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S2 2024 Jun Q2

A Level / Edexcel / S2

IAL 2024 June Paper · Question 2

题目

Problem

The continuous random variable HH has cumulative distribution function given by

F(h)=F(h)= {0,h⩽0,h248,0<h⩽4,h6−13,4<h⩽5,3h10−h275−23,5<h⩽d,1,h>d.\begin{cases} \begin{aligned} &0,\\ &h\leqslant0,\end{aligned}\\[2mm] \begin{aligned} &\dfrac{h^2}{48},\\ &0<h\leqslant4,\end{aligned}\\[2mm] \begin{aligned} &\dfrac{h}{6}-\dfrac13,\\ &4<h\leqslant5,\end{aligned}\\[2mm] \begin{aligned} &\dfrac{3h}{10}-\dfrac{h^2}{75}-\dfrac23,\\ &5<h\leqslant d,\end{aligned}\\[2mm] \begin{aligned} &1,\\ &h>d.\end{aligned} \end{cases}

where dd is a constant.

(a) Show that

2d2−45d+250=02d^2-45d+250=0
(2)

(b) Find

 P(H<1.5 ∣1<H<4.5)\begin{align*} &\,P(H<1.5\\ &\,\hspace{2pt}\mid1<H<4.5) \end{align*}
(4)

(c) Find the probability density function f(h)f(h)

You may leave the limits of hh in terms of dd where necessary.

(3)

(Total for Question 2 is 9 marks)

题目中文翻译

连续随机变量 HH 的累积分布函数为 F(h)F(h)，其中

F(h)=F(h)= {0,h⩽0,h248,0<h⩽4,h6−13,4<h⩽5,3h10−h275−23,5<h⩽d,1,h>d.\begin{cases} \begin{aligned} &0,\\ &h\leqslant0,\end{aligned}\\[2mm] \begin{aligned} &\dfrac{h^2}{48},\\ &0<h\leqslant4,\end{aligned}\\[2mm] \begin{aligned} &\dfrac{h}{6}-\dfrac13,\\ &4<h\leqslant5,\end{aligned}\\[2mm] \begin{aligned} &\dfrac{3h}{10}-\dfrac{h^2}{75}-\dfrac23,\\ &5<h\leqslant d,\end{aligned}\\[2mm] \begin{aligned} &1,\\ &h>d.\end{aligned} \end{cases}

其中 dd 是常数。

(a) 证明

2d2−45d+250=0.2d^2-45d+250=0.
(2)

(b) 求

P(H<1.5∣1<H<4.5).P(H<1.5\mid 1<H<4.5).
(4)

(c) 求概率密度函数 f(h)f(h)。

如有必要，你可以将 hh 的界限用 dd 表示。

(3)

（第 2 题共 9 分）

解答

(a)

解法一

思路

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HH 是连续型随机变量，所以其累积分布函数连续。由于 h>dh>d 时 F(h)=1F(h)=1，令第三段表达式在 h=dh=d 时等于 1。清除分母并整理，即可自然得到题目要求的二次方程。

答题过程

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Since HH is continuous and F(h)=1F(h)=1 for h>dh>d, continuity gives F(d)=1F(d)=1. Thus

3d10−d275−23=1.\frac{3d}{10}-\frac{d^2}{75}-\frac23=1.

Multiplying by 150 gives

 45d−2d2−100 =150, 2d2−45d+250=0,\begin{align*} &\,45d-2d^2-100\\[2mm] &\,=150,\\ &\,2d^2-45d+250=0, \end{align*}

as required.

(b)

解法一

思路

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在条件 1<H<4.51<H<4.5 下，事件 H<1.5H<1.5 实际变成 1<H<1.51<H<1.5。分别用累积分布函数之差求分子与分母，再按条件概率公式相除。

答题过程

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First,

 P(1<H<4.5) =F(4.5)−F(1) =(4.56−13)−148 =(34−13)−148 =2048−148 =1948.\begin{align*} &\,P(1<H<4.5)\\[2mm] &\,=F(4.5)-F(1)\\ &\,=\Big(\frac{4.5}{6}-\frac13\Big)-\frac1{48}\\[2mm] &\,=\Big(\frac34-\frac13\Big)-\frac1{48}\\[2mm] &\,=\frac{20}{48}-\frac1{48}\\ &\,=\frac{19}{48}. \end{align*}

Also,

 P(1<H<1.5) =F(1.5)−F(1) =1.5248−148 =9/4−148 =5/448 =5192.\begin{align*} &\,P(1<H<1.5)\\[2mm] &\,=F(1.5)-F(1)\\ &\,=\frac{1.5^2}{48}-\frac1{48}\\[2mm] &\,=\frac{9/4-1}{48}\\[2mm] &\,=\frac{5/4}{48}\\ &\,=\frac5{192}. \end{align*}

Therefore,

 P(H<1.5∣1<H<4.5) =5/19219/48 =576 ≈0.0658.\begin{align*} &\,P(H<1.5\mid 1<H<4.5)\\[2mm] &\,=\frac{5/192}{19/48}\\ &\,=\boxed{\frac5{76}}\\ &\,\approx0.0658. \end{align*}

(c)

解法一

思路

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概率密度函数是累积分布函数在各段内部的导数。逐段求导，常数段的导数为 0，并保留与原累积分布函数对应的区间。

答题过程

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Differentiating each non-constant part of F(h)F(h) gives

f(h)={h24,0<h⩽4,16,4<h⩽5,310−2h75,5<h⩽d,0,otherwise.f(h)= \begin{cases} \begin{aligned} &\dfrac h{24},\\[-1mm] &\quad0<h\leqslant4, \end{aligned}\\[2mm] \begin{aligned} &\dfrac16,\\[-1mm] &\quad4<h\leqslant5, \end{aligned}\\[2mm] \begin{aligned} &\dfrac3{10}-\dfrac{2h}{75},\\[-1mm] &\quad5<h\leqslant d, \end{aligned}\\[2mm] \begin{aligned} &0,\\[-1mm] &\quad\text{otherwise}. \end{aligned} \end{cases}