题目

Problem

The heights of buildings in a large city are normally distributed with mean $18.3\,\mathrm{m}$ and standard deviation $2.5\,\mathrm{m}$ .

(a) Find the probability that the total height of $5$ randomly chosen buildings in the city is more than $95\,\mathrm{m}$ .

(4)

(b) Find the probability that the difference between the heights of two randomly chosen buildings in the city is less than $1\,\mathrm{m}$ .

(5)

题目中文翻译

某大城市的建筑物高度服从均值为 $18.3\,\mathrm{m}$ 、标准差为 $2.5\,\mathrm{m}$ 的正态分布。

(a) 在该市随机选取 $5$ 栋建筑物，求它们的高度之和大于 $95\,\mathrm{m}$ 的概率。

(4)

(b) 在该市随机选取两栋建筑物，求它们的高度相差小于 $1\,\mathrm{m}$ 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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随机抽取的五幢建筑按独立抽样模型，总高度的均值与方差分别乘5。

答题过程

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Let $T$ be the total height of the five independently sampled buildings.

\begin{align*} T\sim&\,\mathrm{N}\bigl(5(18.3),\,5(2.5^2)\bigr)\\[4mm] =&\,\mathrm{N}(91.5,31.25). \end{align*}

Therefore

\begin{align*} \mathrm{P}(T>95)=&\,1-\Phi\bigg(\frac{95-91.5}{\sqrt{31.25}}\bigg)\\[4mm] =&\,\boxed{0.266}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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高度差小于1指绝对差小于1；有向差的均值0，方差为两个方差之和。

答题过程

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For two independently sampled heights, let $D=H_1-H_2$ . Then

D\sim\mathrm{N}(0,\,2(2.5^2))=\mathrm{N}(0,12.5).

The required probability is

\begin{align*} \mathrm{P}(|D|<1) =&\,\mathrm{P}(-1<D<1)\\[4mm] =&\,\Phi\bigg(\frac{1}{\sqrt{12.5}}\bigg)\\[4mm] &\,-\Phi\bigg(\frac{-1}{\sqrt{12.5}}\bigg)\\[4mm] =&\,\boxed{0.223}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2022 March Paper 62 Q5

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程