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CIE 9709 2026 June Paper 55 Q4

A Level / CIE / S1

CIE 9709 2026 June Paper 55 Paper · Question 4

题目

Problem

In the village of Torombo, 80% of the residents shop at the local store.

A random sample of 9 residents of Torombo is selected.

(a) Find the probability that fewer than 7 of these 9 residents shop at the local store.

(3)

A random sample of 120 residents of Torombo is selected.

(b) Use an approximation to find the probability that more than 90 of these 120 residents shop at the local store.

(5)
题目中文翻译

在 Torombo 村，80% 的居民会在当地商店购物。

从 Torombo 村的居民中随机抽取 9 人。

(a) 求这 9 人中在当地商店购物的人数少于 7 的概率。

(3)

从 Torombo 村的居民中随机抽取 120 人。

(b) 利用近似方法，求这 120 人中在当地商店购物的人数多于 90 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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二项分布中少于7即0至6次。补集只有7、8、9三项，用补集计算更简洁。

答题过程

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Let XX be the number of residents who shop at the local store. Then X∼B(9,0.8)X\sim\mathrm{B}(9,0.8), so

P(X<7)= 1−P(X≥7)= 1−[(97)(0.8)7(0.2)2 +(98)(0.8)8(0.2)+(0.8)9]= 0.261802496≈ 0.262.\begin{align*} \mathrm{P}(X<7)=&\,1-\mathrm{P}(X\geq7)\\[4mm] =&\,1-\bigl[\binom97(0.8)^7(0.2)^2\\[4mm] &\,\hspace{2pt}+\binom98(0.8)^8(0.2)+(0.8)^9\bigr]\\[4mm] =&\,0.261802496\\[4mm] \approx&\,\boxed{0.262}. \end{align*}

(b)

解法一

思路

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用正态分布近似120次的二项分布。超过90从91开始，连续性修正后的下界是90.5。

答题过程

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Let X∼B(120,0.8)X\sim\mathrm{B}(120,0.8). Since np=96np=96 and n(1−p)=24n(1-p)=24 are both greater than 5, a normal approximation is suitable.

The approximating variable has distribution

Y∼N(96,19.2),Y\sim\mathrm{N}(96,19.2),

where 19.2=120(0.8)(0.2)19.2=120(0.8)(0.2) is the variance. Applying a continuity correction,

P(X>90)= P(X≥91)≈ P(Y>90.5)= P(Z>90.5−9619.2)= P(Z>−1.2552…)= 0.895296468…≈ 0.895.\begin{align*} \mathrm{P}(X>90)=&\,\mathrm{P}(X\geq91)\\[4mm] \approx&\,\mathrm{P}(Y>90.5)\\[4mm] =&\,\mathrm{P}\biggl(Z>\frac{90.5-96}{\sqrt{19.2}}\biggr)\\[4mm] =&\,\mathrm{P}(Z>-1.2552\ldots)\\[4mm] =&\,0.895296468\ldots\\[4mm] \approx&\,\boxed{0.895}. \end{align*}