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CIE 9231 2024 June Paper 43 Q5

A Level / CIE / FS

CIE 9231 2024 June Paper 43 Paper · Question 5

题目

Problem

The continuous random variable XX has cumulative distribution function FF given by

F(x)={0,x<2,(x−2)212,2≤x<4,1−(8−x)224,4≤x≤8,1,x>8.F(x) = \begin{cases} 0, & x < 2, \\ \dfrac{(x - 2)^2}{12}, & 2 \leq x < 4, \\ 1 - \dfrac{(8 - x)^2}{24}, & 4 \leq x \leq 8, \\ 1, & x > 8. \end{cases}

(a) Sketch the graph of the probability density function of XX.

[3]

(b) Find E(X)E(X).

[3]

(c) Find the exact value of the interquartile range of XX.

[4]
题目中文翻译

连续随机变量 XX 的累积分布函数 FF 为

F(x)={0,x<2,(x−2)212,2≤x<4,1−(8−x)224,4≤x≤8,1,x>8.F(x) = \begin{cases} 0, & x < 2, \\ \dfrac{(x - 2)^2}{12}, & 2 \leq x < 4, \\ 1 - \dfrac{(8 - x)^2}{24}, & 4 \leq x \leq 8, \\ 1, & x > 8. \end{cases}

(a) 画出 XX 的概率密度函数的图像草图。

(3)

(b) 求 E(X)E(X)。

(3)

(c) 求 XX 的四分位距的精确值。

(4)
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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对累积分布函数分段求导。算出两条线段的端点以及它们在 x=4x=4 的共同高度。

答题过程

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On 2<x<42<x<4, the density is (x−2)/6(x-2)/6. On 4<x<84<x<8, it is (8−x)/12(8-x)/12. It is zero outside the support. The sketch consists of straight segments joining (2,0)(2,0) to (4,1/3)(4,1/3) and then to (8,0)(8,0).

(b)

解法一

思路

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对 xfX(x)xf_X(x) 分段积分,两段分别使用 (a) 的一次式。

答题过程

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The mean is

 E⁡[X]= 16∫24x(x−2) dx+112∫48x(8−x) dx= 16[x33−x2]24+ 112[4x2−x33]48= 109+329=143.\begin{align*} &\,\operatorname E[X]\\ =&\,\frac16\int_2^4 x(x-2)\,\mathrm{d}x+\frac1{12}\int_4^8 x(8-x)\,\mathrm{d}x\\ =&\,\frac16\Big[\frac{x^3}{3}-x^2\Big]_2^4\\ +&\,\frac1{12}\Big[4x^2-\frac{x^3}{3}\Big]_4^8\\ =&\,\frac{10}{9}+\frac{32}{9}=\frac{14}{3}. \end{align*}

(c)

解法一

思路

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FX(4)=1/3F_X(4)=1/3,因此两个四分位数落在不同分段。分别求解并相减。

答题过程

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The lower quartile satisfies (x−2)2/12=1/4(x-2)^2/12=1/4 on [2,4][2,4], giving 2+32+\sqrt3. The upper quartile satisfies 1−(8−x)2/24=3/41-(8-x)^2/24=3/4 on [4,8][4,8], giving 8−68-\sqrt6. The interquartile range is 6−6−36-\sqrt6-\sqrt3.