Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2024 June Paper 62 Q4

A Level / CIE / S2

CIE 9709 2024 June Paper 62 Paper · Question 4

题目

Problem

A random variable XX has the distribution N(10,12)\mathrm{N}(10,12). Two independent values of XX, denoted by X1X_1 and X2X_2, are chosen at random.

(a) Write down the value of P(X1>X2)\mathrm{P}(X_1>X_2).

(1)

(b) Find P(X1>2X2−3)\mathrm{P}(X_1>2X_2-3).

(5)
题目中文翻译

随机变量 XX 服从 N(10,12)\mathrm{N}(10,12) 分布。从 XX 中独立地随机选取两个值,分别记为 X1X_1 和 X2X_2。

(a) 写出 P(X1>X2)\mathrm{P}(X_1>X_2) 的值。

(1)

(b) 求 P(X1>2X2−3)\mathrm{P}(X_1>2X_2-3)。

(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

展开

两个独立同分布连续变量,任一个较大的概率由对称性均为一半。

答题过程

展开

By symmetry, since the variables are identically distributed and continuous,

P(X1>X2)=12.\boxed{\mathrm{P}(X_1>X_2)=\frac12}.

(b)

解法一

思路

展开

把不等式移为X1减2X2加3大于0,计算这个正态变量的均值与方差后标准化。

答题过程

展开

Let D=X1−2X2+3D=X_1-2X_2+3. By independence,

D∼ N(10−2(10)+3, 12+22(12))= N(−7,60).\begin{align*} D\sim&\,\mathrm{N}\bigl(10-2(10)+3,\ 12+2^2(12)\bigr)\\[4mm] =&\,\mathrm{N}(-7,60). \end{align*}

Therefore

 P(X1>2X2−3)= P(D>0)= 1−Φ(760)= 0.183078…= 0.183(3 s.f.).\begin{align*} &\,\mathrm{P}(X_1>2X_2-3)\\[4mm] =&\,\mathrm{P}(D>0)\\[4mm] =&\,1-\Phi\bigg(\frac7{\sqrt{60}}\bigg)\\[4mm] =&\,0.183078\ldots\\[4mm] =&\,\boxed{0.183}\quad\text{(3 s.f.).} \end{align*}