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CIE 9709 2022 November Paper 51 Q2

A Level / CIE / S1

CIE 9709 2022 Nov Paper 51 Paper · Question 2

题目

Problem

The residents of Persham were surveyed about the reliability of their internet service. 12% rated the service as ‘poor’, 36% rated it as ‘satisfactory’ and 52% rated it as ‘good’.

A random sample of 8 residents of Persham is chosen.

(a) Find the probability that more than 2 and fewer than 8 of them rate their internet service as poor or satisfactory.

(3)

A random sample of 125 residents of Persham is now chosen.

(b) Use an approximation to find the probability that more than 72 of these residents rate their internet service as good.

(5)

题目中文翻译

对 Persham 的居民进行调查，询问他们对网络服务可靠性的评价。12% 的居民评价为“差”，36% 评价为“尚可”，52% 评价为“好”。

随机抽取 Persham 的 8 名居民作为样本。

(a) 求其中评价网络服务为差或尚可的人数多于 2 人且少于 8 人的概率。

(3)

现随机抽取 Persham 的 125 名居民作为样本。

(b) 用近似方法求其中超过 72 人评价网络服务为好的概率。

(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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差或尚可合为概率0.48，所求人数3至7。补集包含0、1、2以及8，不能漏掉8。

答题过程

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Let X∼B(8,0.48)X\sim\mathrm{B}(8,0.48) count residents rating the service poor or satisfactory. Then

P(2<X<8)= 1−[(0.52)8+8(0.48)(0.52)7 +(82)(0.48)2(0.52)6+(0.48)8]= 0.824813988…≈0.825.\begin{align*} \mathrm{P}(2<X<8) =&\,1-\bigl[(0.52)^8+8(0.48)(0.52)^7\\[4mm] &\,\hspace{2pt}+\binom82(0.48)^2(0.52)^6+(0.48)^8\bigr]\\[4mm] =&\,0.824813988\ldots\approx\boxed{0.825}. \end{align*}

(b)

解法一

思路

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评价好的人数使用p0.52，超过72修正为72.5。

答题过程

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Let G∼B(125,0.52)G\sim\mathrm{B}(125,0.52). Since np=65np=65 and n(1−p)=60n(1-p)=60 are both greater than 5, use Y∼N(65,31.2)Y\sim\mathrm{N}(65,31.2).

With a continuity correction,

P(G>72)= P(G≥73)≈ P(Y>72.5)= 1−Φ(72.5−6531.2)= 0.089682074…≈0.0897.\begin{align*} \mathrm{P}(G>72) =&\,\mathrm{P}(G\geq73)\\[4mm] \approx&\,\mathrm{P}(Y>72.5)\\[4mm] =&\,1-\Phi\biggl(\frac{72.5-65}{\sqrt{31.2}}\biggr)\\[4mm] =&\,0.089682074\ldots\approx\boxed{0.0897}. \end{align*}