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CIE 9709 2023 June Paper 63 Q6

A Level / CIE / S2

CIE 9709 2023 June Paper 63 Paper · Question 6

题目

Problem

It is known that 11 in 50005000 people in Atalia have a certain condition. A random sample of 12 50012\,500 people from Atalia is chosen for a medical trial. The number having the condition is denoted by XX.

(a) Use an appropriate approximating distribution to find P(X≤3)\mathrm{P}(X\leq3).

(3)

(b) Find the values of E(X)\mathrm{E}(X) and Var(X)\mathrm{Var}(X), and explain how your answers suggest that the approximating distribution used in (a) is likely to be appropriate.

(2)
题目中文翻译

已知 Atalia 每 50005000 人中有 11 人患有某种疾病。从 Atalia 随机抽取 12 50012\,500 人参加一项医学试验。患有这种疾病的人数记为 XX。

(a) 用适当的近似分布求 P(X≤3)\mathrm{P}(X\leq3)。

(3)

(b) 求 E(X)\mathrm{E}(X) 和 Var(X)\mathrm{Var}(X) 的值,并说明这些结果为何表明 (a) 中采用的近似分布可能是合适的。

(2)
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

展开

样本量大、患病率低,采用参数2.5的泊松近似并累加0至3。

答题过程

展开

The binomial model is X∼B(12500,15000)X\sim\mathrm{B}(12500,\frac1{5000}). Since the sample size is large and the probability is small, use

X≈Po(1250015000)=Po(2.5).X\approx\mathrm{Po}\bigg(12500\frac1{5000}\bigg)=\mathrm{Po}(2.5).

Then

P(X≤3)≈ e−2.5(1+2.5) +e−2.5(2.522!+2.533!)= 0.757576…= 0.758(3 s.f.).\begin{align*} \mathrm{P}(X\leq3)\approx&\,e^{-2.5}(1+2.5)\\[4mm] &\,+e^{-2.5}\bigg(\frac{2.5^2}{2!}+\frac{2.5^3}{3!}\bigg)\\[4mm] =&\,0.757576\ldots\\[4mm] =&\,\boxed{0.758}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

展开

本问求原始二项模型的均值与方差,不能把近似泊松的方差2.5当成精确原值。

答题过程

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For the binomial distribution,

E(X)=1250015000=2.5,\boxed{\mathrm{E}(X)=12500\frac1{5000}=2.5},

and

Var(X)= 1250015000(1−15000)= 2.4995.\begin{align*} \mathrm{Var}(X)=&\,12500\frac1{5000}\bigg(1-\frac1{5000}\bigg)\\[4mm] =&\,\boxed{2.4995}. \end{align*}

The mean and variance are almost equal, which is consistent with a Poisson approximation.