题目

Problem

The heights of a certain species of deer are known to have standard deviation $0.35\,\mathrm{m}$ . A zoologist takes a random sample of $150$ of these deer and finds that the mean height of the deer in the sample is $1.42\,\mathrm{m}$ .

(a) Calculate a $96\%$ confidence interval for the population mean height.

(3)

(b) Bubay says that $96\%$ of deer of this species are likely to have heights that are within this confidence interval.

Explain briefly whether Bubay is correct.

(1)

题目中文翻译

已知某种鹿的身高的标准差为 $0.35\,\mathrm{m}$ 。一名动物学家随机抽取了 $150$ 只这种鹿，发现样本中鹿的平均身高为 $1.42\,\mathrm{m}$ 。

(a) 计算总体平均身高的 $96\%$ 置信区间。

(3)

(b) Bubay 说，这种鹿中很可能有 $96\%$ 的个体的身高落在这个置信区间内。

简要说明 Bubay 的说法是否正确。

(1)

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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总体标准差已知，用样本均值加减正态临界值乘标准误；96%中央概率对应两端各2%。

答题过程

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For a 96% confidence interval, $z=\Phi^{-1}(0.98)=2.053749\ldots$ . Hence the interval for the population mean height is

\begin{align*} 1.42\pm z\frac{0.35}{\sqrt{150}} =&\,1.42\pm0.0586908\ldots. \end{align*}

Thus the interval is

\boxed{(1.36,\ 1.48)\text{ m}}\quad\text{(3 s.f.).}

(b)

解法一

思路

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置信区间估计的是总体均值，而非覆盖固定比例个体身高的区间。

答题过程

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Bubay is incorrect. This is a confidence interval for the population mean height, rather than an interval intended to contain 96% of individual deer heights.

CIE 9709 2024 November Paper 61 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程