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CIE 9709 2022 November Paper 52 Q6

A Level / CIE / S1

CIE 9709 2022 Nov Paper 52 Paper · Question 6

题目

Problem

At a company’s call centre, 90% of callers are connected immediately to a representative.

A random sample of 12 callers is chosen.

(a) Find the probability that fewer than 10 of these callers are connected immediately.

(3)

A random sample of 80 callers is chosen.

(b) Use an approximation to find the probability that more than 69 of these callers are connected immediately.

(5)

(c) Justify the use of your approximation in part (b).

(1)

题目中文翻译

在某公司的呼叫中心，90% 的来电者能立即接通一名客服人员。

随机抽取 12 名来电者作为样本。

(a) 求这些来电者中能立即接通的人数少于 10 人的概率。

(3)

随机抽取 80 名来电者作为样本。

(b) 用近似方法求这些来电者中能立即接通的人数超过 69 人的概率。

(5)

(c) 说明你在 (b) 中使用该近似方法的合理性。

(1)

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

12人少于10人立即接通，取10、11、12三项补集。

答题过程

展开

Let X∼B(12,0.9)X\sim\mathrm{B}(12,0.9). Then

P(X<10)= 1−[(1210)(0.9)10(0.1)2 +12(0.9)11(0.1)+(0.9)12]= 0.110869977…≈0.111.\begin{align*} \mathrm{P}(X<10) =&\,1-\bigl[\binom{12}{10}(0.9)^{10}(0.1)^2\\[4mm] &\,\hspace{2pt}+12(0.9)^{11}(0.1)+(0.9)^{12}\bigr]\\[4mm] =&\,0.110869977\ldots\approx\boxed{0.111}. \end{align*}

(b)

解法一

思路

展开

80人的二项分布用正态近似，超过69即至少70，修正为69.5。

答题过程

展开

Let X∼B(80,0.9)X\sim\mathrm{B}(80,0.9). Use Y∼N(72,7.2)Y\sim\mathrm{N}(72,7.2), since

E(X)=80(0.9)=72,Var⁡(X)=80(0.9)(0.1)=7.2.\mathrm{E}(X)=80(0.9)=72,\qquad \operatorname{Var}(X)=80(0.9)(0.1)=7.2.

With a continuity correction,

P(X>69)= P(X≥70)≈ P(Y>69.5)= 1−Φ(69.5−727.2)= 0.824252911…≈0.824.\begin{align*} \mathrm{P}(X>69) =&\,\mathrm{P}(X\geq70)\\[4mm] \approx&\,\mathrm{P}(Y>69.5)\\[4mm] =&\,1-\Phi\biggl(\frac{69.5-72}{\sqrt{7.2}}\biggr)\\[4mm] =&\,0.824252911\ldots\approx\boxed{0.824}. \end{align*}

(c)

解法一

思路

展开

必须同时检查预期成功数与预期失败数，且都大于5。

答题过程

展开 np=80(0.9)=72>5,n(1−p)=80(0.1)=8>5.np=80(0.9)=72>5,\qquad n(1-p)=80(0.1)=8>5.

Thus a normal approximation to the binomial distribution is appropriate.