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S2 2026 Jan A Q6

A Level / Edexcel / S2

IAL 2026 Jan A Paper · Question 6

题目

Problem

The continuous random variable XX has probability density function f(x)f(x) given by

f(x)={x320,0⩽x⩽2,6−x10,2<x⩽6,0,otherwise.f(x)=\begin{cases} \begin{aligned}\dfrac{x^3}{20},\\[-2pt]0\leqslant x\leqslant2,\end{aligned}\\[6pt] \begin{aligned}\dfrac{6-x}{10},\\[-2pt]2<x\leqslant6,\end{aligned}\\[6pt] \begin{aligned}0,\\[-2pt]\text{otherwise}.\end{aligned} \end{cases}

(a) Sketch the graph of f(x)f(x) for all values of xx

(3)

(b) Write down the mode of XX

(1)

(c) Show that P(X>2)=0.8P(X > 2) = 0.8

(2)

(d) Define fully the cumulative distribution function F(x)F(x)

(4)

Given that

P(X<a∣X>2)=58P(X<a\mid X>2)=\dfrac58

(e) find the value of F(a)F(a)

(2)

(f) Hence, or otherwise, find the value of aa. Give your answer to 3 significant figures.

(3)

(Total for Question 6 is 15 marks)

题目中文翻译

连续随机变量 XX 的概率密度函数 f(x)f(x) 为

f(x)={x320,0⩽x⩽2,6−x10,2<x⩽6,0,其他情况.f(x)=\begin{cases} \begin{aligned}\dfrac{x^3}{20},\\[-2pt]0\leqslant x\leqslant2,\end{aligned}\\[6pt] \begin{aligned}\dfrac{6-x}{10},\\[-2pt]2<x\leqslant6,\end{aligned}\\[6pt] \begin{aligned}0,\\[-2pt]\text{其他情况}.\end{aligned} \end{cases}

(a) 画出 f(x)f(x) 对所有 xx 值的图像。

(3)

(b) 写出 XX 的众数。

(1)

(c) 证明 P(X>2)=0.8P(X > 2) = 0.8。

(2)

(d) 完整定义累积分布函数 F(x)F(x)。

(4)

已知

P(X<a∣X>2)=58P(X<a\mid X>2)=\dfrac58

(e) 求 F(a)F(a) 的值。

(2)

(f) 由此或其他方法,求 aa 的值。答案保留 3 位有效数字。

(3)

(第 6 题共 15 分)

解答

(a)

解法一

思路

展开

在 0≤x≤20\leq x\leq2 上,图像是从原点上升的三次曲线,并到达 (2,0.4)(2,0.4);在 2<x≤62<x\leq6 上,图像是从 (2,0.4)(2,0.4) 下降到 (6,0)(6,0) 的直线。其余位置的密度为零。

答题过程

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For 0≤x≤20\leq x\leq2, the graph is the increasing cubic

f(x)=x320,f(x)=\frac{x^3}{20},

from (0,0)(0,0) to (2,0.4)(2,0.4). For 2<x≤62<x\leq6, it is the straight line

f(x)=6−x10,f(x)=\frac{6-x}{10},

from (2,0.4)(2,0.4) to (6,0)(6,0). Also, f(x)=0f(x)=0 outside [0,6][0,6].

(b)

解法一

思路

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众数是概率密度达到最大值的位置。图像在 x=2x=2 处达到高度 0.40.4,两侧都从该点向下,因此众数为 22。

答题过程

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The maximum value of the probability density occurs at x=2x=2. Therefore,

Mode⁡(X)=2.\boxed{\operatorname{Mode}(X)=2}.

(c)

解法一

思路

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X>2X>2 对应密度图像右侧的区域。将第二段密度积分,写出原函数并代入两个端点,即可求出这部分概率。

答题过程

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Using the second part of the density function,

 P(X>2)= ∫266−x10 dx= 110[6x−x22]26= 110[(36−18) −(12−2)]= 18−1010= 0.8,\begin{align*} &\,P(X>2)\\[4mm] =&\,\int_2^6\frac{6-x}{10}\,\mathrm{d}x\\ =&\,\frac1{10}\Big[6x-\frac{x^2}{2}\Big]_2^6\\[4mm] =&\,\frac1{10}\big[(36-18)\\[4mm] &\,\hspace{2pt}-(12-2)\big]\\[4mm] =&\,\frac{18-10}{10}\\[4mm] =&\,0.8, \end{align*}

as required.

