题目

Problem

The widths, $w\,\mathrm{cm}$ , of a random sample of $150$ leaves of a certain kind were measured. The sample mean of $w$ was found to be $3.12\,\mathrm{cm}$ .

Using this sample, an approximate $95\%$ confidence interval for the population mean of the widths in centimetres was found to be $[3.01,3.23]$ .

(a) Calculate an estimate of the population standard deviation.

(3)

(b) Explain whether it was necessary to use the Central Limit theorem in your answer to part (a).

(1)

题目中文翻译

从某类叶片中随机抽取 $150$ 片，测量其宽度 $w\,\mathrm{cm}$ 。测得 $w$ 的样本均值为 $3.12\,\mathrm{cm}$ 。

利用这个样本，得到宽度总体均值（以厘米为单位）的近似 $95\%$ 置信区间为 $[3.01,3.23]$ 。

(a) 估算总体标准差。

(3)

(b) 说明解答 (a) 时是否有必要使用中心极限定理。

(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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95%区间半宽为0.11，等于1.96倍标准误，反求总体标准差估计。

答题过程

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Let $s$ estimate the population standard deviation. The margin of error is

\frac{3.23-3.01}{2}=0.11\text{ cm}.

Thus

1.96\frac{s}{\sqrt{150}}=0.11.

Hence

\begin{align*} s=&\,\frac{0.11\sqrt{150}}{1.96}\\[4mm] \approx&\,\boxed{0.687\text{ cm}}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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总体宽度分布未知，需用中心极限定理得到大样本均值的近似正态分布。

答题过程

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Yes. The population distribution of leaf widths is not given to be normal, so the Central Limit Theorem is needed to approximate the distribution of the sample mean by a normal distribution.

CIE 9709 2024 June Paper 63 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程