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CIE 9231 2024 June Paper 42 Q1

A Level / CIE / FS

CIE 9231 2024 June Paper 42 Paper · Question 1

题目

Problem

The times taken by members of a large cycling club to complete a cross-country circuit have a normal distribution with mean μ\mu minutes. The times taken, xx minutes, are recorded for a random sample of 14 members of the club. The results are summarised as follows, where xˉ\bar{x} is the sample mean.

xˉ=42.8∑(x−xˉ)2=941.5\bar{x} = 42.8 \qquad \sum (x - \bar{x})^2 = 941.5

Find a 95% confidence interval for μ\mu.

[4]
题目中文翻译

一个大型自行车俱乐部的会员完成一次越野骑行路线所用的时间服从均值为 μ\mu 分钟的正态分布。从该俱乐部会员中随机抽取 14 人,记录他们所用的时间 xx 分钟。结果概括如下,其中 xˉ\bar{x} 是样本均值。

xˉ=42.8∑(x−xˉ)2=941.5\bar{x} = 42.8 \qquad \sum (x - \bar{x})^2 = 941.5

求 μ\mu 的 95% 置信区间。

(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

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样本来自正态总体,方差未知。用题给平方离差和求无偏方差,再用 tt 临界值。

答题过程

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The sample variance is s2=941.5/(14−1)=72.42307…s^2=941.5/(14-1)=72.42307\ldots. The two-sided 95%95\% critical value with 1313 degrees of freedom is 2.1602.160. The interval is

 42.8±2.160941.5/1314= (37.8872…,47.7127…).\begin{align*} &\,42.8\pm2.160\sqrt{\frac{941.5/13}{14}}\\ =&\,(37.8872\ldots,47.7127\ldots). \end{align*}

Thus the 95%95\% confidence interval is (37.9,47.7)(37.9,47.7) minutes.