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CIE 9709 2025 June Paper 62 Q2

A Level / CIE / S2

CIE 9709 2025 June Paper 62 Paper · Question 2

题目

Problem

The height of a certain species of plant is denoted by H cmH\,\mathrm{cm}. The heights of a random sample of 100100 plants were measured, and the following results were found.

  • The mean, h‾\overline{h}, for the sample was 80.280.2.
  • An unbiased estimate of the population variance of HH was 15.615.6.

Calculate the value of ∑h2\sum h^2.

(3)
题目中文翻译

某种植物的高度记为 H cmH\,\mathrm{cm}。随机抽取 100100 株植物,测量其高度,得到下列结果。

  • 样本均值 h‾\overline{h} 为 80.280.2。
  • HH 的总体方差的一个无偏估计值为 15.615.6。

计算 ∑h2\sum h^2 的值。

(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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反用无偏方差公式,把已知的样本均值与方差代入,再解出平方和。

答题过程

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Using s2=(∑h2−nh‾2)/(n−1)s^2=(\sum h^2-n\overline h^2)/(n-1),

15.6=∑h2−100(80.2)299.15.6=\frac{\sum h^2-100(80.2)^2}{99}.

Therefore

∑h2= 99(15.6)+100(80.2)2= 1544.4+643204= 644748.4.\begin{align*} \sum h^2 =&\,99(15.6)+100(80.2)^2\\[4mm] =&\,1544.4+643204\\[4mm] =&\,\boxed{644748.4}. \end{align*}