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CIE 9709 2022 June Paper 52 Q5

A Level / CIE / S1

CIE 9709 2022 June Paper 52 Paper · Question 5

题目

Problem

In a large college, 28% of the students do not play any musical instrument, 52% play exactly one musical instrument and the remainder play two or more musical instruments.

A random sample of 12 students from the college is chosen.

(a) Find the probability that more than 9 of these students play at least one musical instrument.

(3)

A random sample of 90 students from the college is now chosen.

(b) Use an approximation to find the probability that fewer than 40 of these students play exactly one musical instrument.

(5)

题目中文翻译

在一所大型学院中，28% 的学生不会演奏任何乐器，52% 的学生恰好会演奏一种乐器，其余学生会演奏两种或以上的乐器。

从该学院随机抽取 12 名学生作为样本。

(a) 求这些学生中至少会演奏一种乐器的人数多于 9 的概率。

(3)

现在从该学院随机抽取 90 名学生作为样本。

(b) 使用近似方法，求这些学生中恰好会演奏一种乐器的人数少于 40 的概率。

(5)

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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至少一种的成功率是0.72；多于9就是10、11、12三项。

答题过程

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Let XX count students who play at least one instrument. Then X∼B(12,0.72)X\sim\mathrm{B}(12,0.72), since 1−0.28=0.721-0.28=0.72.

P(X>9)= (1210)(0.72)10(0.28)2 +(1211)(0.72)11(0.28) +(0.72)12= 0.303705674…≈0.304.\begin{align*} \mathrm{P}(X>9)=&\,\binom{12}{10}(0.72)^{10}(0.28)^2\\[4mm] &\,\hspace{2pt}+\binom{12}{11}(0.72)^{11}(0.28)\\[4mm] &\,\hspace{4pt}+(0.72)^{12}\\[4mm] =&\,0.303705674\ldots\approx\boxed{0.304}. \end{align*}

(b)

解法一

思路

展开

这次恰好一种的成功率为0.52。少于40对应至39，连续性修正39.5。

答题过程

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Let Y∼B(90,0.52)Y\sim\mathrm{B}(90,0.52) count students who play exactly one instrument. Since np=46.8np=46.8 and n(1−p)=43.2n(1-p)=43.2 are both greater than 5, use the normal approximation

V∼N(90(0.52),90(0.52)(0.48))=N(46.8,22.464).V\sim\mathrm{N}\bigl(90(0.52),90(0.52)(0.48)\bigr) =\mathrm{N}(46.8,22.464).

With a continuity correction,

P(Y<40)≈ P(V<39.5)= Φ(39.5−46.822.464)= 0.061754856…≈0.0618.\begin{align*} \mathrm{P}(Y<40)\approx&\,\mathrm{P}(V<39.5)\\[4mm] =&\,\Phi\biggl(\frac{39.5-46.8}{\sqrt{22.464}}\biggr)\\[4mm] =&\,0.061754856\ldots\approx\boxed{0.0618}. \end{align*}