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CIE 9709 2024 November Paper 62 Q1

A Level / CIE / S2

CIE 9709 2024 Nov Paper 62 Paper · Question 1

题目

Problem

A random variable XX has the distribution B(4 500 000,11 000 000)\mathrm{B}\left(4\,500\,000,\dfrac{1}{1\,000\,000}\right).

Use a Poisson distribution to calculate an estimate of P(X≥4)\mathrm{P}(X\geq 4).

(3)
题目中文翻译

随机变量 XX 服从 B(4 500 000,11 000 000)\mathrm{B}\left(4\,500\,000,\dfrac{1}{1\,000\,000}\right) 分布。

用泊松分布估算 P(X≥4)\mathrm{P}(X\geq 4)。

(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

解法一

思路

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二项分布的试验次数很大、成功概率很小,使用参数np的泊松近似。至少4次取不超过3次的补集。

答题过程

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Since nn is large and pp is small,

X≈ Po(np)= Po(450000011000000)= Po(4.5).\begin{align*} X\approx&\,\mathrm{Po}(np)\\[4mm] =&\,\mathrm{Po}\bigg(4500000\frac{1}{1000000}\bigg)\\[4mm] =&\,\mathrm{Po}(4.5). \end{align*}

Therefore

P(X≥4)≈ 1−e−4.5(1+4.5) −e−4.5(4.522!+4.533!)= 0.657704…= 0.658(3 s.f.).\begin{align*} \mathrm{P}(X\geq4)\approx&\,1-e^{-4.5}(1+4.5)\\[4mm] &\,-e^{-4.5}\bigg(\frac{4.5^2}{2!}+\frac{4.5^3}{3!}\bigg)\\[4mm] =&\,0.657704\ldots\\[4mm] =&\,\boxed{0.658}\quad\text{(3 s.f.).} \end{align*}