题目
Problem
The continuous random variable X X X
f ( x ) = { 1 2 k ( x − 1 ) for 1 ⩽ x ⩽ 3 k for 3 < x ⩽ 6 1 4 k ( 10 − x ) for 6 < x ⩽ 10 0 otherwise \begin{align*}
&\,f(x)=\\[2mm]
&\,\begin{cases}
\begin{aligned}
&\,\dfrac12k(x-1)\\[-1mm]
&\,\text{for }1\leqslant x\leqslant3
\end{aligned}\\[3mm]
\begin{aligned}
&\,k\\[-1mm]
&\,\text{for }3<x\leqslant6
\end{aligned}\\[3mm]
\begin{aligned}
&\,\dfrac14k(10-x)\\[-1mm]
&\,\text{for }6<x\leqslant10
\end{aligned}\\[3mm]
\begin{aligned}
&\,0\\[-1mm]
&\,\text{otherwise}
\end{aligned}
\end{cases}
\end{align*} f ( x ) = ⎩ ⎨ ⎧ 2 1 k ( x − 1 ) for 1 ⩽ x ⩽ 3 k for 3 < x ⩽ 6 4 1 k ( 10 − x ) for 6 < x ⩽ 10 0 otherwise where k k k
(a) Sketch f ( x ) f(x) f ( x ) x x x
(2)
(b) Show that k = 1 6 k = \dfrac{1}{6} k = 6 1
(2)
(c) Specify fully the cumulative distribution function F ( x ) F(x) F ( x ) X X X
(7)
Given that E ( X ) = 61 12 E(X) = \dfrac{61}{12} E ( X ) = 12 61
(d) find P ( X > E ( X ) ) P(X > E(X)) P ( X > E ( X ))
(2)
(e) Describe the skewness of the distribution, giving a reason for your answer.
(2)
(Total for Question 4 is 15 marks)
题目中文翻译
连续随机变量 X X X
f ( x ) = { 1 2 k ( x − 1 ) for 1 ⩽ x ⩽ 3 k for 3 < x ⩽ 6 1 4 k ( 10 − x ) for 6 < x ⩽ 10 0 otherwise \begin{align*}
&\,f(x)=\\[2mm]
&\,\begin{cases}
\begin{aligned}
&\,\dfrac12k(x-1)\\[-1mm]
&\,\text{for }1\leqslant x\leqslant3
\end{aligned}\\[3mm]
\begin{aligned}
&\,k\\[-1mm]
&\,\text{for }3<x\leqslant6
\end{aligned}\\[3mm]
\begin{aligned}
&\,\dfrac14k(10-x)\\[-1mm]
&\,\text{for }6<x\leqslant10
\end{aligned}\\[3mm]
\begin{aligned}
&\,0\\[-1mm]
&\,\text{otherwise}
\end{aligned}
\end{cases}
\end{align*} f ( x ) = ⎩ ⎨ ⎧ 2 1 k ( x − 1 ) for 1 ⩽ x ⩽ 3 k for 3 < x ⩽ 6 4 1 k ( 10 − x ) for 6 < x ⩽ 10 0 otherwise 其中 k k k
(a) 画出所有 x x x f ( x ) f(x) f ( x )
(2)
(b) 证明 k = 1 6 k = \dfrac{1}{6} k = 6 1
(2)
(c) 完整写出 X X X F ( x ) F(x) F ( x )
(7)
已知 E ( X ) = 61 12 E(X) = \dfrac{61}{12} E ( X ) = 12 61
(d) 求 P ( X > E ( X ) ) P(X > E(X)) P ( X > E ( X ))
(2)
(e) 描述该分布的偏度，并给出理由。
(2)
（第 4 题共 15 分）
解答
(a)
解法一
思路
展开
逐段读取密度函数：从 ( 1 , 0 ) (1,0) ( 1 , 0 ) ( 3 , k ) (3,k) ( 3 , k ) 3 < x ⩽ 6 3<x\leqslant6 3 < x ⩽ 6 k k k ( 10 , 0 ) (10,0) ( 10 , 0 ) 0 0 0 1 , 3 , 6 , 10 1,3,6,10 1 , 3 , 6 , 10 k k k
答题过程
展开
The graph consists of straight-line segments joining
( 1 , 0 ) , ( 3 , k ) , ( 6 , k ) , ( 10 , 0 ) , \begin{gathered}
(1,0),\quad(3,k),\\[2mm]
(6,k),\quad(10,0),
\end{gathered} ( 1 , 0 ) , ( 3 , k ) , ( 6 , k ) , ( 10 , 0 ) , with f ( x ) = 0 f(x)=0 f ( x ) = 0 x < 1 x<1 x < 1 x > 10 x>10 x > 10
(b)
解法一
思路
展开
概率密度曲线下总面积为 1 1 1
答题过程
展开
The total area under a probability density function is 1 1 1
1 2 ( 3 − 1 ) k + ( 6 − 3 ) k + 1 2 ( 10 − 6 ) k = 1. \begin{align*}
&\,\frac12(3-1)k+(6-3)k\\[2mm]
&\,\hspace{2pt}+\frac12(10-6)k\\[2mm]
=&\,1.
