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CIE 9709 2025 June Paper 55 Q3

A Level / CIE / S1

CIE 9709 2025 June Paper 55 Paper · Question 3

题目

Problem

In a certain large school, on average, two pupils in five have music lessons.

A random sample of 80 pupils from this school is chosen.

(a) Use an approximation to find the probability that fewer than 27 pupils have music lessons.

(5)

A random sample of 10 pupils from this school is now chosen.

(b) Find the probability that no more than 2 pupils have music lessons.

(3)
题目中文翻译

在某所规模很大的学校中,平均每 5 名学生中有 2 人上音乐课。

从这所学校中随机抽取 80 名学生作为样本。

(a) 使用近似方法,求其中上音乐课的学生少于 27 人的概率。

(5)

现在从这所学校中随机抽取 10 名学生作为样本。

(b) 求其中上音乐课的学生不超过 2 人的概率。

(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

展开

人数服从二项分布,用正态分布近似。少于27即26及以下,连续性修正后的上界为26.5。

答题过程

展开

Let X∼B(80,0.4)X\sim\mathrm{B}(80,0.4) be the number taking music lessons.

Since np=32np=32 and n(1−p)=48n(1-p)=48 are both greater than 5, use Y∼N(32,19.2)Y\sim\mathrm{N}(32,19.2), where 19.2=80(0.4)(0.6)19.2=80(0.4)(0.6).

Applying a continuity correction,

P(X<27)= P(X≤26)≈ P(Y<26.5)= Φ(26.5−3219.2)= Φ(−1.255197…)= 0.104703…≈ 0.105.\begin{align*} \mathrm{P}(X<27)=&\,\mathrm{P}(X\leq26)\\[4mm] \approx&\,\mathrm{P}(Y<26.5)\\[4mm] =&\,\Phi\biggl(\frac{26.5-32}{\sqrt{19.2}}\biggr)\\[4mm] =&\,\Phi(-1.255197\ldots)\\[4mm] =&\,0.104703\ldots\\[4mm] \approx&\,\boxed{0.105}. \end{align*}

(b)

解法一

思路

展开

10人中不超过2人,直接用二项分布将0、1、2人的概率相加。

答题过程

展开

Let X∼B(10,0.4)X\sim\mathrm{B}(10,0.4). Then

P(X≤2)= (0.6)10+10(0.4)(0.6)9 +45(0.4)2(0.6)8= 0.1672897536≈ 0.167.\begin{align*} \mathrm{P}(X\leq2)=&\,(0.6)^{10}+10(0.4)(0.6)^9\\[4mm] &\,\hspace{2pt}+45(0.4)^2(0.6)^8\\[4mm] =&\,0.1672897536\\[4mm] \approx&\,\boxed{0.167}. \end{align*}