解法二

思路

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右侧密度是一条直线,下面的区域是底为 6−26-2、高为 0.40.4 的三角形,因此可直接用面积求概率。

答题过程

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The region under the density for 2<X<62<X<6 is a triangle. Hence,

P(X>2)= 12(6−2)(0.4)= 0.8,\begin{align*} P(X>2)=&\,\frac12(6-2)(0.4)\\[4mm] =&\,0.8, \end{align*}

as required.

解法三

思路

展开

也可先积分第一段密度求 P(X≤2)P(X\leq2),再从总概率 11 中减去它。

答题过程

展开  P(X>2)= 1−∫02x320 dx= 1−[x480]02= 1−16−080= 0.8,\begin{align*} &\,P(X>2)\\[4mm] =&\,1-\int_0^2\frac{x^3}{20}\,\mathrm{d}x\\[4mm] =&\,1-\Big[\frac{x^4}{80}\Big]_0^2\\[4mm] =&\,1-\frac{16-0}{80}\\[4mm] =&\,0.8, \end{align*}

as required.

(d)

解法一

思路

展开

CDF 定义为从负无穷累积到 xx 的概率。依密度函数的分段点 0,2,60,2,6 分成四段:中间两段分别积分;在支撑集左侧为 00,右侧为 11。

答题过程

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For 0≤x≤20\leq x\leq2,

F(x)= ∫0xt320 dt= [t480]0x= x4−080= x480.\begin{align*} F(x) =&\,\int_0^x\frac{t^3}{20}\,\mathrm{d}t\\ =&\,\Big[\frac{t^4}{80}\Big]_0^x\\[4mm] =&\,\frac{x^4-0}{80}\\[4mm] =&\,\frac{x^4}{80}. \end{align*}

In particular, F(2)=1680=0.2F(2)=\frac{16}{80}=0.2. For 2<x≤62<x\leq6,

 F(x)= F(2)+∫2x6−t10 dt= 15+110[6t−t22]2x= 15+110(6x−x22−10)= 12x−x2−1620= 20−(x2−12x+36)20= 1−(6−x)220.\begin{align*} &\,F(x)\\[4mm] =&\,F(2)+\int_2^x \frac{6-t}{10}\,\mathrm{d}t\\ =&\,\frac15+\frac1{10}\Big[6t-\frac{t^2}{2}\Big]_2^x\\[4mm] =&\,\frac15+\frac1{10}\Big(6x-\frac{x^2}{2}-10\Big)\\[4mm] =&\,\frac{12x-x^2-16}{20}\\[4mm] =&\,\frac{20-(x^2-12x+36)}{20}\\[4mm] =&\,1-\frac{(6-x)^2}{20}. \end{align*}

Therefore, the cumulative distribution function is

 F(x)= {0,x<0,x480,0≤x≤2,1−(6−x)220,2<x≤6,1,x>6.\begin{align*} &\,F(x)=\\[4mm] &\,\begin{cases} \begin{aligned}0,\\[-2pt]x<0,\end{aligned}\\[6pt] \begin{aligned}\dfrac{x^4}{80},\\[-2pt]0\leq x\leq2,\end{aligned}\\[6pt] \begin{aligned}1-\dfrac{(6-x)^2}{20},\\[-2pt]2<x\leq6,\end{aligned}\\[6pt] \begin{aligned}1,\\[-2pt]x>6.\end{aligned} \end{cases} \end{align*}

解法二

思路

展开

分别对两段密度求不定积分,再用 CDF 在支撑区间两端的值确定积分常数。第一段用 F(0)=0F(0)=0,第二段用 F(6)=1F(6)=1。

答题过程

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For 0≤x≤20\leq x\leq2,

F(x)= ∫x320 dx= x480+C1.\begin{align*} F(x)=&\,\int\frac{x^3}{20}\,\mathrm{d}x\\[4mm] =&\,\frac{x^4}{80}+C_1. \end{align*}

Since F(0)=0F(0)=0, C1=0C_1=0. Thus F(x)=x4/80F(x)=x^4/80 on this interval.

For 2<x≤62<x\leq6,

F(x)= ∫6−x10 dx= −(6−x)220+C2.\begin{align*} F(x)=&\,\int\frac{6-x}{10}\,\mathrm{d}x\\[4mm] =&\,-\frac{(6-x)^2}{20}+C_2. \end{align*}

Since F(6)=1F(6)=1,

1=−(6−6)220+C2,1=-\frac{(6-6)^2}{20}+C_2,

so C2=1C_2=1. Hence,

F(x)=1−(6−x)220.F(x)=1-\frac{(6-x)^2}{20}.

