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CIE 9709 2024 June Paper 63 Q1

A Level / CIE / S2

CIE 9709 2024 June Paper 63 Paper · Question 1

题目

Problem

The random variable XX has the distribution B(4000,0.001)\mathrm{B}(4000,0.001).

(a) Use a suitable approximating distribution to find P(2≤X<5)\mathrm{P}(2\leq X<5).

(3)

(b) Justify your approximating distribution in this case.

(1)
题目中文翻译

随机变量 XX 服从 B(4000,0.001)\mathrm{B}(4000,0.001) 分布。

(a) 用适当的近似分布求 P(2≤X<5)\mathrm{P}(2\leq X<5)。

(3)

(b) 说明这里采用该近似分布的理由。

(1)
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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试验次数大且成功概率小，用参数np等于4的泊松近似，范围只包含2、3、4。

答题过程

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Using X≈Po(4000(0.001))=Po(4)X\approx\mathrm{Po}(4000(0.001))=\mathrm{Po}(4),

P(2≤X<5)≈ e−4(422!+433!) +e−4444!= 0.537259…= 0.537(3 s.f.).\begin{align*} \mathrm{P}(2\leq X<5)\approx&\,e^{-4}\bigg(\frac{4^2}{2!}+\frac{4^3}{3!}\bigg)\\[4mm] &\,+e^{-4}\frac{4^4}{4!}\\[4mm] =&\,0.537259\ldots\\[4mm] =&\,\boxed{0.537}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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用本题具体参数说明泊松近似条件。

答题过程

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The number of trials is large, n=4000>50n=4000>50, and the success probability is small, p=0.001<0.1p=0.001<0.1, so the Poisson approximation is suitable.