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CIE 9709 2024 June Paper 63 Q7

A Level / CIE / S2

CIE 9709 2024 June Paper 63 Paper · Question 7

题目

Problem

The independent random variables XX and YY have the distributions Po(1.9)\mathrm{Po}(1.9) and Po(2.2)\mathrm{Po}(2.2) respectively.

(a) Find P(X+Y<4)\mathrm{P}(X+Y<4).

(3)

(b) Find the probability that X=2X=2 given that X+Y<4X+Y<4.

(4)

(c) A sample of 6060 randomly chosen pairs of values of XX and YY is taken, and the value of X+YX+Y is calculated for each pair. The sample mean of these 6060 values is found.

Find the probability that the sample mean of X+YX+Y is less than 4.04.0.

(6)
题目中文翻译

相互独立的随机变量 XX 和 YY 分别服从 Po(1.9)\mathrm{Po}(1.9) 和 Po(2.2)\mathrm{Po}(2.2) 分布。

(a) 求 P(X+Y<4)\mathrm{P}(X+Y<4)。

(3)

(b) 已知 X+Y<4X+Y<4，求 X=2X=2 的概率。

(4)

(c) 随机抽取 6060 对 XX 和 YY 的取值作为样本，计算每对的 X+YX+Y，再求这 6060 个和的样本均值。

求 X+YX+Y 的样本均值小于 4.04.0 的概率。

(6)
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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独立泊松变量和的参数为4.1，少于4即不超过3。

答题过程

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By independence,

S=X+Y∼Po(1.9+2.2)=Po(4.1).S=X+Y\sim\mathrm{Po}(1.9+2.2)=\mathrm{Po}(4.1).

Hence

P(S<4)= e−4.1(1+4.1) +e−4.1(4.122!+4.133!)= 0.4141815…= 0.414(3 s.f.).\begin{align*} \mathrm{P}(S<4)=&\,e^{-4.1}(1+4.1)\\[4mm] &\,+e^{-4.1}\bigg(\frac{4.1^2}{2!}+\frac{4.1^3}{3!}\bigg)\\[4mm] =&\,0.4141815\ldots\\[4mm] =&\,\boxed{0.414}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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条件事件与X等于2同时发生时，Y只能为0或1。利用独立性求分子，再除以前问未舍入概率。

答题过程

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The event {X=2, X+Y<4}\{X=2,\ X+Y<4\} is {X=2, Y≤1}\{X=2,\ Y\leq1\}. By independence,

 P(X=2, X+Y<4)= (e−1.91.922!)e−2.2(1+2.2).\begin{align*} &\,\mathrm{P}(X=2,\ X+Y<4)\\[4mm] =&\,\bigg(e^{-1.9}\frac{1.9^2}{2!}\bigg)e^{-2.2}(1+2.2). \end{align*}

Thus

 P(X=2∣X+Y<4)= (e−1.91.922!)e−2.2(1+2.2)0.4141815…= 0.231115…= 0.231(3 s.f.).\begin{align*} &\,\mathrm{P}(X=2\mid X+Y<4)\\[4mm] =&\,\frac{\bigg(e^{-1.9}\frac{1.9^2}{2!}\bigg)e^{-2.2}(1+2.2)} {0.4141815\ldots}\\[4mm] =&\,0.231115\ldots\\[4mm] =&\,\boxed{0.231}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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把60个和的样本均值转成总数，总数服从参数246的泊松分布。均值严格小于4相当于整数总数不超过239，连续性修正边界为239.5。

答题过程

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Let Si=Xi+YiS_i=X_i+Y_i for the iith randomly chosen pair, and put T=∑i=160SiT=\sum_{i=1}^{60}S_i. Then

T∼ Po(60(4.1))=Po(246)≈ N(246,246).\begin{align*} T\sim&\,\mathrm{Po}(60(4.1))=\mathrm{Po}(246)\\[4mm] \approx&\,\mathrm{N}(246,246). \end{align*}

Since

S‾<4.0⟺ T<240⟺ T≤239,\begin{align*} \overline S<4.0\quad\Longleftrightarrow&\,T<240\\[4mm] \Longleftrightarrow&\,T\leq239, \end{align*}

a continuity correction gives

P(S‾<4.0)≈ Φ(239.5−246246)= 0.339281…= 0.339(3 s.f.).\begin{align*} \mathrm{P}(\overline S<4.0)\approx&\,\Phi\bigg(\frac{239.5-246}{\sqrt{246}}\bigg)\\[4mm] =&\,0.339281\ldots\\[4mm] =&\,\boxed{0.339}\quad\text{(3 s.f.).} \end{align*}