题目

Problem

The heights, $H$ centimetres, of a random sample of $8$ one-year-old plants of a certain variety were measured, with the following results.

104,\quad 96,\quad 110,\quad 122,\quad 96,\quad 101,\quad 99,\quad 113

(a) Calculate unbiased estimates of the population mean and variance of $H$ .

(3)

It is now given that the true value of the population variance is $85.0$ .

(b) Stating a necessary assumption, find a $99\%$ confidence interval for the population mean of $H$ , giving the limits of the interval correct to $1$ decimal place.

(4)

题目中文翻译

从某品种的一年生植物中随机抽取 $8$ 株，测量其高度 $H$ （以厘米为单位），得到下列结果。

104,\quad 96,\quad 110,\quad 122,\quad 96,\quad 101,\quad 99,\quad 113

(a) 计算 $H$ 的总体均值和总体方差的无偏估计值。

(3)

现已知总体方差的真实值为 $85.0$ 。

(b) 写出一个必要的假设，并求 $H$ 的总体均值的一个 $99\%$ 置信区间，区间端点保留 $1$ 位小数。

(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

(a)

解法一

思路

展开

从八个数分别求和与平方和，无偏方差使用分母 $7$ 。

答题过程

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The data give

n=8,\qquad\sum h=841,\qquad\sum h^2=89003.

An unbiased estimate of the population mean is

\boxed{\overline h=\frac{841}{8}=105.125\ \text{cm}}.

An unbiased estimate of the population variance is

\begin{align*} s^2=&\,\frac1{7}\bigg(89003-\frac{841^2}{8}\bigg)\\[4mm] =&\,\frac{592.875}{7}\\[4mm] =&\,\frac{4743}{56}\approx\boxed{84.7\ \text{cm}^2}. \end{align*}

(b)

解法一

思路

展开

样本只有八个，需要假设总体正态。此问已经给出真实总体方差 $85$ ，不能再用(a)的估计值代替。

答题过程

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Assume that the population distribution of plant heights $H$ is normal. Using the known population variance $85.0$ , a $99\%$ confidence interval for the population mean is

\overline h\pm z_{0.995}\sqrt{\frac{85}{8}}.

With $z_{0.995}\approx2.576$ ,

105.125\pm2.576\sqrt{\frac{85}{8}} \approx105.125\pm8.397.

Hence the interval, with limits to $1$ decimal place, is

\boxed{(96.7,\ 113.5)\ \text{cm}}.

CIE 9709 2026 March Paper 62 Q1

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程