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CIE 9709 2024 June Paper 52 Q6

A Level / CIE / S1

CIE 9709 2024 June Paper 52 Paper · Question 6

题目

Problem

The residents of Mahjing were asked to classify their local bus service:

  • 25%25\% of residents classified their service as good.
  • 60%60\% of residents classified their service as satisfactory.
  • 15%15\% of residents classified their service as poor.

(a) A random sample of 110 residents of Mahjing is chosen.

Use a suitable approximation to find the probability that fewer than 22 residents classified their bus service as good.

(5)

(b) For a random sample of 10 residents of Mahjing, find the probability that fewer than 8 classified their bus service as good or satisfactory.

(3)

(c) Three residents of Mahjing are selected at random.

Find the probability that one resident classified the bus service as good, one as satisfactory and one as poor.

(2)
题目中文翻译

请 Mahjing 的居民对当地公交服务进行评价：

  • 25%25\% 的居民将服务评为良好。
  • 60%60\% 的居民将服务评为满意。
  • 15%15\% 的居民将服务评为差。

(a) 从 Mahjing 的居民中随机抽取一个容量为 110 的样本。

使用适当的近似方法，求其中少于 22 名居民将公交服务评为良好的概率。

(5)

(b) 对于从 Mahjing 的居民中随机抽取的一个容量为 10 的样本，求其中少于 8 人将公交服务评为良好或满意的概率。

(3)

(c) 从 Mahjing 的居民中随机选出三人。

求其中一人将公交服务评为良好、一人评为满意、另一人评为差的概率。

(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

良好人数服从二项分布，少于22即至多21，修正边界为21.5。

答题过程

展开

Let G∼B(110,0.25)G\sim\mathrm{B}(110,0.25). Since np=27.5np=27.5 and n(1−p)=82.5n(1-p)=82.5 are both greater than 5, use Y∼N(27.5,20.625)Y\sim\mathrm{N}(27.5,20.625).

With a continuity correction,

P(G<22)= P(G≤21)≈ P(Y<21.5)= Φ(21.5−27.520.625)= 0.0932245…≈ 0.0932.\begin{align*} \mathrm{P}(G<22) =&\,\mathrm{P}(G\leq21)\\[4mm] \approx&\,\mathrm{P}(Y<21.5)\\[4mm] =&\,\Phi\biggl(\frac{21.5-27.5}{\sqrt{20.625}}\biggr)\\[4mm] =&\,0.0932245\ldots\\[4mm] \approx&\,\boxed{0.0932}. \end{align*}

(b)

解法一

思路

展开

良好或满意是互斥事件，概率合为0.85。用8、9、10人的三项补集。

答题过程

展开

For each resident, the probability of good or satisfactory is 0.25+0.60=0.850.25+0.60=0.85. Let X∼B(10,0.85)X\sim\mathrm{B}(10,0.85). Then

P(X<8)= 1−[(108)(0.85)8(0.15)2 +10(0.85)9(0.15)+(0.85)10]= 0.179803519…≈ 0.180.\begin{align*} \mathrm{P}(X<8) =&\,1-\bigl[\binom{10}{8}(0.85)^8(0.15)^2\\[4mm] &\,\hspace{2pt}+10(0.85)^9(0.15)+(0.85)^{10}\bigr]\\[4mm] =&\,0.179803519\ldots\\[4mm] \approx&\,\boxed{0.180}. \end{align*}

(c)

解法一

思路

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一良好、一满意、一差共有3!种互斥次序。

答题过程

展开 P=3!(0.25)(0.60)(0.15)=0.135.\mathrm{P}=3!(0.25)(0.60)(0.15)=\boxed{0.135}.