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CIE 9231 2022 June Paper 43 Q4

A Level / CIE / FS

CIE 9231 2022 June Paper 43 Paper · Question 4

题目

Problem

The continuous random variable XX has probability density function f given by

f(x)={38(1+1x2),1≤x≤3,0,otherwise.f(x) = \begin{cases} \dfrac{3}{8}\bigg(1 + \dfrac{1}{x^2}\bigg), & 1 \leq x \leq 3,\\ 0, & \text{otherwise.} \end{cases}

(a) Find E(X)\mathrm{E}(\sqrt{X}).

[3]

The random variable YY is given by Y=X2Y = X^2.

(b) Find the probability density function of YY.

[4]

(c) Find the 40th percentile of YY.

[3]
题目中文翻译

连续随机变量 XX 的概率密度函数 f 定义如下:

f(x)={38(1+1x2),1≤x≤3,0,otherwise.f(x) = \begin{cases} \dfrac{3}{8}\bigg(1 + \dfrac{1}{x^2}\bigg), & 1 \leq x \leq 3,\\ 0, & \text{otherwise.} \end{cases}

(a) 求 E(X)\mathrm{E}(\sqrt{X})。

(3)

随机变量 YY 定义为 Y=X2Y = X^2。

(b) 求 YY 的概率密度函数。

(4)

(c) 求 YY 的第 40 百分位数。

(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

将 x\sqrt x 乘密度后化成两个幂函数,逐项积分。

答题过程

展开

The expectation is

 E⁡[X]= 38∫13(x1/2+x−3/2) dx= 38[23x3/2−2x−1/2]13= 1+32.\begin{align*} &\,\operatorname E[\sqrt X]\\ =&\,\frac38\int_1^3(x^{1/2}+x^{-3/2})\,\mathrm{d}x\\ =&\,\frac38\Big[\frac23x^{3/2}-2x^{-1/2}\Big]_1^3\\ =&\,\frac{1+\sqrt3}{2}. \end{align*}

(b)

解法一

思路

展开

平方在支撑区间上递增。将 XX 的 CDF 代入 y\sqrt y,再求导得到 YY 的密度。

答题过程

展开

For 1≤x≤31\le x\le3, the CDF of XX is 38[u−1/u]1x=38(x−1/x)\frac38[u-1/u]_{1}^{x}=\frac38(x-1/x). Thus on 1≤y≤91\le y\le9,

FY(y)=FX(y)=38(y1/2−y−1/2).F_Y(y)=F_X(\sqrt y)=\frac38(y^{1/2}-y^{-1/2}).

The CDF is constant outside this interval. Therefore a density is

fY(y)={316(y−1/2+y−3/2),1≤y≤9,0,otherwise.f_Y(y)= \begin{cases} \frac3{16}(y^{-1/2}+y^{-3/2}), & 1\le y\le9,\\ 0, & \text{otherwise}. \end{cases}

(c)

解法一

思路

展开

将 YY 的累计概率设为 0.40.4。用 u=yu=\sqrt y 把所得方程化成二次方程。

答题过程

展开

For the percentile yy, 38(u−1/u)=0.4\frac38(u-1/u)=0.4 with u=y∈[1,3]u=\sqrt y\in[1,3]. Hence 15u2−16u−15=(3u−5)(5u+3)=015u^2-16u-15=(3u-5)(5u+3)=0. Only u=5/3u=5/3 is valid, so the fortieth percentile is y=25/9y=25/9.