题目

Problem

The length, in minutes, of mathematics lectures at a certain college has mean $\mu$ and standard deviation $8.3$ .

(a) The total length of a random sample of $85$ lectures was $4590$ minutes.

Calculate a $95\%$ confidence interval for $\mu$ .

(3)

The length, in minutes, of history lectures at the college has mean $m$ and standard deviation $s$ .

(b) Using a random sample of $100$ history lectures, a $95\%$ confidence interval for $m$ was found to have width $2.8$ minutes.

Find the value of $s$ .

(2)

题目中文翻译

某学院的数学课时长（以分钟为单位）的均值为 $\mu$ ，标准差为 $8.3$ 。

(a) 随机抽取 $85$ 节数学课，总时长为 $4590$ 分钟。

计算 $\mu$ 的 $95\%$ 置信区间。

(3)

该学院的历史课时长（以分钟为单位）的均值为 $m$ ，标准差为 $s$ 。

(b) 用随机抽取的 $100$ 节历史课的数据求得 $m$ 的一个 $95\%$ 置信区间，其宽度为 $2.8$ 分钟。

求 $s$ 的值。

(2)

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

总时长除以85为样本均值，已知总体标准差8.3，建立95%均值区间。

答题过程

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The sample mean is $\bar x=\frac{4590}{85}=54$ minutes. Since the sample size is large, the Central Limit Theorem gives an approximately normal distribution for the sample mean. The 95% confidence interval is

54\pm1.96\frac{8.3}{\sqrt{85}} =54\pm1.76451\ldots.

Hence

\boxed{(52.2,\ 55.8)\text{ minutes}}\quad\text{(3 s.f.).}

(b)

解法一

思路

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区间宽度为两个标准误边距之和，令2乘1.96乘s/10等于2.8。

答题过程

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The width of the interval is

2(1.96)\frac{s}{\sqrt{100}}=2.8.

Therefore

\begin{align*} s=&\,\frac{2.8(10)}{2(1.96)}=\frac{50}{7}\\[4mm] \approx&\,\boxed{7.14\text{ minutes}}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2023 November Paper 62 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程