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CIE 9709 2022 November Paper 61 Q1

A Level / CIE / S2

CIE 9709 2022 Nov Paper 61 Paper · Question 1

题目

Problem

The heights, in metres, of a random sample of 1010 mature trees of a certain variety are given below.

5.96.56.75.96.96.06.46.25.85.85.9\quad6.5\quad6.7\quad5.9\quad6.9\quad6.0\quad6.4\quad6.2\quad5.8\quad5.8

Find unbiased estimates of the population mean and variance of the heights of all mature trees of this variety.

(3)
题目中文翻译

随机抽取某品种的 1010 棵成熟树木，其高度（以米为单位）如下。

5.96.56.75.96.96.06.46.25.85.85.9\quad6.5\quad6.7\quad5.9\quad6.9\quad6.0\quad6.4\quad6.2\quad5.8\quad5.8

求该品种所有成熟树木高度的总体均值和方差的无偏估计值。

(3)
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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先求样本总和与平方和，用分母9的无偏方差公式。

答题过程

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For the ten tree heights,

∑x=62.1,∑x2=387.05.\sum x=62.1,\qquad\sum x^2=387.05.

The unbiased estimate of the population mean is

xˉ=62.110=6.21 m.\bar x=\frac{62.1}{10}=\boxed{6.21\text{ m}}.

The unbiased estimate of the population variance is

s2= 387.05−62.12/109= 14099000≈ 0.157 m2(3 s.f.).\begin{align*} s^2=&\,\frac{387.05-62.1^2/10}{9}\\[4mm] =&\,\frac{1409}{9000}\\[4mm] \approx&\,\boxed{0.157\text{ m}^2}\quad\text{(3 s.f.).} \end{align*}