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CIE 9709 2023 June Paper 63 Q7

A Level / CIE / S2

CIE 9709 2023 June Paper 63 Paper · Question 7

题目

Problem

A random variable XX has probability density function f\mathrm{f}, where the graph of y=f(x)y=\mathrm{f}(x) is a semicircle with centre (0,0)(0,0) and radius 2π\sqrt{\dfrac2\pi}, entirely above the xx-axis. Elsewhere f(x)=0\mathrm{f}(x)=0 (see diagram).

(a) Verify that f\mathrm{f} can be a probability density function.

(2)

AA and BB are the points where the line x=1πx=\sqrt{\dfrac1\pi} meets the xx-axis and the semicircle respectively.

(b) Show that angle AOBAOB is 14π\dfrac14\pi radians and hence find P(X>1π)\mathrm{P}\left(X>\sqrt{\dfrac1\pi}\right).

(6)
题目中文翻译

随机变量 XX 的概率密度函数为 f\mathrm{f}。y=f(x)y=\mathrm{f}(x) 的图像是一个整个位于 xx 轴上方的半圆，圆心为 (0,0)(0,0)，半径为 2π\sqrt{\dfrac2\pi}。在其余各处，f(x)=0\mathrm{f}(x)=0（见图）。

(a) 验证 f\mathrm{f} 可以是一个概率密度函数。

(2)

直线 x=1πx=\sqrt{\dfrac1\pi} 与 xx 轴和半圆的交点分别为 AA 和 BB。

(b) 证明角 AOBAOB 为 14π\dfrac14\pi 弧度，并由此求 P(X>1π)\mathrm{P}\left(X>\sqrt{\dfrac1\pi}\right)。

(6)
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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验证密度非负，且半圆总面积等于1。

答题过程

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The semicircle lies above the xx-axis, so f(x)≥0f(x)\geq0. Its area is

12π(2π)2=1.\frac12\pi\bigg(\sqrt{\frac2\pi}\bigg)^2=1.

Thus ff satisfies the two requirements for a probability density function.

(b)

解法一

思路

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由直角三角形求角度，再用扇形面积减去三角形面积，得到右侧圆弓形的概率。

答题过程

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In the right-angled triangle OABOAB,

cos⁡∠AOB= OAOB= 1/π2/π=12.\begin{align*} \cos\angle AOB=&\,\frac{OA}{OB}\\[4mm] =&\,\frac{\sqrt{1/\pi}}{\sqrt{2/\pi}}=\frac1{\sqrt2}. \end{align*}

Since ∠AOB\angle AOB is acute,

∠AOB=π4.\boxed{\angle AOB=\frac\pi4}.

Also,

AB=OB2−OA2=2π−1π=1π.AB=\sqrt{OB^2-OA^2} =\sqrt{\frac2\pi-\frac1\pi} =\frac1{\sqrt\pi}.

The region to the right of x=1πx=\frac1{\sqrt\pi} is the sector with angle π4\frac\pi4 minus triangle OABOAB. Therefore

 P(X>1π)= 12(2π)π4−12(1π)2= 14−12π= 0.0908(3 s.f.).\begin{align*} &\,\mathrm{P}\bigg(X>\frac1{\sqrt\pi}\bigg)\\[4mm] =&\,\frac12\bigg(\frac2\pi\bigg)\frac\pi4 -\frac12\bigg(\frac1{\sqrt\pi}\bigg)^2\\[4mm] =&\,\boxed{\frac14-\frac1{2\pi}}\\[4mm] =&\,0.0908\quad\text{(3 s.f.).} \end{align*}

解法二

思路

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先证明题目要求的角度，再从圆方程写出密度。以正弦代换求圆弓形面积的积分，展示与几何法不同的计算路线。

答题过程

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As OA/OB=1/2OA/OB=1/\sqrt2 and ∠AOB\angle AOB is acute, ∠AOB=π4\angle AOB=\frac\pi4.

Put R=2/πR=\sqrt{2/\pi}. The upper semicircle has equation

f(x)=R2−x2.f(x)=\sqrt{R^2-x^2}.

Hence

P(X>1π)=∫1/πRR2−x2 dx.\mathrm{P}\bigg(X>\frac1{\sqrt\pi}\bigg) =\int_{1/\sqrt\pi}^R\sqrt{R^2-x^2}\,\mathrm{d}x.

Use x=Rsin⁡θx=R\sin\theta. The limits become θ=π4\theta=\frac\pi4 and θ=π2\theta=\frac\pi2, and dx=Rcos⁡θ dθ\mathrm{d}x=R\cos\theta\,\mathrm{d}\theta. Therefore

 P(X>1π)= R2∫π/4π/2cos⁡2θ dθ= 2π[θ2+sin⁡2θ4]π/4π/2= 2π(π4−π8−14)= 14−12π.\begin{align*} &\,\mathrm{P}\bigg(X>\frac1{\sqrt\pi}\bigg)\\[4mm] =&\,R^2\int_{\pi/4}^{\pi/2}\cos^2\theta\,\mathrm{d}\theta\\[4mm] =&\,\frac2\pi\bigg[\frac\theta2+\frac{\sin2\theta}{4}\bigg]_{\pi/4}^{\pi/2}\\[4mm] =&\,\frac2\pi\bigg(\frac\pi4-\frac\pi8-\frac14\bigg)\\[4mm] =&\,\boxed{\frac14-\frac1{2\pi}}. \end{align*}