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CIE 9709 2024 June Paper 61 Q3

A Level / CIE / S2

CIE 9709 2024 June Paper 61 Paper · Question 3

题目

Problem

The time taken in minutes for a certain daily train journey has a normal distribution with standard deviation 5.85.8. For a random sample of 2020 days the journey times were noted and the mean journey time was found to be 81.581.5 minutes.

(a) Calculate a 98%98\% confidence interval for the population mean journey time.

(3)

A student was asked for the meaning of this confidence interval. The student replied as follows.

‘The times for 98%98\% of these journeys are likely to be within the confidence interval.’

(b) Explain briefly whether this statement is true or not.

(1)

Two independent 98%98\% confidence intervals are found.

(c) Given that at least one of these intervals contains the population mean, find the probability that both intervals contain the population mean.

(2)
题目中文翻译

某趟每天运行的列车的行程用时（以分钟为单位）服从标准差为 5.85.8 的正态分布。随机选取 2020 天，记录每天的行程用时，得到平均行程用时为 81.581.5 分钟。

(a) 计算总体平均行程用时的 98%98\% 置信区间。

(3)

有人问一名学生这个置信区间的含义。学生回答如下。

“这些行程中，很可能有 98%98\% 的行程用时落在该置信区间内。”

(b) 简要说明这句话是否正确。

(1)

现求得两个独立的 98%98\% 置信区间。

(c) 已知其中至少一个区间包含总体均值，求两个区间都包含总体均值的概率。

(2)
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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已知总体标准差，98%置信区间的正态临界值为累计概率0.99所对应的值。

答题过程

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With z=Φ−1(0.99)=2.326348…z=\Phi^{-1}(0.99)=2.326348\ldots, the 98% confidence interval is

81.5±z5.820=81.5±3.01709….81.5\pm z\frac{5.8}{\sqrt{20}} =81.5\pm3.01709\ldots.

Therefore the interval for the population mean journey time is

(78.5, 84.5) minutes(3 s.f.).\boxed{(78.5,\ 84.5)\text{ minutes}}\quad\text{(3 s.f.).}

(b)

解法一

思路

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置信区间针对总体平均旅程时间，并非针对个别旅程时间的分布。

答题过程

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The statement is not true. The interval estimates the population mean journey time; it is not an interval intended to contain 98% of individual journey times.

(c)

解法一

思路

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题目明确给出两个区间独立，因此同时包含均值的概率为0.98平方；至少一个包含的概率用两个都不包含的补集求出。

答题过程

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Let AA and BB be the events that the first and second intervals contain the population mean. By the stated independence,

P(A∩B)= 0.982,P(A∪B)= 1−0.022.\begin{align*} \mathrm{P}(A\cap B)=&\,0.98^2,\\[4mm] \mathrm{P}(A\cup B)=&\,1-0.02^2. \end{align*}

Since A∩B⊆A∪BA\cap B\subseteq A\cup B,

 P(A∩B∣A∪B)= 0.9821−0.022= 0.960784…≈ 0.961(3 s.f.).\begin{align*} &\,\mathrm{P}(A\cap B\mid A\cup B)\\[4mm] =&\,\frac{0.98^2}{1-0.02^2}\\[4mm] =&\,0.960784\ldots\\[4mm] \approx&\,\boxed{0.961}\quad\text{(3 s.f.).} \end{align*}