题目
The distance, in metres, a novice tightrope artist, walking on a wire, walks before falling is modelled by the random variable W with cumulative distribution function
0 & w<0 \\ \dfrac{1}{3}\left(w-\dfrac{w^4}{256}\right) & 0 \leqslant w \leqslant 4 \\ 1 & w>4 \end{cases}$$ (a) Find the probability that a novice tightrope artist, walking on the wire, walks at least 3.5 metres before falling. <div style="text-align: right;">(2)</div> A random sample of 30 novice tightrope artists is taken. (b) Find the probability that more than 1 of these novice tightrope artists, walking on the wire, walks at least 3.5 metres before falling. <div style="text-align: right;">(3)</div> Given E(W) = 1.6 (c) use algebraic integration to find Var(W) <div style="text-align: right;">(5)</div> (Total for Question 2 is 10 marks)题目中文翻译
一名在钢丝上行走的新手走绳艺人，在跌落之前所走的距离（单位：米）由随机变量 建模，其累计分布函数为
0 & w<0 \\ \dfrac{1}{3}\left(w-\dfrac{w^4}{256}\right) & 0 \leqslant w \leqslant 4 \\ 1 & w>4 \end{cases}$$ (a) 求一位新手走绳艺人在跌落前至少走 3.5 米的概率。 (b) 随机抽取 30 位新手走绳艺人。求其中多于 1 人在跌落前至少走 3.5 米的概率。 已知 $E(W)=1.6$。 (c) 使用代数积分求 $\mathrm{Var}(W)$。 （第 2 题共 10 分） </details> # 解答 ## (a) ### 解法一 #### 思路 <details> <summary>展开</summary> 累计分布函数给出 $P(W\leqslant w)$。由于 $W$ 是连续随机变量，“至少 3.5 米”的概率可用补事件 $1-F(3.5)$ 求得。 </details> #### 答题过程 <details> <summary>展开</summary> Since $W$ is continuous,\begin{align*} P(W\geqslant3.5) =&,1-F(3.5) \[4mm] =&,1-\frac13 \left(3.5-\frac{3.5^4}{256}\right) \[4mm] =&,0.0287272\ldots \[4mm] =&,\boxed{0.0287}\quad\text{(3 s.f.)}. \end{align*}
</details> ## (b) ### 解法一 #### 思路 <details> <summary>展开</summary> 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。 </details> #### 答题过程 <details> <summary>展开</summary> Let $Y$ be the number of the 30 artists who walk at least 3.5 metres. Using the unrounded probability from part (a),Y\sim\operatorname{B}(30,0.0287272\ldots).
\begin{align*} P(Y>1) =&,1-P(Y=0)-P(Y=1) \[4mm] =&,1-(1-p)^{30} \[4mm] &,\hspace{2pt}-{30\choose1}p(1-p)^{29}, \end{align*}
where $p=0.0287272\ldots$. HenceP(Y>1)=0.212811\ldots =\boxed{0.213}\quad\text{(3 s.f.)}.
</details> ## (c) ### 解法一 #### 思路 <details> <summary>展开</summary> 先对累计分布函数 $F(w)$ 求导，得到概率密度函数 $f(w)$。再用代数积分求 $E(W^2)$，最后代入 $\operatorname{Var}(W)=E(W^2)-[E(W)]^2$。 </details> #### 答题过程 <details> <summary>展开</summary> For $0\leqslant w\leqslant4$, the probability density function is\begin{align*} f(w)=\frac{\mathrm{d}F}{\mathrm{d}w} =&,\frac13\left(1-\frac{4w^3}{256}\right) \[4mm] =&,\frac13\left(1-\frac{w^3}{64}\right). \end{align*}
\begin{align*} E(W^2) =&,\int_0^4 w^2f(w),\mathrm{d}w \[4mm] =&,\frac13\int_0^4 \left(w^2-\frac{w^5}{64}\right),\mathrm{d}w \[4mm] =&,\frac13\left[\frac{w^3}{3} -\frac{w^6}{384}\right]_0^4 \[4mm] =&,\frac{32}{9}. \end{align*}
Since $E(W)=1.6$,\begin{align*} \operatorname{Var}(W) =&,E(W^2)-[E(W)]^2 \[4mm] =&,\frac{32}{9}-(1.6)^2 \[4mm] =&,\frac{224}{225} \[4mm] =&,\boxed{0.996}\quad\text{(3 s.f.)}. \end{align*}