CalcGospel 國際數學圖譜
返回

S2 2025 Oct Q7

A Level / Edexcel / S2

IAL 2025 Oct Paper · Question 7

题目

Problem

The continuous random variable XX has probability density function

 f(x)= { 14x−164x3 0⩽x⩽4 0 otherwise\begin{align*} &\,f(x)=\\ &\,\begin{cases} \begin{aligned} &\,\dfrac{1}{4}x - \dfrac{1}{64}x^3\\[-1mm] &\,\quad 0 \leqslant x \leqslant 4 \end{aligned}\\[3mm] \begin{aligned} &\,0\\[-1mm] &\,\quad \text{otherwise} \end{aligned} \end{cases} \end{align*}

(a) Show that the median of XX is 2.165 to 3 decimal places.

(5)

The continuous random variable YY has probability density function

 f(y)= { 4255y3 1⩽y⩽4 0 otherwise\begin{align*} &\,f(y)=\\ &\,\begin{cases} \begin{aligned} &\,\dfrac{4}{255}y^3\\[-1mm] &\,\quad 1 \leqslant y \leqslant 4 \end{aligned}\\[3mm] \begin{aligned} &\,0\\[-1mm] &\,\quad \text{otherwise} \end{aligned} \end{cases} \end{align*}

(b) Show that Var(1Y)=667225\text{Var}\left(\dfrac{1}{Y}\right) = \dfrac{66}{7225}

(Solutions relying entirely on calculator technology are not acceptable.)

(5)

(c) Hence find Var(4−5Y)\text{Var}\left(4 - \dfrac{5}{Y}\right)

(2)

(Total for Question 7 is 12 marks)

题目中文翻译

连续随机变量 XX 的概率密度函数为

 f(x)= { 14x−164x3 0⩽x⩽4 0 其他情况\begin{align*} &\,f(x)=\\ &\,\begin{cases} \begin{aligned} &\,\dfrac{1}{4}x - \dfrac{1}{64}x^3\\[-1mm] &\,\quad 0 \leqslant x \leqslant 4 \end{aligned}\\[3mm] \begin{aligned} &\,0\\[-1mm] &\,\quad \text{其他情况} \end{aligned} \end{cases} \end{align*}

(a) 证明 XX 的中位数为 2.165（保留 3 位小数）。

(5)

连续随机变量 YY 的概率密度函数为

 f(y)= { 4255y3 1⩽y⩽4 0 其他情况\begin{align*} &\,f(y)=\\ &\,\begin{cases} \begin{aligned} &\,\dfrac{4}{255}y^3\\[-1mm] &\,\quad 1 \leqslant y \leqslant 4 \end{aligned}\\[3mm] \begin{aligned} &\,0\\[-1mm] &\,\quad \text{其他情况} \end{aligned} \end{cases} \end{align*}

(b) 证明 Var(1Y)=667225\text{Var}\left(\dfrac{1}{Y}\right) = \dfrac{66}{7225}。

（完全依赖计算器技术的解答不可接受。）

(5)

(c) 由此求 Var(4−5Y)\text{Var}\left(4 - \dfrac{5}{Y}\right)。

(2)

（第 7 题共 12 分）

解答

(a)

解法一

思路

展开

设中位数为 mm。连续分布的中位数满足 P(X≤m)=1/2P(X\leq m)=1/2，所以从 00 到 mm 积分密度并令结果等于 1/21/2。所得方程可视为关于 m2m^2 的二次方程；再根据支撑范围 0≤m≤40\leq m\leq4 舍去不合适的根。

答题过程

展开

Let the median be mm. Then

∫0m(x4−x364) dx=12.\int_0^m\left(\frac{x}{4}-\frac{x^3}{64}\right) \,\mathrm{d}x=\frac12.

Therefore,

[x28−x4256]0m= 12m28−m4256= 12m4−32m2+128= 0.\begin{align*} \left[\frac{x^2}{8}-\frac{x^4}{256}\right]_0^m =&\,\frac12\\ \frac{m^2}{8}-\frac{m^4}{256} =&\,\frac12\\ m^4-32m^2+128=&\,0. \end{align*}

Treating this as a quadratic in m2m^2 gives

m2=16±128.m^2=16\pm\sqrt{128}.

Since 0≤m≤40\leq m\leq4, we require m2≤16m^2\leq16. Hence,

m= 16−128= 2.16478…≈ 2.165,\begin{align*} m=&\,\sqrt{16-\sqrt{128}}\\ =&\,2.16478\ldots\\ \approx&\,\boxed{2.165}, \end{align*}

to 3 decimal places, as required.

