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CIE 9709 2022 June Paper 63 Q1

A Level / CIE / S2

CIE 9709 2022 June Paper 63 Paper · Question 1

题目

Problem

The number of characters in emails sent by a particular company is modelled by the distribution N(1250,4802)\mathrm{N}(1250,480^2).

Find the probability that the mean number of characters in a random sample of 100100 emails sent by the company is more than 13001300.

(3)
题目中文翻译

某公司发出的电子邮件的字符数用 N(1250,4802)\mathrm{N}(1250,480^2) 分布描述。

从该公司发出的邮件中随机抽取 100100 封，求样本的平均字符数大于 13001300 的概率。

(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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总体按正态模型给出，样本均值的标准差为480/10。

答题过程

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The sample mean has distribution

X‾∼N(1250,4802100).\overline X\sim\mathrm{N}\bigg(1250,\frac{480^2}{100}\bigg).

Hence

P(X‾>1300)= 1−Φ(1300−1250480/100)= 0.149(3 s.f.).\begin{align*} \mathrm{P}(\overline X>1300)=&\,1-\Phi\bigg(\frac{1300-1250}{480/\sqrt{100}}\bigg)\\[4mm] =&\,\boxed{0.149}\quad\text{(3 s.f.).} \end{align*}