题目

Problem

Candidates for a certain diploma take two tests. Their marks for the first test and the second test are modelled by the independent variables with distributions $\mathrm{N}(38.1,3.8^2)$ and $\mathrm{N}(64.0,6.1^2)$ respectively. The final mark, $F$ , for each candidate is found by doubling the mark in the first test and adding the result to the mark in the second test.

Find the probability that the mean, $\overline{F}$ , of the final marks of a random sample of $25$ candidates is greater than $143$ .

(5)

题目中文翻译

申请某文凭的考生参加两项测试。他们在第一项和第二项测试中的分数用相互独立的变量建模，分别服从分布 $\mathrm{N}(38.1,3.8^2)$ 和 $\mathrm{N}(64.0,6.1^2)$ 。每名考生的最终分数 $F$ 等于第一项测试分数的两倍加上第二项测试的分数。

求从考生中随机抽取 $25$ 人，所得最终分数的平均值 $\overline{F}$ 大于 $143$ 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

解法一

思路

展开

先求每名考生加权总分的均值和方差，再把方差除以样本人数。

答题过程

展开

Let $A,B$ be the two test marks. Their independence gives

\begin{align*} \mathrm E(F)=&\,2(38.1)+64.0=140.2,\\[4mm] \operatorname{Var}(F)=&\,2^2(3.8^2)+6.1^2\\[4mm] =&\,94.97. \end{align*}

As a linear combination of independent normal variables, $F=2A+B$ is normal. For the sample of $25$ independent candidates,

\overline F\sim\mathrm N\bigg(140.2,\frac{94.97}{25}\bigg).

Therefore

\begin{align*} \mathrm P(\overline F>143) =&\,1-\Phi\bigg(\frac{143-140.2}{\sqrt{94.97/25}}\bigg)\\[4mm] \approx&\,\boxed{0.0754}. \end{align*}

CIE 9709 2025 June Paper 65 Q5

题目

解答

解法一

思路

答题过程