\end{align*} = 2 1 ( 3 − 1 ) k + ( 6 − 3 ) k + 2 1 ( 10 − 6 ) k 1. Thus,
k + 3 k + 2 k = 1 , 6 k = 1. \begin{align*}
k+3k+2k=&\,1,\\[2mm]
6k=&\,1.
\end{align*} k + 3 k + 2 k = 6 k = 1 , 1. Therefore,
k = 1 6 . \boxed{k=\frac16}. k = 6 1 . Equivalently, treating the whole region as one trapezium gives the same area equation:
1 2 ( 3 + 9 ) k = 1. \frac12(3+9)k=1. 2 1 ( 3 + 9 ) k = 1.
解法二
思路
展开
也可以直接对三段密度函数积分。分别使用各段正确的上下限，并令总积分等于 1 1 1
答题过程
展开
Using the total probability,
1 = ∫ 1 3 1 2 k ( x − 1 ) d x + ∫ 3 6 k d x + ∫ 6 10 1 4 k ( 10 − x ) d x = k + 3 k + 2 k = 6 k . \begin{align*}
1=&\,\int_1^3\frac12k(x-1)\,\mathrm{d}x\\[2mm]
&\,\hspace{2pt}+\int_3^6k\,\mathrm{d}x\\[2mm]
&\,\hspace{4pt}+\int_6^{10}\frac14k(10-x)\,\mathrm{d}x\\[2mm]
=&\,k+3k+2k\\[2mm]
=&\,6k.
\end{align*} 1 = = = ∫ 1 3 2 1 k ( x − 1 ) d x + ∫ 3 6 k d x + ∫ 6 10 4 1 k ( 10 − x ) d x k + 3 k + 2 k 6 k . Thus,
k = 1 6 . \boxed{k=\frac16}. k = 6 1 .
(c)
解法一
思路
展开
累积分布函数为 F ( x ) = P ( X ⩽ x ) F(x)=P(X\leqslant x) F ( x ) = P ( X ⩽ x ) F ( x ) F(x) F ( x )
答题过程
展开
Using k = 1 6 k=\frac16 k = 6 1 1 ⩽ x ⩽ 3 1\leqslant x\leqslant3 1 ⩽ x ⩽ 3
F ( x ) = ∫ 1 x 1 12 ( t − 1 ) d t = ( x − 1 ) 2 24 . \begin{align*}
&\,F(x)\\[2mm]
=&\,\int_1^x\frac1{12}(t-1)\,\mathrm{d}t\\[2mm]
=&\,\frac{(x-1)^2}{24}.
\end{align*} = = F ( x ) ∫ 1 x 12 1 ( t − 1 ) d t 24 ( x − 1 ) 2 . In particular,
F ( 3 ) = 1 6 . F(3)=\frac16. F ( 3 ) = 6 1 . For 3 < x ⩽ 6 3<x\leqslant6 3 < x ⩽ 6
F ( x ) = F ( 3 ) + ∫ 3 x 1 6 d t = 1 6 + x − 3 6 = x − 2 6 . \begin{align*}
&\,F(x)\\[2mm]
=&\,F(3)+\int_3^x\frac16\,\mathrm{d}t\\[2mm]
=&\,\frac16+\frac{x-3}{6}\\[2mm]
=&\,\frac{x-2}{6}.
