CIE 9709 2023 June Paper 62 Paper · Question 4

题目

Problem

The number, XX, of books received at a charity shop has a constant mean of 5.15.1 per day.

(a) State, in context, one condition for XX to be modelled by a Poisson distribution.

(1)

Assume now that XX can be modelled by a Poisson distribution.

(b) Find the probability that exactly 1010 books are received in a 33-day period.

(2)

(c) Use a suitable approximating distribution to find the probability that more than 180180 books are received in a 3030-day period.

(4)

The number of DVDs received at the same shop is modelled by an independent Poisson distribution with mean 2.52.5 per day.

(d) Find the probability that the total number of books and DVDs that are received at the shop in 11 day is more than 33.

(3)
题目中文翻译

某慈善商店收到的图书册数记为 XX，其均值恒为每天 5.15.1 册。

(a) 结合题意写出一个可以用泊松分布描述 XX 的条件。

(1)

现假设可以用泊松分布描述 XX。

(b) 求一段为期 33 天的时间内恰好收到 1010 册图书的概率。

(2)

(c) 用适当的近似分布求一段为期 3030 天的时间内收到的图书超过 180180 册的概率。

(4)

该店收到的 DVD 张数服从独立于 XX 的泊松分布，均值为每天 2.52.5 张。

(d) 求该店在 11 天内收到的图书和 DVD 的总件数超过 33 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

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平均速率已知恒定，再给出独立到达这一条件并结合书本捐赠情境。

答题过程

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Books must be received independently of one another.

(b)

解法一

思路

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三天参数为15.3，求泊松恰好10本的概率。

答题过程

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The number B3B_3 of books received in three days has distribution Po(3(5.1))=Po(15.3)\mathrm{Po}(3(5.1))=\mathrm{Po}(15.3). Hence

P(B3=10)= e−15.315.31010!≈ 0.0439(3 s.f.).\begin{align*} \mathrm{P}(B_3=10)=&\,e^{-15.3}\frac{15.3^{10}}{10!}\\[4mm] \approx&\,\boxed{0.0439}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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30天总数参数为153，使用正态近似并在超过180的边界作连续性修正。

答题过程

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For the 30-day total,

B30∼Po(153)≈N(153,153).B_{30}\sim\mathrm{Po}(153)\approx\mathrm{N}(153,153).

Using a continuity correction,

P(B30>180)≈ 1−Φ(180.5−153153)= 0.0130997…= 0.0131(3 s.f.).\begin{align*} \mathrm{P}(B_{30}>180)\approx&\,1-\Phi\bigg(\frac{180.5-153}{\sqrt{153}}\bigg)\\[4mm] =&\,0.0130997\ldots\\[4mm] =&\,\boxed{0.0131}\quad\text{(3 s.f.).} \end{align*}

(d)

解法一

思路

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两类独立泊松变量相加参数为7.6，超过3取0至3的补集。

答题过程

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The daily total SS of books and DVDs has distribution Po(5.1+2.5)=Po(7.6)\mathrm{Po}(5.1+2.5)=\mathrm{Po}(7.6). Thus

P(S>3)= 1−e−7.6(1+7.6) −e−7.6(7.622!+7.633!)= 0.944629…= 0.945(3 s.f.).\begin{align*} \mathrm{P}(S>3)=&\,1-e^{-7.6}(1+7.6)\\[4mm] &\,-e^{-7.6}\bigg(\frac{7.6^2}{2!}+\frac{7.6^3}{3!}\bigg)\\[4mm] =&\,0.944629\ldots\\[4mm] =&\,\boxed{0.945}\quad\text{(3 s.f.).} \end{align*}