题目

Problem

The lengths, in centimetres, of two types of insect, $A$ and $B$ , are modelled by the random variables $X\sim\mathrm{N}(6.2,0.36)$ and $Y\sim\mathrm{N}(2.4,0.25)$ respectively.

Find the probability that the length of a randomly chosen type $A$ insect is greater than the sum of the lengths of $3$ randomly chosen type $B$ insects.

(5)

题目中文翻译

$A$ 、 $B$ 两类昆虫的体长（以厘米为单位）分别用随机变量 $X\sim\mathrm{N}(6.2,0.36)$ 和 $Y\sim\mathrm{N}(2.4,0.25)$ 描述。

随机选取 $1$ 只 $A$ 类昆虫和 $3$ 只 $B$ 类昆虫，求这只 $A$ 类昆虫的体长大于 $3$ 只 $B$ 类昆虫的体长之和的概率。

(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

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三个B类昆虫长度是三个随机值之和，方差乘3。建立A类长度减去该总和的正态分布。

答题过程

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Let $Y_1,Y_2,Y_3$ be the lengths of the three randomly chosen type $B$ insects. Adding the independent sampled lengths,

D=X-(Y_1+Y_2+Y_3).

Then

\begin{align*} D\sim&\,\mathrm{N}(6.2-3(2.4),\ 0.36+3(0.25))\\[4mm] =&\,\mathrm{N}(-1,1.11). \end{align*}

Therefore

\begin{align*} &\,\mathrm{P}(X>Y_1+Y_2+Y_3)\\[4mm] =&\,1-\Phi\bigg(\frac{0-(-1)}{\sqrt{1.11}}\bigg)\\[4mm] =&\,0.171270\ldots\\[4mm] =&\,\boxed{0.171}\quad\text{(3 s.f.).} \end{align*}

CIE 9709 2022 June Paper 61 Q3

题目

解答

解法一

思路

答题过程