题目
Problem
The diagram shows the continuous random variable X X X f \mathrm{f} f
f ( x ) = { a ( x − 1 ) , 1 ≤ x ≤ 5 , b ( x − 7 ) 2 , 5 ≤ x ≤ 7 , 0 , otherwise. \mathrm{f}(x)=\begin{cases}
a(x-1),&1\le x\le5,\\[2mm]
b(x-7)^2,&5\le x\le7,\\[2mm]
0,&\text{otherwise.}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ a ( x − 1 ) , b ( x − 7 ) 2 , 0 , 1 ≤ x ≤ 5 , 5 ≤ x ≤ 7 , otherwise. The upper quartile of X X X 5 5 5
(a) Show that a = 3 32 a=\frac{3}{32} a = 32 3 b = 3 32 b=\frac{3}{32} b = 32 3
(3)
(b) Find the cumulative distribution function of X X X
(3)
(c) The random variable Y Y Y Y = 10 − X Y=10-X Y = 10 − X
Find the cumulative distribution function of Y Y Y
(4)
题目中文翻译
图中所示的连续随机变量 X X X f \mathrm{f} f
f ( x ) = { a ( x − 1 ) , 1 ≤ x ≤ 5 , b ( x − 7 ) 2 , 5 ≤ x ≤ 7 , 0 , 其他情况。 \mathrm{f}(x)=\begin{cases}
a(x-1),&1\le x\le5,\\[2mm]
b(x-7)^2,&5\le x\le7,\\[2mm]
0,&\text{其他情况。}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ a ( x − 1 ) , b ( x − 7 ) 2 , 0 , 1 ≤ x ≤ 5 , 5 ≤ x ≤ 7 , 其他情况。 X X X 5 5 5
(a) 证明 a = 3 32 a=\frac{3}{32} a = 32 3 b = 3 32 b=\frac{3}{32} b = 32 3
(3)
(b) 求 X X X
(3)
(c) 随机变量 Y Y Y Y = 10 − X Y=10-X Y = 10 − X
求 Y Y Y
(4)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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上四分位数为 5 5 5 3 / 4 3/4 3/4 1 / 4 1/4 1/4
答题过程
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The area from 1 1 1 5 5 5
a ∫ 1 5 ( x − 1 ) d x = a [ ( x − 1 ) 2 2 ] 1 5 = 8 a = 3 4 , \begin{align*}
a\int_1^5(x-1)\,\mathrm{d}x
=&\,a\Big[\frac{(x-1)^2}{2}\Big]_1^5\\
=&\,8a=\frac34,
\end{align*} a ∫ 1 5 ( x − 1 ) d x = = a [ 2 ( x − 1 ) 2 ] 1 5 8 a = 4 3 , so a = 3 / 32 a=3/32 a = 3/32 5 5 5 7 7 7
b ∫ 5 7 ( x − 7 ) 2 d x = b [ ( x − 7 ) 3 3 ] 5 7 = 8 3 b = 1 4 , \begin{align*}
b\int_5^7(x-7)^2\,\mathrm{d}x
=&\,b\Big[\frac{(x-7)^3}{3}\Big]_5^7\\
=&\,\frac83b=\frac14,
\end{align*} b ∫ 5 7 ( x − 7 ) 2 d x = = b [ 3 ( x − 7 ) 3 ] 5 7 3 8 b = 4 1 , so b = 3 / 32 b=3/32 b = 3/32
解法二
思路
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利用图中两段在 x = 5 x=5 x = 5 a = b a=b a = b 3 / 4 3/4 3/4
答题过程
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The two parts of the graph meet at x = 5 x=5 x = 5 4 a 4a 4 a 4 b 4b 4 b a = b a=b a = b
Since the upper quartile is 5 5 5 5 5 5 3 / 4 3/4 3/4
1 2 ( 5 − 1 ) ( 4 a ) = 3 4 . \frac12(5-1)(4a)=\frac34. 2 1 ( 5 − 1 ) ( 4 a ) = 4 3 . Thus 8 a = 3 / 4 8a=3/4 8 a = 3/4 a = 3 / 32 a=3/32 a = 3/32 a = b a=b a = b b = 3 / 32 b=3/32 b = 3/32
(b)
解法一
思路
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左段从 1 1 1 1 1 1 x x x
答题过程
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For 1 ≤ x ≤ 5 1\le x\le5 1 ≤ x ≤ 5
F X ( x ) = 3 32 ∫ 1 x ( u − 1 ) d u = 3 64 [ ( u − 1 ) 2 ] 1 x = 3 ( x − 1 ) 2 64 . \begin{align*}
F_X(x)=&\,\frac3{32}\int_1^x(u-1)\,\mathrm{d}u\\
=&\,\frac3{64}\big[(u-1)^2\big]_1^x\\
=&\,\frac{3(x-1)^2}{64}.
