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CIE 9231 2025 Nov Paper 43 Q3

A Level / CIE / FS

CIE 9231 2025 Nov Paper 43 Paper · Question 3

题目

Problem

A random sample of 10 newborn baby boys is taken and their masses in kg are recorded. From this sample, the population standard deviation of all newborn baby boys is estimated as 0.6 kg0.6\text{ kg}. A random sample of 5 newborn baby girls is taken and their masses in kg are recorded as follows.

3.9, 3.1, 2.9, 3.1, 3.63.9,\ 3.1,\ 2.9,\ 3.1,\ 3.6

It is assumed that the masses of newborn baby boys and girls have the same population standard deviation, σ kg\sigma\text{ kg}.

By pooling the two samples, calculate an estimate of σ\sigma.

[4]
题目中文翻译

随机抽取 10 名新生男婴,并记录他们的体重(单位为 kg)。由该样本估计所有新生男婴总体的标准差为 0.6 kg0.6\text{ kg}。随机抽取 5 名新生女婴,并记录她们的体重(单位为 kg)如下。

3.9, 3.1, 2.9, 3.1, 3.63.9,\ 3.1,\ 2.9,\ 3.1,\ 3.6

假设新生男婴和女婴的体重具有相同的总体标准差 σ kg\sigma\text{ kg}。

通过合并两个样本,计算 σ\sigma 的估计值。

(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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合并的是各组绕本组均值的平方离差和。用每组的自由度将方差换回平方离差和。

答题过程

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The boys contribute (10−1)(0.6)2=3.24(10-1)(0.6)^2=3.24. For the girls, ∑x=16.6\sum x=16.6 and ∑x2=55.8\sum x^2=55.8, so their within-sample sum of squares is 55.8−16.62/5=0.68855.8-16.6^2/5=0.688. The pooled variance estimate is

sp2=3.24+0.688(10−1)+(5−1)=0.302153….s_p^2=\frac{3.24+0.688}{(10-1)+(5-1)}=0.302153\ldots.

The estimate of σ\sigma is sp2=0.550\sqrt{s_p^2}=0.550 kg to three significant figures.