题目
Problem
The random variable T T T f ( t ) f(t) f ( t )
f ( t ) = { 1 3 for 1 ⩽ t < 2 k ( 4 t 2 − t 3 ) for 2 ⩽ t ⩽ 4 0 otherwise \begin{align*}
&\,f(t)=\\
&\,\begin{cases}
\begin{aligned}
&\,\dfrac{1}{3}\\[-1mm]
&\,\quad \text{for }1 \leqslant t < 2
\end{aligned}\\[3mm]
\begin{aligned}
&\,k(4t^2-t^3)\\[-1mm]
&\,\quad \text{for }2 \leqslant t \leqslant 4
\end{aligned}\\[3mm]
\begin{aligned}
&\,0\\[-1mm]
&\,\quad \text{otherwise}
\end{aligned}
\end{cases}
\end{align*} f ( t ) = ⎩ ⎨ ⎧ 3 1 for 1 ⩽ t < 2 k ( 4 t 2 − t 3 ) for 2 ⩽ t ⩽ 4 0 otherwise (a) Show that k = 1 22 k=\dfrac1{22} k = 22 1
(3)
(b) Sketch the probability density function.
(2)
(c) Find the mode of T T T
(3)
(d) Find the cumulative distribution function F ( t ) F(t) F ( t ) t t t
(5)
(e) Find the probability that T T T
(2)
(Total for Question 4 is 15 marks)
题目中文翻译
随机变量 T T T f ( t ) f(t) f ( t )
f ( t ) = { 1 3 for 1 ⩽ t < 2 k ( 4 t 2 − t 3 ) for 2 ⩽ t ⩽ 4 0 otherwise \begin{align*}
&\,f(t)=\\
&\,\begin{cases}
\begin{aligned}
&\,\dfrac{1}{3}\\[-1mm]
&\,\quad \text{for }1 \leqslant t < 2
\end{aligned}\\[3mm]
\begin{aligned}
&\,k(4t^2-t^3)\\[-1mm]
&\,\quad \text{for }2 \leqslant t \leqslant 4
\end{aligned}\\[3mm]
\begin{aligned}
&\,0\\[-1mm]
&\,\quad \text{otherwise}
\end{aligned}
\end{cases}
\end{align*} f ( t ) = ⎩ ⎨ ⎧ 3 1 for 1 ⩽ t < 2 k ( 4 t 2 − t 3 ) for 2 ⩽ t ⩽ 4 0 otherwise (a) 证明 k = 1 22 k=\dfrac{1}{22} k = 22 1
(3)
(b) 画出概率密度函数的草图。
(2)
(c) 求 T T T
(3)
(d) 求所有取值下的累积分布函数 F ( t ) F(t) F ( t )
(5)
(e) 求 T T T
(2)
（第 4 题共 15 分）
解答
(a)
解法一
思路
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概率密度函数在整个定义域上的积分必须等于 1。分别对两个区间积分并相加，即可解出 k k k
答题过程
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Since the total area under a probability density function is 1,
∫ 1 2 1 3 d t + ∫ 2 4 k ( 4 t 2 − t 3 ) d t = 1. \begin{align*}
&\,\int_1^2\frac13\,\mathrm{d}t\\
&\,\hspace{2pt}+\int_2^4k(4t^2-t^3)\,\mathrm{d}t=1.
\end{align*} ∫ 1 2 3 1 d t + ∫ 2 4 k ( 4 t 2 − t 3 ) d t = 1. Therefore,
1 3 + k [ 4 t 3 3 − t 4 4 ] 2 4 = 1 1 3 + 44 k 3 = 1 44 k = 2. \begin{align*}
\frac13
+k\left[\frac{4t^3}{3}-\frac{t^4}{4}\right]_2^4
=&\,1 \\[4mm]
\frac13+\frac{44k}{3}
=&\,1 \\[4mm]
44k=&\,2.
\end{align*} 3 1 + k [ 3 4 t 3 − 4 t 4 ] 2 4 = 3 1 + 3 44 k = 44 k = 1 1 2. Hence,
k = 1 22 , \boxed{k=\frac1{22}}, k = 22 1 , as required.
