题目

Problem

(a) One of a group of three students is to be chosen at random.

Explain how a single throw of a fair six-sided dice could be used to make the choice.

(1)

The times, in minutes, taken by students to complete a test are normally distributed with mean $125$ and variance $50$ . Two students are chosen at random.

(b) Find the probability that the difference between the times taken by these two students to complete the test is more than $12$ minutes.

(5)

题目中文翻译

(a) 要从一组三名学生中随机选出一人。

说明如何只掷一次均匀的六面骰子来进行选择。

(1)

学生完成一项测试所需的时间以分钟为单位，服从均值为 $125$ 、方差为 $50$ 的正态分布。随机选取两名学生。

(b) 求这两名学生完成测试所用时间相差超过 $12$ 分钟的概率。

(5)

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

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把六个等可能点数分成三组，每组两个点数，一次投掷即可保证三人被选概率相同。

答题过程

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Label the students $1,2,3$ . Throw the fair die once: choose student $1$ for a score of $1$ or $2$ , student $2$ for $3$ or $4$ , and student $3$ for $5$ or $6$ . Each student has probability $2/6=1/3$ of selection.

(b)

解法一

思路

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没有指定谁更快，所以“差超过十二分钟”包含两个方向。先求带符号的差的分布，再利用对称性合并两尾。

答题过程

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Let $T_1,T_2$ be the completion times of the independently selected students. Each is $\mathrm N(125,50)$ , so

\begin{align*} D=T_1-T_2\sim&\,\mathrm N(125-125,\,50+50)\\[4mm] =&\,\mathrm N(0,100). \end{align*}

The required event is $|D|>12$ . By symmetry,

\begin{align*} \mathrm P(|D|>12) =&\,2\mathrm P(D>12)\\[4mm] =&\,2\bigg[1-\Phi\bigg(\frac{12}{10}\bigg)\bigg]\\[4mm] =&\,2[1-\Phi(1.2)]\\[4mm] \approx&\,\boxed{0.230}. \end{align*}

CIE 9709 2025 June 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