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CIE 9709 2022 June Paper 61 Q4

A Level / CIE / S2

CIE 9709 2022 June Paper 61 Paper · Question 4

题目

Problem

The independent random variables XX and YY have distributions Po(2)\mathrm{Po}(2) and B(20,14)\mathrm{B}\left(20,\dfrac14\right) respectively.

(a) Find the mean and standard deviation of X−3YX-3Y.

(5)

(b) Find P(Y=15X)\mathrm{P}(Y=15X).

(3)
题目中文翻译

相互独立的随机变量 XX 和 YY 分别服从 Po(2)\mathrm{Po}(2) 和 B(20,14)\mathrm{B}\left(20,\dfrac14\right) 分布。

(a) 求 X−3YX-3Y 的均值和标准差。

(5)

(b) 求 P(Y=15X)\mathrm{P}(Y=15X)。

(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

(a)

解法一

思路

展开

先求泊松与二项变量的均值方差。线性组合的均值带符号,独立方差按系数平方相加。

答题过程

展开

For X∼Po(2)X\sim\mathrm{Po}(2),

E(X)=Var(X)=2.\mathrm{E}(X)=\mathrm{Var}(X)=2.

For Y∼B(20,14)Y\sim\mathrm{B}(20,\frac14),

E(Y)= 5,Var(Y)= 20(14)(34)=154.\begin{align*} \mathrm{E}(Y)=&\,5,\\[4mm] \mathrm{Var}(Y)=&\,20\bigg(\frac14\bigg)\bigg(\frac34\bigg)=\frac{15}{4}. \end{align*}

Hence

E(X−3Y)=2−3(5)=−13.\mathrm{E}(X-3Y)=2-3(5)=\boxed{-13}.

By independence,

Var(X−3Y)=2+32154=1434,\mathrm{Var}(X-3Y)=2+3^2\frac{15}{4}=\frac{143}{4},

so its standard deviation is 1432=5.98\boxed{\frac{\sqrt{143}}2=5.98} (3 s.f.).

(b)

解法一

思路

展开

二项变量Y只能取0至20,因此Y等于15X只可能是X0且Y0,或X1且Y15。用独立性分别求联合概率后相加。

答题过程

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Since 0≤Y≤200\leq Y\leq20, the only possible pairs (X,Y)(X,Y) satisfying Y=15XY=15X are (0,0)(0,0) and (1,15)(1,15). By independence,

P(Y=15X)= P(X=0)P(Y=0) +P(X=1)P(Y=15)= e−2(34)20 +2e−2(2015)(14)15(34)5= 0.000430104…= 0.000430(3 s.f.).\begin{align*} \mathrm{P}(Y=15X)=&\,\mathrm{P}(X=0)\mathrm{P}(Y=0)\\[4mm] &\,+\mathrm{P}(X=1)\mathrm{P}(Y=15)\\[4mm] =&\,e^{-2}\bigg(\frac34\bigg)^{20}\\[4mm] &\,+2e^{-2}\binom{20}{15}\bigg(\frac14\bigg)^{15}\bigg(\frac34\bigg)^5\\[4mm] =&\,0.000430104\ldots\\[4mm] =&\,\boxed{0.000430}\quad\text{(3 s.f.).} \end{align*}