题目
Problem
The continuous random variable X X X f f f
f ( x ) = { k , 0 ≤ x < 1 , k x , 1 ≤ x ≤ 2 , 0 , otherwise , f(x)=
\begin{cases}
k, & 0 \leq x < 1,\\
kx, & 1 \leq x \leq 2,\\
0, & \text{otherwise},
\end{cases} f ( x ) = ⎩ ⎨ ⎧ k , k x , 0 , 0 ≤ x < 1 , 1 ≤ x ≤ 2 , otherwise , where k k k
(a) Show that k = 2 5 k=\frac{2}{5} k = 5 2
[1]
(b) Find the interquartile range of X X X
[5]
(c) Find Var ( X ) \operatorname{Var}(X) Var ( X )
[4]
题目中文翻译
连续随机变量 X X X f f f
f ( x ) = { k , 0 ≤ x < 1 , k x , 1 ≤ x ≤ 2 , 0 , otherwise , f(x)=
\begin{cases}
k, & 0 \leq x < 1,\\
kx, & 1 \leq x \leq 2,\\
0, & \text{otherwise},
\end{cases} f ( x ) = ⎩ ⎨ ⎧ k , k x , 0 , 0 ≤ x < 1 , 1 ≤ x ≤ 2 , otherwise , 其中 k k k
(a) 证明 k = 2 5 k=\frac{2}{5} k = 5 2
(1)
(b) 求 X X X
(5)
(c) 求 Var ( X ) \operatorname{Var}(X) Var ( 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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将两段密度分别积分，并令总面积为 1 1 1
答题过程
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The total integral is
∫ 0 1 k d x + ∫ 1 2 k x d x = k + 3 k 2 = 5 k 2 . \begin{align*}
&\,\int_0^1 k\,\mathrm{d}x+\int_1^2kx\,\mathrm{d}x\\
=&\,k+\frac{3k}{2}=\frac{5k}{2}.
\end{align*} = ∫ 0 1 k d x + ∫ 1 2 k x d x k + 2 3 k = 2 5 k . Thus 5 k / 2 = 1 5k/2=1 5 k /2 = 1 k = 2 / 5 k=2/5 k = 2/5
(b)
解法一
思路
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分界点 x = 1 x=1 x = 1 2 / 5 2/5 2/5
答题过程
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On 0 ≤ x < 1 0\le x<1 0 ≤ x < 1 F X ( x ) = 2 x / 5 F_X(x)=2x/5 F X ( x ) = 2 x /5 5 / 8 5/8 5/8 1 ≤ x ≤ 2 1\le x\le2 1 ≤ x ≤ 2
2 5 + ∫ 1 x 2 5 u d u = x 2 + 1 5 . \frac25+\int_1^x\frac25u\,\mathrm{d}u=\frac{x^2+1}{5}. 5 2 + ∫ 1 x 5 2 u d u = 5 x 2 + 1 . Setting this equal to 3 / 4 3/4 3/4 x 2 = 11 / 4 x^2=11/4 x 2 = 11/4 11 / 2 \sqrt{11}/2 11 /2 11 / 2 − 5 / 8 ≈ 1.03 \sqrt{11}/2-5/8\approx1.03 11 /2 − 5/8 ≈ 1.03
(c)
解法一
思路
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分别积分 x f ( x ) x f(x) x f ( x ) x 2 f ( x ) x^2 f(x) x 2 f ( x )
答题过程
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The mean is
E [ X ] = 2 5 ( ∫ 0 1 x d x + ∫ 1 2 x 2 d x ) = 2 5 ( [ x 2 2 ] 0 1 + [ x 3 3 ] 1 2 ) = 2 5 ( 1 2 + 7 3 ) = 17 15 . \begin{align*}
&\,\operatorname E[X]\\
=&\,\frac25\Big(\int_0^1x\,\mathrm{d}x+\int_1^2x^2\,\mathrm{d}x\Big)\\
=&\,\frac25\Big(\Big[\frac{x^2}{2}\Big]_0^1+\Big[\frac{x^3}{3}\Big]_1^2\Big)\\
=&\,\frac25\Big(\frac12+\frac73\Big)=\frac{17}{15}.
\end{align*} = = = E [ X ] 5 2 ( ∫ 0 1 x d x + ∫ 1 2 x 2 d x ) 5 2 ( [ 2 x 2 ] 0 1 + [ 3 x 3 ] 1 2 ) 5 2 ( 2 1 + 3 7 ) = 15 17 . The second moment is
E [ X 2 ] = 2 5 ( ∫ 0 1 x 2 d x + ∫ 1 2 x 3 d x ) = 2 5 ( [ x 3 3 ] 0 1 + [ x 4 4 ] 1 2 ) = 2 5 ( 1 3 + 15 4 ) = 49 30 . \begin{align*}
&\,\operatorname E[X^2]\\
=&\,\frac25\Big(\int_0^1x^2\,\mathrm{d}x+\int_1^2x^3\,\mathrm{d}x\Big)\\
=&\,\frac25\Big(\Big[\frac{x^3}{3}\Big]_0^1+\Big[\frac{x^4}{4}\Big]_1^2\Big)\\
=&\,\frac25\Big(\frac13+\frac{15}{4}\Big)=\frac{49}{30}.
\end{align*} = = = E [ X 2 ] 5 2 ( ∫ 0 1 x 2 d x + ∫ 1 2 x 3 d x ) 5 2 ( [ 3 x 3 ] 0 1 + [ 4 x 4 ] 1 2 ) 5 2 ( 3 1 + 4 15 ) = 30 49 . Thus Var ( X ) = 49 / 30 − ( 17 / 15 ) 2 = 157 / 450 \operatorname{Var}(X)=49/30-(17/15)^2=157/450 Var ( X ) = 49/30 − ( 17/15 ) 2 = 157/450