(b)
解法一
思路
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在 1 ⩽ t < 2 1\leqslant t<2 1 ⩽ t < 2 1 / 3 1/3 1/3 t = 2 t=2 t = 2 4 / 11 4/11 4/11 t = 4 t=4 t = 4
答题过程
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The sketch has:
a horizontal section f ( t ) = 1 / 3 f(t)=1/3 f ( t ) = 1/3 1 ⩽ t < 2 1\leqslant t<2 1 ⩽ t < 2
a jump to f ( 2 ) = 4 / 11 f(2)=4/11 f ( 2 ) = 4/11
a curved section that first rises and then falls to f ( 4 ) = 0 f(4)=0 f ( 4 ) = 0
the values t = 1 t=1 t = 1 t = 2 t=2 t = 2 t = 4 t=4 t = 4
(c)
解法一
思路
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第一段密度恒为 1 / 3 1/3 1/3
答题过程
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For 2 ⩽ t ⩽ 4 2\leqslant t\leqslant4 2 ⩽ t ⩽ 4
f ′ ( t ) = 1 22 ( 8 t − 3 t 2 ) . f'(t)=\frac1{22}(8t-3t^2). f ′ ( t ) = 22 1 ( 8 t − 3 t 2 ) . At a stationary point,
8 t − 3 t 2 = 0 t ( 8 − 3 t ) = 0. \begin{align*}
8t-3t^2=&\,0 \\[4mm]
t(8-3t)=&\,0.
\end{align*} 8 t − 3 t 2 = t ( 8 − 3 t ) = 0 0. The solution in [ 2 , 4 ] [2,4] [ 2 , 4 ] t = 8 / 3 t=8/3 t = 8/3 1 / 3 1/3 1/3
8 3 . \boxed{\frac83}. 3 8 .
(d)
解法一
思路
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累积分布函数是从最左端累积到 t t t 1 ⩽ t < 2 1\leqslant t<2 1 ⩽ t < 2 2 ⩽ t ⩽ 4 2\leqslant t\leqslant4 2 ⩽ t ⩽ 4 F ( 2 ) = 1 / 3 F(2)=1/3 F ( 2 ) = 1/3 t t t
答题过程
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For 1 ⩽ t < 2 1\leqslant t<2 1 ⩽ t < 2
F ( t ) = ∫ 1 t 1 3 d x = t 3 − 1 3 . F(t)=\int_1^t\frac13\,\mathrm{d}x
=\frac{t}{3}-\frac13. F ( t ) = ∫ 1 t 3 1 d x = 3 t − 3 1 . Also, F ( 2 ) = 1 / 3 F(2)=1/3 F ( 2 ) = 1/3 2 ⩽ t ⩽ 4 2\leqslant t\leqslant4 2 ⩽ t ⩽ 4
F ( t ) = 1 3 + ∫ 2 t 1 22 ( 4 x 2 − x 3 ) d x = 1 3 + 1 22 [ 4 x 3 3 − x 4 4 ] 2 t = 2 t 3 33 − t 4 88 + 1 33 . \begin{align*}
&\,F(t)\\
=&\,\frac13
+\int_2^t\frac1{22}(4x^2-x^3)\,\mathrm{d}x \\[4mm]
=&\,\frac13
+\frac1{22}\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]_2^t \\[4mm]
=&\,\frac{2t^3}{33}-\frac{t^4}{88}+\frac1{33}.
\end{align*} = = = F ( t ) 3 1 + ∫ 2 t 22 1 ( 4 x 2 − x 3 ) d x 3 1 + 22 1 [ 3 4 x 3 − 4 x 4 ] 2 t 33 2 t 3 − 88 t 4 + 33 1 . Therefore,
F ( t ) = { 0 for t < 1 t 3 − 1 3 for 1 ⩽ t < 2 2 t 3 33 − t 4 88 + 1 33 for 2 ⩽ t ⩽ 4 1 for t > 4. \begin{align*}
&\,F(t)=\\
&\,\begin{cases}
\begin{aligned}
&\,0\\[-1mm]
&\,\quad \text{for }t<1
\end{aligned}\\[3mm]
\begin{aligned}
&\,\dfrac{t}{3}-\dfrac13\\[-1mm]
&\,\quad \text{for }1\leqslant t<2
\end{aligned}\\[3mm]
\begin{aligned}
&\,\dfrac{2t^3}{33}-\dfrac{t^4}{88}+\dfrac1{33}\\[-1mm]
&\,\quad \text{for }2\leqslant t\leqslant4
\end{aligned}\\[3mm]
\begin{aligned}
&\,1\\[-1mm]
&\,\quad \text{for }t>4.
\end{aligned}
\end{cases}
\end{align*} F ( t ) = ⎩ ⎨ ⎧ 0 for t < 1 3 t − 3 1 for 1 ⩽ t < 2 33 2 t 3 − 88 t 4 + 33 1 for 2 ⩽ t ⩽ 4 1 for t > 4.
