CalcGospel 國際數學圖譜
返回

IAL 2025 Oct S2 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−164x30⩽x⩽40otherwisef(x) = \begin{cases} \dfrac{1}{4}x - \dfrac{1}{64}x^3 & 0 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}

(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)={4255y31⩽y⩽40otherwisef(y) = \begin{cases} \dfrac{4}{255}y^3 & 1 \leqslant y \leqslant 4 \\ 0 & \text{otherwise} \end{cases}

(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−164x30⩽x⩽40otherwisef(x) = \begin{cases} \dfrac{1}{4}x - \dfrac{1}{64}x^3 & 0 \leqslant x \leqslant 4 \\ 0 & \text{otherwise} \end{cases}

(a) 证明 XX 的中位数为 2.165（保留 3 位小数）。

连续随机变量 YY 的概率密度函数为

f(y)={4255y31⩽y⩽40otherwisef(y) = \begin{cases} \dfrac{4}{255}y^3 & 1 \leqslant y \leqslant 4 \\ 0 & \text{otherwise} \end{cases}

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

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

（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(3 d.p.),\begin{align*} m=&\,\sqrt{16-\sqrt{128}}\\ =&\,2.16478\ldots\\ =&\,\boxed{2.165}\quad\text{(3 d.p.)}, \end{align*}

as required.

解法二

思路

展开

官方评分资料也允许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…<0.5F(2.1645)=0.499891\ldots<0.5

and

F(2.1655)=0.500273…>0.5.F(2.1655)=0.500273\ldots>0.5.

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

2.165to 3 decimal places.\boxed{2.165}\quad\text{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\operatorname{Var}(1/Y)=E(1/Y^2)-[E(1/Y)]^2。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(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}\left(\frac{63}{3}\right)\\ =&\,\frac{28}{85}. \end{align*}

Similarly,

E(1Y2)= ∫141y2(4y3255) dy= 4255∫14y dy= 4255[y22]14= 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}\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) -\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)\operatorname{Var}(a+bZ)=b^2\operatorname{Var}(Z)。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(3 s.f.).\begin{align*} \operatorname{Var}\left(4-\frac5Y\right) =&\,(-5)^2\operatorname{Var}\left(\frac1Y\right)\\ =&\,25\left(\frac{66}{7225}\right)\\ =&\,\frac{66}{289}\\ =&\,\boxed{0.228}\quad\text{(3 s.f.)}. \end{align*}