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CIE 9709 2025 November Paper 62 Q2

A Level / CIE / S2

CIE 9709 2025 Nov Paper 62 Paper · Question 2

题目

Problem

The random variable XX has a normal distribution with mean 1010 and standard deviation 33. The independent random variable YY has a Poisson distribution with mean 44.

(a) Find the standard deviation of X+YX+Y.

(3)

(b) Find the standard deviation of 5X−Y5X-Y.

(3)
题目中文翻译

随机变量 XX 服从均值为 1010、标准差为 33 的正态分布。与 XX 相互独立的随机变量 YY 服从均值为 44 的泊松分布。

(a) 求 X+YX+Y 的标准差。

(3)

(b) 求 5X−Y5X-Y 的标准差。

(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

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求方差不要求两变量属于同一类分布;利用独立性,正态变量和 Poisson 变量的方差仍可相加。

答题过程

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Since XX has standard deviation 33, Var⁡(X)=9\operatorname{Var}(X)=9. Since Y∼Po(4)Y\sim\mathrm{Po}(4), Var⁡(Y)=4\operatorname{Var}(Y)=4.

By independence,

Var⁡(X+Y)=9+4=13.\operatorname{Var}(X+Y)=9+4=13.

Therefore sd⁡(X+Y)=13≈3.61\boxed{\operatorname{sd}(X+Y)=\sqrt{13}\approx3.61}.

(b)

解法一

思路

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线性组合的方差中系数要平方,减去独立变量也会增加总方差。

答题过程

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Using independence,

 Var⁡(5X−Y)= 52Var⁡(X) +(−1)2Var⁡(Y)= 25(9)+4=229.\begin{align*} &\,\operatorname{Var}(5X-Y)\\[4mm] =&\,5^2\operatorname{Var}(X)\\[4mm] &\,+(-1)^2\operatorname{Var}(Y)\\[4mm] =&\,25(9)+4=229. \end{align*}

Thus sd⁡(5X−Y)=229≈15.1\boxed{\operatorname{sd}(5X-Y)=\sqrt{229}\approx15.1}.