At x=2x=2, both expressions give F(2)=1/5F(2)=1/5, so the pieces agree. Together with F(x)=0F(x)=0 for x<0x<0 and F(x)=1F(x)=1 for x>6x>6, this gives the full CDF above.

(e)

解法一

思路

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条件 X>2X>2 已知其概率为 0.80.8。事件 2<X<a2<X<a 的概率是 F(a)−F(2)F(a)-F(2),所以把条件概率写成分式,并使用 (d) 中的 F(2)=0.2F(2)=0.2 求 F(a)F(a)。

答题过程

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Since P(X>2)=0.8P(X>2)=0.8 and F(2)=0.2F(2)=0.2,

 P(X<a∣X>2)= P(2<X<a)P(X>2)= F(a)−0.20.8.\begin{align*} &\,P(X<a\mid X>2)\\[4mm] =&\,\frac{P(2<X<a)}{P(X>2)}\\ =&\,\frac{F(a)-0.2}{0.8}. \end{align*}

Therefore,

F(a)−0.20.8= 58,F(a)−0.2= 0.5.\begin{align*} \frac{F(a)-0.2}{0.8} =&\,\frac58,\\ F(a)-0.2 =&\,0.5. \end{align*}

Hence,

F(a)=0.7.\boxed{F(a)=0.7}.

(f)

解法一

思路

展开

由 (e) 得 F(a)=0.7F(a)=0.7。由于 F(2)=0.2F(2)=0.2、F(6)=1F(6)=1,可知 2<a<62<a<6,因此使用 CDF 的第三段。解平方方程时,6−a>06-a>0,所以只取正平方根。

答题过程

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Since 2<a<62<a<6, use the third part of the cumulative distribution function:

1−(6−a)220=0.7.1-\frac{(6-a)^2}{20}=0.7.

Thus,

(6−a)2= 6,6−a= 6,\begin{align*} (6-a)^2 =&\,6,\\ 6-a =&\,\sqrt6, \end{align*}

because 6−a>06-a>0. Therefore,

a= 6−6= 3.550510…≈ 3.55.\begin{align*} a =&\,6-\sqrt6\\ =&\,3.550510\ldots\\ \approx&\,\boxed{3.55} . \end{align*}

The answer is given to 3 significant figures.

解法二

思路

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由 F(a)=0.7F(a)=0.7,可知右侧尾概率为 0.30.3。在 2<a<62<a<6 内,尾部图形是底为 6−a6-a、高为 (6−a)/10(6-a)/10 的三角形,用面积建立方程。

答题过程

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From part (e),

P(X>a)= 1−F(a)= 0.3.\begin{align*} P(X>a)=&\,1-F(a)\\[4mm] =&\,0.3. \end{align*}

Since F(2)=0.2F(2)=0.2, F(a)=0.7F(a)=0.7 and F(6)=1F(6)=1, we have 2<a<62<a<6. The triangular tail has base 6−a6-a and height f(a)=(6−a)/10f(a)=(6-a)/10. Therefore,

12(6−a)6−a10= 0.3,(6−a)2= 6.\begin{align*} \frac12(6-a)\frac{6-a}{10}=&\,0.3,\\[4mm] (6-a)^2=&\,6. \end{align*}

As 6−a>06-a>0, we take the positive square root:

6−a= 6,a= 6−6≈ 3.55.\begin{align*} 6-a=&\,\sqrt6,\\[4mm] a=&\,6-\sqrt6\\[4mm] \approx&\,\boxed{3.55}. \end{align*}

The answer is given to 3 significant figures.

解法三

思路

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尾部小三角形与 22 到 66 下面的整个三角形相似。面积比等于对应边长比的平方;大三角形面积是 (c) 的 0.80.8,小三角形面积是 1−F(a)=0.31-F(a)=0.3。

答题过程

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The triangular tail for a<X<6a<X<6 is similar to the triangle for 2<X<62<X<6. Their respective areas are 0.30.3 and 0.80.8, while their bases are 6−a6-a and 6−2=46-2=4.

Hence,

(6−a4)2= 0.30.8,(6−a)2= 16(38)= 6.\begin{align*} \Big(\frac{6-a}{4}\Big)^2=&\,\frac{0.3}{0.8},\\[4mm] (6-a)^2=&\,16\Big(\frac38\Big)\\[4mm] =&\,6. \end{align*}

Since 2<a<62<a<6, 6−a>06-a>0. Thus,

a=6−6≈3.55,a=6-\sqrt6\approx\boxed{3.55},

to 3 significant figures.