解法二
思路
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也可以先对每一段密度函数作不定积分，再利用端点的累积概率确定常数。左段用 F ( 1 ) = 0 F(1)=0 F ( 1 ) = 0 F ( 4 ) = 1 F(4)=1 F ( 4 ) = 1 t = 2 t=2 t = 2 1 / 3 1/3 1/3
答题过程
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For 1 ⩽ t < 2 1\leqslant t<2 1 ⩽ t < 2
F ( t ) = t 3 + C 1 . F(t)=\frac{t}{3}+C_1. F ( t ) = 3 t + C 1 . Since F ( 1 ) = 0 F(1)=0 F ( 1 ) = 0
0 = 1 3 + C 1 , C 1 = − 1 3 . \begin{align*}
0=&\,\frac13+C_1,\\[4mm]
C_1=&\,-\frac13.
\end{align*} 0 = C 1 = 3 1 + C 1 , − 3 1 . Thus
F ( t ) = t 3 − 1 3 . F(t)=\frac{t}{3}-\frac13. F ( t ) = 3 t − 3 1 . For 2 ⩽ t ⩽ 4 2\leqslant t\leqslant4 2 ⩽ t ⩽ 4
F ( t ) = ∫ 1 22 ( 4 t 2 − t 3 ) d t = 2 t 3 33 − t 4 88 + C 2 . \begin{align*}
&\,F(t)\\[4mm]
=&\,\int\frac1{22}(4t^2-t^3)\,\mathrm{d}t\\[4mm]
=&\,\frac{2t^3}{33}-\frac{t^4}{88}+C_2.
\end{align*} = = F ( t ) ∫ 22 1 ( 4 t 2 − t 3 ) d t 33 2 t 3 − 88 t 4 + C 2 . Using F ( 4 ) = 1 F(4)=1 F ( 4 ) = 1
1 = 128 33 − 32 11 + C 2 = 32 33 + C 2 , C 2 = 1 33 . \begin{align*}
1=&\,\frac{128}{33}-\frac{32}{11}\\[4mm]
&\,\hspace{2pt}+C_2\\[4mm]
=&\,\frac{32}{33}+C_2,\\[4mm]
C_2=&\,\frac1{33}.
\end{align*} 1 = = C 2 = 33 128 − 11 32 + C 2 33 32 + C 2 , 33 1 . Hence, on this interval,
F ( t ) = 2 t 3 33 − t 4 88 + 1 33 . \begin{align*}
&\,F(t)\\[4mm]
=&\,\frac{2t^3}{33}-\frac{t^4}{88}+\frac1{33}.
\end{align*} = F ( t ) 33 2 t 3 − 88 t 4 + 33 1 . Both pieces give F ( 2 ) = 1 / 3 F(2)=1/3 F ( 2 ) = 1/3 F ( t ) = 0 F(t)=0 F ( t ) = 0 t < 1 t<1 t < 1 F ( t ) = 1 F(t)=1 F ( t ) = 1 t > 4 t>4 t > 4
(e)
解法一
思路
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利用上一小题求得的累积分布函数，P ( T > 3 ) P(T>3) P ( T > 3 ) 1 − F ( 3 ) 1-F(3) 1 − F ( 3 )
答题过程
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Using the result from part (d),
P ( T > 3 ) = 1 − F ( 3 ) = 1 − ( 2 ( 3 ) 3 33 − 3 4 88 + 1 33 ) = 67 264 = 0.254 (3 s.f.) . \begin{align*}
&\,P(T>3)\\
=&\,1-F(3) \\[4mm]
=&\,1-\left(
\frac{2(3)^3}{33}-\frac{3^4}{88}+\frac1{33}
\right) \\[4mm]
=&\,\frac{67}{264} \\[4mm]
=&\,\boxed{0.254}\quad\text{(3 s.f.)}.
\end{align*} = = = = P ( T > 3 ) 1 − F ( 3 ) 1 − ( 33 2 ( 3 ) 3 − 88 3 4 + 33 1 ) 264 67 0.254 (3 s.f.) .
解法二
思路
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也可以直接把概率密度函数从 3 积到 4。这条路线不依赖 (d) 的累积分布函数，可作为独立核对。
答题过程
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Alternatively,
P ( T > 3 ) = ∫ 3 4 1 22 ( 4 t 2 − t 3 ) d t = 1 22 [ 4 t 3 3 − t 4 4 ] 3 4 = 67 264 = 0.254 (3 s.f.) . \begin{align*}
&\,P(T>3)\\
=&\,\int_3^4\frac1{22}(4t^2-t^3)\,\mathrm{d}t \\[4mm]
=&\,\frac1{22}
\left[\frac{4t^3}{3}-\frac{t^4}{4}\right]_3^4 \\[4mm]
=&\,\frac{67}{264} \\[4mm]
=&\,\boxed{0.254}\quad\text{(3 s.f.)}.
\end{align*} = = = = P ( T > 3 ) ∫ 3 4 22 1 ( 4 t 2 − t 3 ) d t 22 1 [ 3 4 t 3 − 4 t 4 ] 3 4 264 67 0.254 (3 s.f.) . 
