题目
Problem
The distance, in metres, a novice tightrope artist, walking on a wire, walks before falling is modelled by the random variable W W W
F ( w ) = F(w)= F ( w ) = { 0 , w < 0 , 1 3 ( w − w 4 256 ) , 0 ⩽ w ⩽ 4 , 1 , w > 4. \begin{cases}
0,\\[-1mm]
\quad w<0,\\[3mm]
\dfrac13\Big(w-\dfrac{w^4}{256}\Big),\\[-1mm]
\quad 0\leqslant w\leqslant4,\\[3mm]
1,\\[-1mm]
\quad w>4.
\end{cases} ⎩ ⎨ ⎧ 0 , w < 0 , 3 1 ( w − 256 w 4 ) , 0 ⩽ w ⩽ 4 , 1 , w > 4. (a) Find the probability that a novice tightrope artist, walking on the wire, walks at least 3.5 metres before falling.
(2)
A random sample of 30 novice tightrope artists is taken.
(b) Find the probability that more than 1 of these novice tightrope artists, walking on the wire, walks at least 3.5 metres before falling.
(3)
Given E ( W ) = 1.6 E(W)=1.6 E ( W ) = 1.6
(c) use algebraic integration to find Var ( W ) \operatorname{Var}(W) Var ( W )
(5)
(Total for Question 2 is 10 marks)
题目中文翻译
一名在钢丝上行走的新手走绳艺人，在跌落之前所走的距离（单位：米）由随机变量 W W W
F ( w ) = F(w)= F ( w ) = { 0 , w < 0 , 1 3 ( w − w 4 256 ) , 0 ⩽ w ⩽ 4 , 1 , w > 4. \begin{cases}
0,\\[-1mm]
\quad w<0,\\[3mm]
\dfrac13\Big(w-\dfrac{w^4}{256}\Big),\\[-1mm]
\quad 0\leqslant w\leqslant4,\\[3mm]
1,\\[-1mm]
\quad w>4.
\end{cases} ⎩ ⎨ ⎧ 0 , w < 0 , 3 1 ( w − 256 w 4 ) , 0 ⩽ w ⩽ 4 , 1 , w > 4. (a) 求一位新手走绳艺人在跌落前至少走 3.5 米的概率。
(2)
随机抽取 30 位新手走绳艺人。
(b) 求其中多于 1 人在跌落前至少走 3.5 米的概率。
(3)
已知 E ( W ) = 1.6 E(W)=1.6 E ( W ) = 1.6
(c) 使用代数积分求 V a r ( W ) \mathrm{Var}(W) Var ( W )
(5)
（第 2 题共 10 分）
解答
(a)
解法一
思路
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累计分布函数给出 P ( W ⩽ w ) P(W\leqslant w) P ( W ⩽ w ) W W W 1 − F ( 3.5 ) 1-F(3.5) 1 − F ( 3.5 )
答题过程
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Evaluating the cumulative distribution function,
F ( 3.5 ) F(3.5) F ( 3.5 ) = 1 3 ( 3.5 − 3.5 4 256 ) = 0.971272786 … . \begin{align*}
=&\,\frac13\Big(3.5-\frac{3.5^4}{256}\Big)\\[4mm]
=&\,0.971272786\ldots.
\end{align*} = = 3 1 ( 3.5 − 256 3. 5 4 ) 0.971272786 … . Since W W W
P ( W ⩾ 3.5 ) P(W\geqslant3.5) P ( W ⩾ 3.5 ) = 1 − F ( 3.5 ) = 0.028727213 … = 0.0287 . \begin{align*}
=&\,1-F(3.5)\\[4mm]
=&\,0.028727213\ldots\\[4mm]
=&\,\boxed{0.0287}.
\end{align*} = = = 1 − F ( 3.5 ) 0.028727213 … 0.0287 .