\end{align*} = = = F ( x ) F ( 3 ) + ∫ 3 x 6 1 d t 6 1 + 6 x − 3 6 x − 2 . Thus F ( 6 ) = 2 3 F(6)=\frac23 F ( 6 ) = 3 2 6 < x ⩽ 10 6<x\leqslant10 6 < x ⩽ 10
F ( x ) = 1 − ∫ x 10 1 24 ( 10 − t ) d t = 1 − ( 10 − x ) 2 48 . \begin{align*}
&\,F(x)\\[2mm]
=&\,1-\int_x^{10}\frac1{24}(10-t)\,\mathrm{d}t\\[2mm]
=&\,1-\frac{(10-x)^2}{48}.
\end{align*} = = F ( x ) 1 − ∫ x 10 24 1 ( 10 − t ) d t 1 − 48 ( 10 − x ) 2 . The same last segment can also be found by accumulating from x = 6 x=6 x = 6
F ( x ) = F ( 6 ) + ∫ 6 x 10 − t 24 d t = 2 3 + [ 10 t 24 − t 2 48 ] 6 x = 2 3 + 10 x 24 − x 2 48 − 7 4 = 20 x − x 2 − 52 48 . \begin{align*}
&\,F(x)\\[2mm]
=&\,F(6)\\[2mm]
&\,\hspace{2pt}+\int_6^x\frac{10-t}{24}\,\mathrm{d}t\\[2mm]
=&\,\frac23+\bigg[\frac{10t}{24}-\frac{t^2}{48}\bigg]_6^x\\[2mm]
=&\,\frac23+\frac{10x}{24}-\frac{x^2}{48}\\[2mm]
&\,\hspace{2pt}-\frac74\\[2mm]
=&\,\frac{20x-x^2-52}{48}.
\end{align*} = = = = F ( x ) F ( 6 ) + ∫ 6 x 24 10 − t d t 3 2 + [ 24 10 t − 48 t 2 ] 6 x 3 2 + 24 10 x − 48 x 2 − 4 7 48 20 x − x 2 − 52 . Therefore,
F ( x ) = { 0 for x < 1 ( x − 1 ) 2 24 for 1 ⩽ x ⩽ 3 x − 2 6 for 3 < x ⩽ 6 1 − ( 10 − x ) 2 48 for 6 < x ⩽ 10 1 for x > 10. \begin{align*}
&\,F(x)=\\[2mm]
&\,\begin{cases}
\begin{aligned}
&\,0\\[-1mm]
&\,\text{for }x<1
\end{aligned}\\[3mm]
\begin{aligned}
&\,\dfrac{(x-1)^2}{24}\\[-1mm]
&\,\text{for }1\leqslant x\leqslant3
\end{aligned}\\[3mm]
\begin{aligned}
&\,\dfrac{x-2}{6}\\[-1mm]
&\,\text{for }3<x\leqslant6
\end{aligned}\\[3mm]
\begin{aligned}
&\,1-\dfrac{(10-x)^2}{48}\\[-1mm]
&\,\text{for }6<x\leqslant10
\end{aligned}\\[3mm]
\begin{aligned}
&\,1\\[-1mm]
&\,\text{for }x>10.
\end{aligned}
\end{cases}
\end{align*} F ( x ) = ⎩ ⎨ ⎧ 0 for x < 1 24 ( x − 1 ) 2 for 1 ⩽ x ⩽ 3 6 x − 2 for 3 < x ⩽ 6 1 − 48 ( 10 − x ) 2 for 6 < x ⩽ 10 1 for x > 10.
解法二
思路
展开
每一段先作不定积分，再用边界处的累积概率确定常数。第一段用 F ( 1 ) = 0 F(1)=0 F ( 1 ) = 0 F ( 3 ) = 1 / 6 F(3)=1/6 F ( 3 ) = 1/6 F ( 10 ) = 1 F(10)=1 F ( 10 ) = 1
答题过程
展开
For 1 ⩽ x ⩽ 3 1\leqslant x\leqslant3 1 ⩽ x ⩽ 3
F ( x ) = ∫ x − 1 12 d x = ( x − 1 ) 2 24 + C 1 . \begin{align*}
F(x)=&\,\int\frac{x-1}{12}\,\mathrm{d}x\\[2mm]
=&\,\frac{(x-1)^2}{24}+C_1.
