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CIE 9709 2024 March Paper 62 Q1

A Level / CIE / S2

CIE 9709 2024 March Paper 62 Paper · Question 1

题目

Problem

The lengths, X cmX\,\mathrm{cm}, of a sample of 100100 insects of a certain type were summarised as follows.

n=100∑x=36.8∑x2=17.34n=100\qquad\sum x=36.8\qquad\sum x^2=17.34

(a) Calculate unbiased estimates for the population mean and variance of XX.

(3)

(b) State a necessary condition for the estimates found in part (a) to be reliable.

(1)
题目中文翻译

从某类昆虫中抽取 100100 只作为样本,其体长 X cmX\,\mathrm{cm} 的数据汇总如下。

n=100∑x=36.8∑x2=17.34n=100\qquad\sum x=36.8\qquad\sum x^2=17.34

(a) 计算 XX 的总体均值和方差的无偏估计值。

(3)

(b) 写出使 (a) 中所得估计值可靠的一个必要条件。

(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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样本均值为总体均值的无偏估计,样本离均差平方和除以99为总体方差的无偏估计。

答题过程

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The unbiased estimate of the population mean is

xˉ=36.8100=0.368 cm.\bar x=\frac{36.8}{100}=\boxed{0.368\text{ cm}}.

The unbiased estimate of the population variance is

s2= 17.34−36.82/10099= 0.0383596…≈ 0.0384 cm2(3 s.f.).\begin{align*} s^2=&\,\frac{17.34-36.8^2/100}{99}\\[4mm] =&\,0.0383596\ldots\\[4mm] \approx&\,\boxed{0.0384\text{ cm}^2}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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可靠估计要求样本选取不偏向总体某类个体。

答题过程

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The insects must have been selected as a random sample from the population.