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CIE 9709 2025 November Paper 63 Q7

A Level / CIE / S2

CIE 9709 2025 Nov Paper 63 Paper · Question 7

题目

Problem

The time, in minutes, taken by students to complete a test is modelled by the random variable XX with probability density function

f(x)={−34(x−3)(x−5)3≤x≤5,0otherwise.f(x)= \begin{cases} -\frac34(x-3)(x-5) & 3\le x\le 5,\\ 0 & \text{otherwise}. \end{cases}

(a) Find the probability that a randomly chosen student takes longer than 4.54.5 minutes to complete the test.

(4)

(b) Write down the median of XX.

(1)

(c) Without performing an integration, use your answer to part (a) to find P(3.5<X<4.5)\mathrm{P}(3.5<X<4.5).

(2)
题目中文翻译

用随机变量 XX 对学生完成一项测试所需的时间建模,时间以分钟为单位,概率密度函数为

f(x)={−34(x−3)(x−5)3≤x≤5,0其他情况.f(x)= \begin{cases} -\frac34(x-3)(x-5) & 3\le x\le 5,\\ 0 & \text{其他情况}. \end{cases}

(a) 求随机选取的一名学生用超过 4.54.5 分钟完成测试的概率。

(4)

(b) 写出 XX 的中位数。

(1)

(c) 不进行积分,用你对 (a) 的答案求 P(3.5<X<4.5)\mathrm{P}(3.5<X<4.5)。

(2)
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

按密度在指定右尾区间积分,先展开因式,并用精确分数代入上下限。

答题过程

展开

Expanding the density,

f(x)=−34(x2−8x+15).f(x)=-\frac34(x^2-8x+15).

This applies for 3≤x≤53\le x\le5.

Therefore

P(X>4.5)= −34∫9/25(x2−8x+15) dx= −34[x33−4x2+15x]9/25= −34(503−1358)= −34(−524)=532.\begin{align*} \mathrm P(X>4.5) =&\,-\frac34\int_{9/2}^{5}(x^2-8x+15)\,\mathrm{d}x\\[4mm] =&\,-\frac34\bigg[\frac{x^3}3-4x^2+15x\bigg]_{9/2}^{5}\\[4mm] =&\,-\frac34\bigg(\frac{50}3-\frac{135}8\bigg)\\[4mm] =&\,-\frac34\bigg(-\frac5{24}\bigg)=\boxed{\frac5{32}}. \end{align*}

(b)

解法一

思路

展开

密度关于支持区间中点 44 对称,所以中位数直接读出,无须再积分。

答题过程

展开

The density is symmetric about x=4x=4, so median=4\boxed{\text{median}=4}.

(c)

解法一

思路

展开

按题目指示不再积分,使用对称性把左尾概率与(a)的右尾概率对应,再从一减去两端。

答题过程

展开

By symmetry about x=4x=4,

P(X<3.5)=P(X>4.5)=532.\mathrm P(X<3.5)=\mathrm P(X>4.5)=\frac5{32}.

Using part (a), without further integration,

P(3.5<X<4.5)= 1−2(532)= 1116.\begin{align*} \mathrm P(3.5<X<4.5)=&\,1-2\bigg(\frac5{32}\bigg)\\[4mm] =&\,\boxed{\frac{11}{16}}. \end{align*}