题目

Problem

In a certain country, $20\,540$ adults out of a population of $6\,012\,300$ have a degree in medicine.

(a) Use an approximating distribution to calculate the probability that, in a random sample of $1000$ adults in this country, there will be fewer than $4$ adults who have a degree in medicine.

(4)

(b) Justify the approximating distribution used in part (a).

(2)

题目中文翻译

某国共有 $6\,012\,300$ 名成年人，其中 $20\,540$ 人拥有医学学位。

(a) 从该国随机抽取 $1000$ 名成年人，用近似分布计算样本中拥有医学学位的人少于 $4$ 人的概率。

(4)

(b) 说明 (a) 中采用该近似分布的理由。

(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

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先求成人中医学学位持有者的比例，大样本低概率适合泊松近似，参数为1000p。

答题过程

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The probability that a randomly chosen adult has a medical degree is

p=\frac{20540}{6012300}=0.0034163299\ldots.

For the number $M$ in the sample, use

M\approx\mathrm{Po}(\lambda),

where

\lambda=1000p=3.4163299\ldots.

Thus

\begin{align*} \mathrm{P}(M<4)\approx&\,e^{-\lambda}(1+\lambda)\\[4mm] &\,+e^{-\lambda}\bigg(\frac{\lambda^2}{2!}+\frac{\lambda^3}{3!}\bigg)\\[4mm] =&\,0.554791\ldots\\[4mm] =&\,\boxed{0.555}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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用本题具体n与np说明泊松近似条件，并在巨大总体中抽取相对很小的样本。

答题过程

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The sample size is large, $n=1000>50$ , and

np=3.4163299\ldots<5,

so the success probability is small and a Poisson approximation is suitable.

CIE 9709 2023 June Paper 61 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程