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CIE 9709 2025 November Paper 55 Q6

A Level / CIE / S1

CIE 9709 2025 Nov Paper 55 Paper · Question 6

题目

Problem

A large number of runners took part in two charity runs to raise money for a new community centre.

In the first run, the times to complete the run were normally distributed with mean 46.3 seconds and standard deviation 6.4 seconds.

(a) Find the probability that a randomly chosen runner took more than 55.1 seconds to complete the run.

(3)

In the second run, the times to complete the run were normally distributed with mean 39.8 seconds and standard deviation σ\sigma seconds. 10% of the runners took more than 48.6 seconds.

(b) Find the value of σ\sigma.

(3)

150 runners are chosen at random from those who took part in the second run.

(c) Use an approximation to find the probability that fewer than 20 of the 150 runners took more than 48.6 seconds to complete the run.

(5)
题目中文翻译

许多跑步者参加了两场慈善跑，为一座新的社区中心筹款。

在第一场慈善跑中，完成赛程所用的时间服从正态分布，平均数为 46.3 秒，标准差为 6.4 秒。

(a) 求随机选取的一名跑步者完成赛程所用时间多于 55.1 秒的概率。

(3)

在第二场慈善跑中，完成赛程所用的时间服从正态分布，平均数为 39.8 秒，标准差为 σ\sigma 秒。10% 的跑步者所用的时间多于 48.6 秒。

(b) 求 σ\sigma 的值。

(3)

从参加第二场慈善跑的人中随机选取 150 人。

(c) 利用近似方法，求这 150 人中完成赛程所用时间多于 48.6 秒的人数少于 20 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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55.1比均值大8.8秒，标准化后取右尾概率。

答题过程

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Let T∼N(46.3,6.42)T\sim\mathrm{N}(46.3,6.4^2) denote the first-run time in seconds. Then

P(T>55.1)= 1−Φ(55.1−46.36.4)= 1−Φ(1.375)= 0.0845657…≈ 0.0846(3 s.f.).\begin{align*} \mathrm{P}(T>55.1) =&\,1-\Phi\biggl(\frac{55.1-46.3}{6.4}\biggr)\\[4mm] =&\,1-\Phi(1.375)\\[4mm] =&\,0.0845657\ldots\\[4mm] \approx&\,\boxed{0.0846}\quad\text{(3 s.f.)}. \end{align*}

(b)

解法一

思路

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超过48.6的概率为0.10，所以48.6对应左侧累计概率0.90。标准化后求标准差。

答题过程

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For the second run, P(T>48.6)=0.10\mathrm{P}(T>48.6)=0.10 implies P(T<48.6)=0.90\mathrm{P}(T<48.6)=0.90. Hence

48.6−39.8σ=Φ−1(0.90)=1.281551….\frac{48.6-39.8}{\sigma} =\Phi^{-1}(0.90)=1.281551\ldots.

Therefore,

σ=8.81.281551…=6.866676…≈6.87.\sigma=\frac{8.8}{1.281551\ldots} =6.866676\ldots \approx\boxed{6.87}.

(c)

解法一

思路

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超过指定时间的人数服从成功概率0.10的二项分布，用正态近似。“少于20”即19及以下，连续性修正边界为19.5。

答题过程

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Let X∼B(150,0.10)X\sim\mathrm{B}(150,0.10) be the number taking more than 48.6 seconds.

Since np=15np=15 and n(1−p)=135n(1-p)=135 are both greater than 5, use

Y∼N(15,13.5),13.5=150(0.1)(0.9).Y\sim\mathrm{N}(15,13.5), \qquad 13.5=150(0.1)(0.9).

Applying a continuity correction,

P(X<20)= P(X≤19)≈ P(Y<19.5)= Φ(19.5−1513.5)= Φ(1.224744…)= 0.889664…≈ 0.890.\begin{align*} \mathrm{P}(X<20)=&\,\mathrm{P}(X\leq19)\\[4mm] \approx&\,\mathrm{P}(Y<19.5)\\[4mm] =&\,\Phi\biggl(\frac{19.5-15}{\sqrt{13.5}}\biggr)\\[4mm] =&\,\Phi(1.224744\ldots)\\[4mm] =&\,0.889664\ldots\\[4mm] \approx&\,\boxed{0.890}. \end{align*}