Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2025 June Paper 63 Q5

A Level / CIE / S2

CIE 9709 2025 June Paper 63 Paper · Question 5

题目

Problem

In Urberia, the masses, in kilograms, of men have the distribution N(70.3,5.92)\mathrm{N}(70.3,5.9^2). A certain footbridge in Urberia can take a maximum safe load of 1500 kg1500\,\mathrm{kg}. When nn men stand on the bridge, the probability that the bridge is unsafe is less than 0.010.01.

Stating a necessary assumption, find the maximum value of nn.

(4)
题目中文翻译

在 Urberia,男性的体重以千克为单位,服从分布 N(70.3,5.92)\mathrm{N}(70.3,5.9^2)。Urberia 的某座人行桥的最大安全荷载为 1500 kg1500\,\mathrm{kg}。当 nn 名男子站在桥上时,桥不安全的概率小于 0.010.01。

写出一个必要的假设,并求 nn 的最大值。

(4)
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

解法一

思路

展开

假设桥上的人可视为总体中的独立随机样本。总质量正态,方差随人数线性增长。把危险概率限制转成正态分位数不等式,再令 t=nt=\sqrt n 解二次不等式。

答题过程

展开

Assume that the men on the bridge form an independent random sample from the stated population. Let MM be their total mass. Then

M∼N(70.3n, 5.92n).M\sim\mathrm N(70.3n,\,5.9^2n).

The bridge is unsafe when M>1500M>1500. The requirement P(M>1500)<0.01\mathrm P(M>1500)<0.01 implies

1500−70.3n5.9n> Φ−1(0.99),Φ−1(0.99)≈ 2.326.\begin{align*} \frac{1500-70.3n}{5.9\sqrt n}>&\,\Phi^{-1}(0.99),\\[4mm] \Phi^{-1}(0.99)\approx&\,2.326. \end{align*}

Putting t=n>0t=\sqrt n>0 gives

70.3t2+13.7234t−1500<0.70.3t^2+13.7234t-1500<0.

The positive boundary root is

t= −13.7234+13.72342+4(70.3)(1500)2(70.3)≈ 4.52264.\begin{align*} t=&\,\frac{-13.7234+\sqrt{13.7234^2+4(70.3)(1500)}}{2(70.3)}\\[4mm] \approx&\,4.52264. \end{align*}

The other root is negative, so for t>0t>0 the inequality requires t<4.52264t<4.52264, or n<20.4543…n<20.4543\ldots. The greatest integer satisfying this is

n=20.\boxed{n=20}.

For confirmation, at n=20n=20 the unsafe probability is about 0.000184<0.010.000184<0.01, whereas at n=21n=21 it is about 0.190>0.010.190>0.01.