\end{align*} F ( x ) = = ∫ 12 x − 1 d x 24 ( x − 1 ) 2 + C 1 . Since F ( 1 ) = 0 F(1)=0 F ( 1 ) = 0 C 1 = 0 C_1=0 C 1 = 0
F ( x ) = ( x − 1 ) 2 24 . F(x)=\frac{(x-1)^2}{24}. F ( x ) = 24 ( x − 1 ) 2 . In particular, F ( 3 ) = 1 / 6 F(3)=1/6 F ( 3 ) = 1/6 3 < x ⩽ 6 3<x\leqslant6 3 < x ⩽ 6
F ( x ) = ∫ 1 6 d x = x 6 + C 2 . F(x)=\int\frac16\,\mathrm{d}x=\frac{x}{6}+C_2. F ( x ) = ∫ 6 1 d x = 6 x + C 2 . Using the value at x = 3 x=3 x = 3
3 6 + C 2 = 1 6 , C 2 = − 1 3 . \begin{align*}
\frac36+C_2=&\,\frac16,\\[2mm]
C_2=&\,-\frac13.
\end{align*} 6 3 + C 2 = C 2 = 6 1 , − 3 1 . Therefore,
F ( x ) = x − 2 6 . F(x)=\frac{x-2}{6}. F ( x ) = 6 x − 2 . For 6 < x ⩽ 10 6<x\leqslant10 6 < x ⩽ 10
F ( x ) = ∫ 10 − x 24 d x = C 3 − ( 10 − x ) 2 48 . \begin{align*}
&\,F(x)\\[2mm]
=&\,\int\frac{10-x}{24}\,\mathrm{d}x\\[2mm]
=&\,C_3-\frac{(10-x)^2}{48}.
\end{align*} = = F ( x ) ∫ 24 10 − x d x C 3 − 48 ( 10 − x ) 2 . Since F ( 10 ) = 1 F(10)=1 F ( 10 ) = 1 C 3 = 1 C_3=1 C 3 = 1
F ( x ) = 1 − ( 10 − x ) 2 48 . \begin{align*}
&\,F(x)\\[2mm]
=&\,1-\frac{(10-x)^2}{48}.
\end{align*} = F ( x ) 1 − 48 ( 10 − x ) 2 . Finally, F ( x ) = 0 F(x)=0 F ( x ) = 0 x < 1 x<1 x < 1 F ( x ) = 1 F(x)=1 F ( x ) = 1 x > 10 x>10 x > 10
(d)
解法一
思路
展开
61 12 \frac{61}{12} 12 61 3 3 3 6 6 6 F ( x ) = x − 2 6 F(x)=\frac{x-2}{6} F ( x ) = 6 x − 2 1 − F ( E ( X ) ) 1-F(E(X)) 1 − F ( E ( X ))
答题过程
展开
Since
3 < 61 12 < 6 , 3<\frac{61}{12}<6, 3 < 12 61 < 6 , we use F ( x ) = x − 2 6 F(x)=\frac{x-2}{6} F ( x ) = 6 x − 2
P ( X > E ( X ) ) = 1 − F ( 61 12 ) = 1 − 61 12 − 2 6 = 1 − 37 72 = 35 72 ≈ 0.486 . \begin{align*}
&\,P\big(X>E(X)\big)\\[2mm]
=&\,1-F\Big(\frac{61}{12}\Big)\\[2mm]
=&\,1-\frac{\frac{61}{12}-2}{6}\\[2mm]
=&\,1-\frac{37}{72}\\[2mm]
=&\,\boxed{\frac{35}{72}\approx0.486}.
\end{align*} = = = = P ( X > E ( X ) ) 1 − F ( 12 61 ) 1 − 6 12 61 − 2 1 − 72 37 72 35 ≈ 0.486 .
(e)
解法一
思路
展开
由 (d)，均值右侧的概率小于 0.5 0.5 0.5 0.5 0.5 0.5 mean > median \text{mean}>\text{median} mean > median
答题过程
展开
From part (d),
P ( X > E ( X ) ) = 35 72 < 0.5. \begin{align*}
&\,P\big(X>E(X)\big)\\[2mm]
=&\,\frac{35}{72}<0.5.
\end{align*} = P ( X > E ( X ) ) 72 35 < 0.5. Therefore more than half of the distribution lies below the mean, so
mean > median . \text{mean}>\text{median}. mean > median . Hence the distribution is positively skewed . This also agrees with the sketch: its right-hand sloping tail extends farther than its left-hand sloping tail.