解法二

思路

展开

也可以通过三位小数的舍入边界验证结果。计算 F(2.1645)F(2.1645) 与 F(2.1655)F(2.1655)；若 0.50.5 落在两者之间，中位数就必定舍入为 2.1652.165。

答题过程

展开

For 0≤x≤40\leq x\leq4,

F(x)=x28−x4256.F(x)=\frac{x^2}{8}-\frac{x^4}{256}.

At the two rounding boundaries,

 F(2.1645) =0.499891…<0.5\begin{align*} &\,F(2.1645)\\ &\,=0.499891\ldots<0.5 \end{align*}

and

 F(2.1655) =0.500273…>0.5.\begin{align*} &\,F(2.1655)\\ &\,=0.500273\ldots>0.5. \end{align*}

Since FF is increasing on the support, the median lies between 2.16452.1645 and 2.16552.1655. Therefore, it is

2.165.\boxed{2.165}.

This is given to 3 decimal places.

(b)

解法一

思路

展开

要计算 Var⁡(1/Y)\operatorname{Var}(1/Y)，分别按连续随机变量期望公式求 E(1/Y)E(1/Y) 和 E(1/Y2)E(1/Y^2)，再使用

 Var⁡(1/Y) =E(1/Y2)−[E(1/Y)]2.\begin{align*} &\,\operatorname{Var}(1/Y)\\ &\,=E(1/Y^2)-[E(1/Y)]^2. \end{align*}

保留精确分数即可自然得到题目所给结果。

答题过程

展开

First,

 E(1Y) =∫141y(4y3255) dy =4255∫14y2 dy =4255[y33]14 =4255(43−133) =4255(633) =2885.\begin{align*} &\,E\left(\frac1Y\right)\\ &\,=\int_1^4\frac1y\left(\frac{4y^3}{255}\right) \,\mathrm{d}y\\ &\,=\frac4{255}\int_1^4y^2\,\mathrm{d}y\\ &\,=\frac4{255}\left[\frac{y^3}{3}\right]_1^4\\ &\,=\frac4{255}\Big(\frac{4^3-1^3}{3}\Big)\\ &\,=\frac4{255}\left(\frac{63}{3}\right)\\ &\,=\frac{28}{85}. \end{align*}

Similarly,

 E(1Y2) =∫141y2(4y3255) dy =4255∫14y dy =4255[y22]14 =4255(42−122) =4255(152) =217.\begin{align*} &\,E\left(\frac1{Y^2}\right)\\ &\,=\int_1^4\frac1{y^2} \left(\frac{4y^3}{255}\right)\,\mathrm{d}y\\ &\,=\frac4{255}\int_1^4y\,\mathrm{d}y\\ &\,=\frac4{255}\left[\frac{y^2}{2}\right]_1^4\\ &\,=\frac4{255}\Big(\frac{4^2-1^2}{2}\Big)\\ &\,=\frac4{255}\left(\frac{15}{2}\right)\\ &\,=\frac2{17}. \end{align*}

Therefore,

 Var⁡(1Y) =E(1Y2) −[E(1Y)]2 =217−(2885)2 =850−7847225 =667225,\begin{align*} &\,\operatorname{Var}\left(\frac1Y\right)\\ &\,=E\left(\frac1{Y^2}\right)\\ &\,\hspace{2pt}-\left[E\left(\frac1Y\right)\right]^2\\ &\,=\frac2{17}-\left(\frac{28}{85}\right)^2\\ &\,=\frac{850-784}{7225}\\ &\,=\boxed{\frac{66}{7225}}, \end{align*}

as required.

(c)

解法一

思路

展开

承接 (b)，使用方差的线性变换性质

 Var⁡(a+bZ) =b2Var⁡(Z).\begin{align*} &\,\operatorname{Var}(a+bZ)\\ &\,=b^2\operatorname{Var}(Z). \end{align*}

常数 44 不影响方差，而系数 −5-5 会使方差乘以 2525。

答题过程

展开

Using the result from part (b),

 Var⁡(4−5Y) =(−5)2Var⁡(1Y) =25(667225) =66289 ≈0.228.\begin{align*} &\,\operatorname{Var}\left(4-\frac5Y\right)\\ &\,=(-5)^2\operatorname{Var}\left(\frac1Y\right)\\ &\,=25\left(\frac{66}{7225}\right)\\ &\,=\frac{66}{289}\\ &\,\approx\boxed{0.228}. \end{align*}

The answer is given to 3 significant figures.