题目
Problem
The continuous random variable X X X f \mathrm{f} f
f ( x ) = { k x ( 4 − x ) , 0 ≤ x < 2 , k ( 6 − x ) , 2 ≤ x ≤ 6 , 0 , otherwise, \mathrm{f}(x)=\begin{cases}
kx(4-x),&0\le x<2,\\[2mm]
k(6-x),&2\le x\le6,\\[2mm]
0,&\text{otherwise,}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ k x ( 4 − x ) , k ( 6 − x ) , 0 , 0 ≤ x < 2 , 2 ≤ x ≤ 6 , otherwise, where k k k
(a) Show that k = 3 40 k=\frac{3}{40} k = 40 3
(1)
(b) Given that E ( X ) = 2.5 \mathrm{E}(X)=2.5 E ( X ) = 2.5 Var ( X ) \operatorname{Var}(X) Var ( X )
(3)
(c) Find the median value of X X X
(4)
题目中文翻译
连续随机变量 X X X f \mathrm{f} f
f ( x ) = { k x ( 4 − x ) , 0 ≤ x < 2 , k ( 6 − x ) , 2 ≤ x ≤ 6 , 0 , 其他情况， \mathrm{f}(x)=\begin{cases}
kx(4-x),&0\le x<2,\\[2mm]
k(6-x),&2\le x\le6,\\[2mm]
0,&\text{其他情况，}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ k x ( 4 − x ) , k ( 6 − x ) , 0 , 0 ≤ x < 2 , 2 ≤ x ≤ 6 , 其他情况， 其中 k k k
(a) 证明 k = 3 40 k=\frac{3}{40} k = 40 3
(1)
(b) 已知 E ( X ) = 2.5 \mathrm{E}(X)=2.5 E ( X ) = 2.5 Var ( X ) \operatorname{Var}(X) Var ( X )
(3)
(c) 求 X X X
(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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分别积分两段密度。两块面积相加为 1 1 1
答题过程
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The total integral is
k [ 2 x 2 − x 3 3 ] 0 2 + k [ 6 x − x 2 2 ] 2 6 . \begin{align*}
&\,k\Big[2x^2-\frac{x^3}{3}\Big]_0^2\\
&\,\hspace{2pt}+k\Big[6x-\frac{x^2}{2}\Big]_2^6.
\end{align*} k [ 2 x 2 − 3 x 3 ] 0 2 + k [ 6 x − 2 x 2 ] 2 6 . Evaluating the two parts,
k ∫ 0 2 ( 4 x − x 2 ) d x + k ∫ 2 6 ( 6 − x ) d x = k ( 16 3 + 8 ) = 40 k 3 . \begin{align*}
&\,k\int_0^2(4x-x^2)\,\mathrm{d}x+k\int_2^6(6-x)\,\mathrm{d}x\\
=&\,k\Big(\frac{16}{3}+8\Big)\\
=&\,\frac{40k}{3}.
\end{align*} = = k ∫ 0 2 ( 4 x − x 2 ) d x + k ∫ 2 6 ( 6 − x ) d x k ( 3 16 + 8 ) 3 40 k . Thus 40 k / 3 = 1 40k/3=1 40 k /3 = 1 k = 3 / 40 k=3/40 k = 3/40
(b)
解法一
思路
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用 x 2 x^2 x 2
答题过程
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The second moment is
E [ X 2 ] = 3 40 ( ∫ 0 2 ( 4 x 3 − x 4 ) d x + ∫ 2 6 ( 6 x 2 − x 3 ) d x ) = 3 40 ( [ x 4 − x 5 5 ] 0 2 + [ 2 x 3 − x 4 4 ] 2 6 ) = 3 40 ( 48 5 + 96 ) = 7.92. \begin{align*}
&\,\operatorname E[X^2]\\
=&\,\frac3{40}\Big(\int_0^2(4x^3-x^4)\,\mathrm{d}x\\
+&\,\int_2^6(6x^2-x^3)\,\mathrm{d}x\Big)\\
=&\,\frac3{40}\Big(\Big[x^4-\frac{x^5}{5}\Big]_0^2\\
&\,\hspace{2pt}+\Big[2x^3-\frac{x^4}{4}\Big]_2^6\Big)\\
=&\,\frac3{40}\Big(\frac{48}{5}+96\Big)=7.92.
\end{align*} = + = = E [ X 2 ] 40 3 ( ∫ 0 2 ( 4 x 3 − x 4 ) d x ∫ 2 6 ( 6 x 2 − x 3 ) d x ) 40 3 ( [ x 4 − 5 x 5 ] 0 2 + [ 2 x 3 − 4 x 4 ] 2 6 ) 40 3 ( 5 48 + 96 ) = 7.92. Therefore Var ( X ) = 7.92 − 2.5 2 = 1.67 \operatorname{Var}(X)=7.92-2.5^2=1.67 Var ( X ) = 7.92 − 2. 5 2 = 1.67
(c)
解法一
思路
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到 x = 2 x=2 x = 2
答题过程
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The probability below 2 2 2 ( 3 / 40 ) ( 16 / 3 ) = 0.4 (3/40)(16/3)=0.4 ( 3/40 ) ( 16/3 ) = 0.4 m m m ( 2 , 6 ) (2,6) ( 2 , 6 )
3 40 ∫ m 6 ( 6 − x ) d x = 3 80 ( 6 − m ) 2 = 1 2 . \begin{align*}
&\,\frac3{40}\int_m^6(6-x)\,\mathrm{d}x\\
=&\,\frac3{80}(6-m)^2=\frac12.
\end{align*} = 40 3 ∫ m 6 ( 6 − x ) d x 80 3 ( 6 − m ) 2 = 2 1 . Thus m = 6 − 40 / 3 = 6 − 2 30 / 3 ≈ 2.35 m=6-\sqrt{40/3}=6-2\sqrt{30}/3\approx2.35 m = 6 − 40/3 = 6 − 2 30 /3 ≈ 2.35
解法二
思路
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累积到 2 2 2 0.4 0.4 0.4 2 2 2 0.1 0.1 0.1
答题过程
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Since P ( X ≤ 2 ) = 0.4 P(X\le2)=0.4 P ( X ≤ 2 ) = 0.4 m m m ( 2 , 6 ) (2,6) ( 2 , 6 )
0.4 + 3 40 ∫ 2 m ( 6 − x ) d x = 1 2 , 3 40 [ 6 x − x 2 2 ] 2 m = 1 10 , 6 m − m 2 2 − 10 = 4 3 , m 2 − 12 m + 68 3 = 0. \begin{align*}
0.4+\frac3{40}\int_2^m(6-x)\,\mathrm{d}x=&\,\frac12,\\
\frac3{40}\Big[6x-\frac{x^2}{2}\Big]_2^m=&\,\frac1{10},\\
6m-\frac{m^2}{2}-10=&\,\frac43,\\
m^2-12m+\frac{68}{3}=&\,0.
\end{align*} 0.4 + 40 3 ∫ 2 m ( 6 − x ) d x = 40 3 [ 6 x − 2 x 2 ] 2 m = 6 m − 2 m 2 − 10 = m 2 − 12 m + 3 68 = 2 1 , 10 1 , 3 4 , 0. Thus m = 6 ± 2 30 / 3 m=6\pm2\sqrt{30}/3 m = 6 ± 2 30 /3 6 6 6
m = 6 − 2 30 3 ≈ 2.35. m=6-\frac{2\sqrt{30}}3\approx2.35. m = 6 − 3 2 30 ≈ 2.35. 