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CIE 9231 2023 Nov Paper 42 Q4

A Level / CIE / FS

CIE 9231 2023 Nov Paper 42 Paper · Question 4

题目

Problem

The diagram shows the continuous random variable XX with probability density function ff given by

f(x)={1128(4ax−bx3),0≤x≤4,c,4≤x≤6,0,otherwise,f(x)= \begin{cases} \frac{1}{128}(4ax-bx^3), & 0 \le x \le 4,\\ c, & 4 \le x \le 6,\\ 0, & \text{otherwise}, \end{cases}

where aa, bb and cc are constants.

The upper quartile of XX is equal to 4.

(a) Show that c=18c=\frac{1}{8} and find the values of aa and bb.

[4]

(b) Find the exact value of the median of XX.

[3]

(c) Find E⁡(X)\operatorname{E}(\sqrt{X}), giving your answer correct to 2 decimal places.

[3]
题目中文翻译

图中显示连续随机变量 XX,其概率密度函数 ff 为

f(x)={1128(4ax−bx3),0≤x≤4,c,4≤x≤6,0,其他情况,f(x)= \begin{cases} \frac{1}{128}(4ax-bx^3), & 0 \le x \le 4,\\ c, & 4 \le x \le 6,\\ 0, & \text{其他情况}, \end{cases}

其中 aa、bb 和 cc 为常数。

XX 的上四分位数等于 4。

(a) 证明 c=18c=\frac{1}{8},并求 aa 和 bb 的值。

(4)

(b) 求 XX 的中位数的精确值。

(3)

(c) 求 E⁡(X)\operatorname{E}(\sqrt{X}),答案精确到 2 位小数。

(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

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上四分位数把右段的概率定为 1/41/4。再用左段的面积和两段在 x=4x=4 的接合条件求系数。

答题过程

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Since 44 is the upper quartile, the area from 44 to 66 is 1/41/4. It equals 2c2c, so c=1/8c=1/8. The area from 00 to 44 gives

 1128[2ax2−b4x4]04= 32a−64b128=34,\begin{align*} &\,\frac1{128}\Big[2ax^2-\frac b4x^4\Big]_0^4\\ =&\,\frac{32a-64b}{128}=\frac34, \end{align*}

so a−2b=3a-2b=3. The two density expressions agree at 44, giving (16a−64b)/128=1/8(16a-64b)/128=1/8, or a−4b=1a-4b=1. Subtraction yields b=1b=1 and a=5a=5.

(b)

解法一

思路

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中位数小于上四分位数 44,所以用左段积分。令 u=m2u=m^2 把方程降成二次方程。

答题过程

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The median mm is between 00 and 44. It satisfies

 1128∫0m(20x−x3) dx= 5m264−m4512=12.\begin{align*} &\,\frac1{128}\int_0^m(20x-x^3)\,\mathrm{d}x\\ =&\,\frac{5m^2}{64}-\frac{m^4}{512}=\frac12. \end{align*}

Thus m4−40m2+256=0m^4-40m^2+256=0, or (m2−8)(m2−32)=0(m^2-8)(m^2-32)=0. Only m2=8m^2=8 gives 0<m<40<m<4. Hence the exact median is m=22m=2\sqrt2.

(c)

解法一

思路

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在两段上分别积分 x\sqrt x 乘密度,最后将结果取到两位小数。

答题过程

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The expectation is

 E⁡[X]= 1128∫04(20x3/2−x7/2) dx+ 18∫46x1/2 dx= 1128[8x5/2−29x9/2]04+ 112[x3/2]46= 109+66−812= 49+62.\begin{align*} &\,\operatorname E[\sqrt X]\\ =&\,\frac1{128}\int_0^4(20x^{3/2}-x^{7/2})\,\mathrm{d}x\\ +&\,\frac18\int_4^6x^{1/2}\,\mathrm{d}x\\ =&\,\frac1{128}\Big[8x^{5/2}-\frac29x^{9/2}\Big]_0^4\\ +&\,\frac1{12}\Big[x^{3/2}\Big]_4^6\\ =&\,\frac{10}{9}+\frac{6\sqrt6-8}{12}\\ =&\,\frac49+\frac{\sqrt6}{2}. \end{align*}

This is 1.671.67 to two decimal places.