CalcGospel 國際數學圖譜
返回

CIE 9709 2023 June Paper 61 Q2

A Level / CIE / S2

CIE 9709 2023 June Paper 61 Paper · Question 2

题目

Problem

(a)

The graph of the function f\mathrm{f} is a straight line segment from (0,0)(0,0) to (2,1)(2,1).

Show that f\mathrm{f} could be a probability density function.

(2)

(b)

The graph of the function g\mathrm{g} is a semicircle, centre (0,0)(0,0), entirely above the xx-axis.

Given that g\mathrm{g} is a probability density function, find the radius of the semicircle.

(2)

(c)

The time, XX minutes, taken by a large number of students to complete a test has probability density function h\mathrm{h}, as shown in the diagram.

(i) Without calculation, use the diagram to explain how you can tell that the median time is less than 1515 minutes.

(1)

It is now given that

h(x)={40x2−11010≤x≤20,0otherwise.\mathrm{h}(x)=\begin{cases} \dfrac{40}{x^2}-\dfrac1{10} & 10\leq x\leq20,\\ 0 & \text{otherwise}. \end{cases}

(ii) Find the mean time.

(3)
题目中文翻译

(a) 函数 f\mathrm{f} 的图像是从 (0,0)(0,0) 到 (2,1)(2,1) 的直线段。

说明 f\mathrm{f} 可以是一个概率密度函数。

(2)

(b) 函数 g\mathrm{g} 的图像是以 (0,0)(0,0) 为圆心、整个位于 xx 轴上方的半圆。

已知 g\mathrm{g} 是一个概率密度函数，求半圆的半径。

(2)

(c) 有很多学生完成了一次测试，所用的时间记为 XX 分钟，其概率密度函数 h\mathrm{h} 如图所示。

(i) 不进行计算，利用图像说明如何判断用时的中位数小于 1515 分钟。

(1)

现已知

h(x)={40x2−11010≤x≤20,0其他情况。\mathrm{h}(x)=\begin{cases} \dfrac{40}{x^2}-\dfrac1{10} & 10\leq x\leq20,\\ 0 & \text{其他情况。} \end{cases}

(ii) 求平均用时。

(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

作为密度，函数须非负且图下总面积为1；用三角形面积验证。

答题过程

展开

The graph is non-negative throughout its support. The total area under it is

12(2)(1)=1.\frac12(2)(1)=1.

Therefore ff could be a probability density function, taking f(x)=0f(x)=0 outside [0,2][0,2].

(b)

解法一

思路

展开

密度下的半圆面积为1，求正半径。

答题过程

展开

Let the radius be r>0r>0. Since the total probability is 1,

12πr2=1⟹r=2π.\frac12\pi r^2=1 \quad\Longrightarrow\quad \boxed{r=\sqrt{\frac2\pi}}.

(c)(i)

解法一

思路

展开

图上15左侧面积大于右侧面积，因此累计概率在15前已超过一半，中位数较小。

答题过程

展开

The area under the curve to the left of 15 is greater than half the total area. Therefore the point with half the probability to its left, the median, is less than 15 minutes.

(c)(ii)

解法一

思路

展开

期望为x乘密度的积分，分别积分40/x和负x/10，代入10与20。

答题过程

展开 E(X)= ∫1020x(40x2−110) dx= [40ln⁡x−x220]1020= 40ln⁡2010−400−10020= 40ln⁡2−15= 12.7 minutes(3 s.f.).\begin{align*} \mathrm{E}(X)=&\,\int_{10}^{20}x\bigg(\frac{40}{x^2}-\frac1{10}\bigg)\,\mathrm{d}x\\[4mm] =&\,\bigg[40\ln x-\frac{x^2}{20}\bigg]_{10}^{20}\\[4mm] =&\,40\ln\frac{20}{10}-\frac{400-100}{20}\\[4mm] =&\,40\ln2-15\\[4mm] =&\,\boxed{12.7\text{ minutes}}\quad\text{(3 s.f.).} \end{align*}