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CIE 9709 2023 November Paper 62 Q5

A Level / CIE / S2

CIE 9709 2023 Nov Paper 62 Paper · Question 5

题目

Problem

The random variable XX has probability density function, f\mathrm{f}, given by

f(x)={1x2a<x<b,0otherwise,\mathrm{f}(x)=\begin{cases} \dfrac1{x^2} & a<x<b,\\ 0 & \text{otherwise}, \end{cases}

where aa and bb are positive constants.

(a) It is given that E(X)=ln⁡2\mathrm{E}(X)=\ln 2.

Show that b=2ab=2a.

(3)

(b) Show that a=12a=\dfrac12.

(3)

(c) Find the median of XX.

(3)
题目中文翻译

随机变量 XX 的概率密度函数 f\mathrm{f} 为

f(x)={1x2a<x<b,0其他情况,\mathrm{f}(x)=\begin{cases} \dfrac1{x^2} & a<x<b,\\ 0 & \text{其他情况,} \end{cases}

其中 aa 和 bb 为正常数。

(a) 已知 E(X)=ln⁡2\mathrm{E}(X)=\ln 2。

证明 b=2ab=2a。

(3)

(b) 证明 a=12a=\dfrac12。

(3)

(c) 求 XX 的中位数。

(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

期望积分为对数b/a,再用已知对数2推出b等于2a。

答题过程

展开 E(X)= ∫abx1x2 dx= [ln⁡x]ab=ln⁡ba.\begin{align*} \mathrm{E}(X)=&\,\int_a^b x\frac1{x^2}\,\mathrm{d}x\\[4mm] =&\,[\ln x]_a^b=\ln\frac ba. \end{align*}

Since E(X)=ln⁡2\mathrm{E}(X)=\ln2 and a,b>0a,b>0,

ln⁡ba= ln⁡2⟹ba= 2⟹b= 2a,\begin{align*} \ln\frac ba=&\,\ln2\\[4mm] \Longrightarrow\quad\frac ba=&\,2\\[4mm] \Longrightarrow\quad b=&\,\boxed{2a}, \end{align*}

as required.

(b)

解法一

思路

展开

密度归一化并代入b等于2a,解出a。

答题过程

展开 ∫a2a1x2 dx= 1⟹[−1x]a2a= 1.\begin{align*} \int_a^{2a}\frac1{x^2}\,\mathrm{d}x=&\,1\\[4mm] \Longrightarrow\quad\bigg[-\frac1x\bigg]_a^{2a}=&\,1. \end{align*}

Thus

1a−12a=1⟹12a=1,\frac1a-\frac1{2a}=1 \quad\Longrightarrow\quad \frac1{2a}=1,

giving a=12\boxed{a=\frac12}, as required. Hence b=1b=1.

(c)

解法一

思路

展开

从支持下限1/2至中位数的概率为一半,积分求解。

答题过程

展开

Let mm be the median. Then

∫1/2m1x2 dx= 12⟹[−1x]1/2m= 12.\begin{align*} \int_{1/2}^m\frac1{x^2}\,\mathrm{d}x=&\,\frac12\\[4mm] \Longrightarrow\quad\bigg[-\frac1x\bigg]_{1/2}^m=&\,\frac12. \end{align*}

Hence 2−1m=122-\frac1m=\frac12, so m=23\boxed{m=\frac23}.