\end{align*} F X ( x ) = = = 32 3 ∫ 1 x ( u − 1 ) d u 64 3 [ ( u − 1 ) 2 ] 1 x 64 3 ( x − 1 ) 2 . For 5 < x ≤ 7 5<x\le7 5 < x ≤ 7 x x x
3 32 ∫ x 7 ( u − 7 ) 2 d u = 1 32 [ ( u − 7 ) 3 ] x 7 = ( 7 − x ) 3 32 . \begin{align*}
\frac3{32}\int_x^7(u-7)^2\,\mathrm{d}u
=&\,\frac1{32}\big[(u-7)^3\big]_x^7\\
=&\,\frac{(7-x)^3}{32}.
\end{align*} 32 3 ∫ x 7 ( u − 7 ) 2 d u = = 32 1 [ ( u − 7 ) 3 ] x 7 32 ( 7 − x ) 3 . Thus
F X ( x ) = { 0 , x < 1 , 3 ( x − 1 ) 2 / 64 , 1 ≤ x ≤ 5 , 1 − ( 7 − x ) 3 / 32 , 5 < x ≤ 7 , 1 , x > 7. F_X(x)=
\begin{cases}
0, & x<1,\\
3(x-1)^2/64, & 1\le x\le5,\\
1-(7-x)^3/32, & 5<x\le7,\\
1, & x>7.
\end{cases} F X ( x ) = ⎩ ⎨ ⎧ 0 , 3 ( x − 1 ) 2 /64 , 1 − ( 7 − x ) 3 /32 , 1 , x < 1 , 1 ≤ x ≤ 5 , 5 < x ≤ 7 , x > 7.
(c)
解法一
思路
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Y = 10 − X Y=10-X Y = 10 − X Y ≤ y Y\le y Y ≤ y X ≥ 10 − y X\ge10-y X ≥ 10 − y
答题过程
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Since X X X F Y ( y ) = 1 − F X ( 10 − y ) F_Y(y)=1-F_X(10-y) F Y ( y ) = 1 − F X ( 10 − y ) 1 , 5 , 7 1,5,7 1 , 5 , 7 X X X 9 , 5 , 3 9,5,3 9 , 5 , 3 Y Y Y
F Y ( y ) = { 0 , y < 3 , ( y − 3 ) 3 / 32 , 3 ≤ y < 5 , 1 − 3 ( 9 − y ) 2 / 64 , 5 ≤ y ≤ 9 , 1 , y > 9. F_Y(y)=
\begin{cases}
0, & y<3,\\
(y-3)^3/32, & 3\le y<5,\\
1-3(9-y)^2/64, & 5\le y\le9,\\
1, & y>9.
\end{cases} F Y ( y ) = ⎩ ⎨ ⎧ 0 , ( y − 3 ) 3 /32 , 1 − 3 ( 9 − y ) 2 /64 , 1 , y < 3 , 3 ≤ y < 5 , 5 ≤ y ≤ 9 , y > 9. 
