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CIE 9231 2023 June Paper 42 Q1

A Level / CIE / FS

CIE 9231 2023 June Paper 42 Paper · Question 1

题目

Problem

The lengths of the leaves of a particular type of tree are normally distributed with mean μ\mu cm. The lengths, xx cm, of a random sample of 12 leaves of this type are recorded. The results are summarised as follows.

∑x=91.2∑x2=695.8\sum x = 91.2 \qquad \sum x^2 = 695.8

Find a 95% confidence interval for μ\mu.

[4]
题目中文翻译

某种树的叶片长度服从正态分布,均值为 μ\mu cm。随机抽取这种树的 12 片叶子,记录其长度 xx cm。结果汇总如下。

∑x=91.2∑x2=695.8\sum x = 91.2 \qquad \sum x^2 = 695.8

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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由汇总值求样本均值和无偏方差。总体正态、方差未知,使用小样本 tt 区间。

答题过程

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The sample mean is 91.2/12=7.691.2/12=7.6, and the unbiased variance estimate is

s2=695.8−91.22/1211=67275.s^2=\frac{695.8-91.2^2/12}{11}=\frac{67}{275}.

The two-sided 95%95\% critical value for 1111 degrees of freedom is 2.2012.201. Hence the interval is

 7.6±2.20167/27512= (7.28638…,7.91361…).\begin{align*} &\,7.6\pm2.201\sqrt{\frac{67/275}{12}}\\ =&\,(7.28638\ldots,7.91361\ldots). \end{align*}

The 95%95\% confidence interval is (7.29,7.91)(7.29,7.91) cm.