(b)
解法一
思路
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把每位艺人至少走 3.5 米视为一次“成功”，30 人中的成功人数服从二项分布。“多于 1 人”用补事件“0 人或 1 人”计算最简便。
答题过程
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Let Y Y Y
p = 0.028727213 … . p=0.028727213\ldots. p = 0.028727213 … . Then
Y ∼ B ( 30 , p ) . Y\sim\operatorname{B}(30,p). Y ∼ B ( 30 , p ) . Therefore,
P ( Y > 1 ) P(Y>1) P ( Y > 1 ) = 1 − P ( Y ⩽ 1 ) = 1 − ( 1 − p ) 30 − 30 p ( 1 − p ) 29 = 0.212811 … = 0.213 . \begin{align*}
=&\,1-P(Y\leqslant1)\\[4mm]
=&\,1-(1-p)^{30}\\
&\,\hspace{2pt}-30p(1-p)^{29}\\[4mm]
=&\,0.212811\ldots\\[4mm]
=&\,\boxed{0.213}.
\end{align*} = = = = 1 − P ( Y ⩽ 1 ) 1 − ( 1 − p ) 30 − 30 p ( 1 − p ) 29 0.212811 … 0.213 .
(c)
解法一
思路
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先对累计分布函数 F ( w ) F(w) F ( w ) f ( w ) f(w) f ( w ) E ( W 2 ) E(W^2) E ( W 2 ) 1.6 1.6 1.6
答题过程
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For 0 ⩽ w ⩽ 4 0\leqslant w\leqslant4 0 ⩽ w ⩽ 4
f ( w ) = d F d w f(w)=\frac{\mathrm{d}F}{\mathrm{d}w} f ( w ) = d w d F = 1 3 ( 1 − 4 w 3 256 ) = 1 3 ( 1 − w 3 64 ) . \begin{align*}
=&\,\frac13\Big(1-\frac{4w^3}{256}\Big)\\[4mm]
=&\,\frac13\Big(1-\frac{w^3}{64}\Big).
\end{align*} = = 3 1 ( 1 − 256 4 w 3 ) 3 1 ( 1 − 64 w 3 ) . Thus,
E ( W 2 ) E(W^2) E ( W 2 ) = ∫ 0 4 w 2 f ( w ) d w = 1 3 ∫ 0 4 ( w 2 − w 5 64 ) d w = 1 3 [ w 3 3 − w 6 384 ] 0 4 = 1 3 ( 64 3 − 4096 384 ) = 1 3 ( 64 3 − 32 3 ) = 32 9 . \begin{align*}
=&\,\int_0^4 w^2f(w)\,\mathrm{d}w\\[4mm]
=&\,\frac13\int_0^4
\Big(w^2-\frac{w^5}{64}\Big)\,\mathrm{d}w\\[4mm]
=&\,\frac13\Big[\frac{w^3}{3}-\frac{w^6}{384}\Big]_0^4\\[4mm]
=&\,\frac13\Big(\frac{64}{3}-\frac{4096}{384}\Big)\\[4mm]
=&\,\frac13\Big(\frac{64}{3}-\frac{32}{3}\Big)\\[4mm]
=&\,\frac{32}{9}.
\end{align*} = = = = = = ∫ 0 4 w 2 f ( w ) d w 3 1 ∫ 0 4 ( w 2 − 64 w 5 ) d w 3 1 [ 3 w 3 − 384 w 6 ] 0 4 3 1 ( 3 64 − 384 4096 ) 3 1 ( 3 64 − 3 32 ) 9 32 . Since E ( W ) = 1.6 E(W)=1.6 E ( W ) = 1.6
Var ( W ) \operatorname{Var}(W) Var ( W ) = E ( W 2 ) − [ E ( W ) ] 2 = 32 9 − 64 25 = 800 − 576 225 = 224 225 = 0.996 . \begin{align*}
=&\,E(W^2)-[E(W)]^2\\[4mm]
=&\,\frac{32}{9}-\frac{64}{25}\\[4mm]
=&\,\frac{800-576}{225}\\[4mm]
=&\,\frac{224}{225}\\[4mm]
=&\,\boxed{0.996}.
\end{align*} = = = = = E ( W 2 ) − [ E ( W ) ] 2 9 32 − 25 64 225 800 − 576 225 224 0.996 . 