题目
Problem
A random variable X X X
f ( x ) = { a − x 0 ≤ x ≤ a , 0 otherwise , f(x)=
\begin{cases}
a-x & 0\le x\le a,\\
0 & \text{otherwise},
\end{cases} f ( x ) = { a − x 0 0 ≤ x ≤ a , otherwise , where a a a
(a) Show that E ( X ) = k 2 \mathrm{E}(X)=k\sqrt{2} E ( X ) = k 2 k k k
(6)
(b) Find the exact value of the median of X X X
(4)
题目中文翻译
随机变量 X X X
f ( x ) = { a − x 0 ≤ x ≤ a , 0 其他情况 , f(x)=
\begin{cases}
a-x & 0\le x\le a,\\
0 & \text{其他情况},
\end{cases} f ( x ) = { a − x 0 0 ≤ x ≤ a , 其他情况 , 其中 a a a
(a) 证明 E ( X ) = k 2 \mathrm{E}(X)=k\sqrt{2} E ( X ) = k 2 k k k
(6)
(b) 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
展开
先用密度总积分为 1 1 1 a a a x f ( x ) xf(x) x f ( x )
答题过程
展开
A valid density requires a > 0 a>0 a > 0
∫ 0 a ( a − x ) d x = 1 , [ a x − x 2 2 ] 0 a = 1 , a 2 2 = 1. \begin{align*}
\int_0^a(a-x)\,\mathrm{d}x=&\,1,\\[4mm]
\bigg[ax-\frac{x^2}2\bigg]_0^a=&\,1,\\[4mm]
\frac{a^2}2=&\,1.
\end{align*} ∫ 0 a ( a − x ) d x = [ a x − 2 x 2 ] 0 a = 2 a 2 = 1 , 1 , 1. Hence a = 2 a=\sqrt2 a = 2
E ( X ) = ∫ 0 2 x ( 2 − x ) d x = [ 2 2 x 2 − x 3 3 ] 0 2 = 2 − 2 2 3 = 2 3 . \begin{align*}
\mathrm E(X)
=&\,\int_0^{\sqrt2}x(\sqrt2-x)\,\mathrm{d}x\\[4mm]
=&\,\bigg[\frac{\sqrt2}2x^2-\frac{x^3}3\bigg]_0^{\sqrt2}\\[4mm]
=&\,\sqrt2-\frac{2\sqrt2}3\\[4mm]
=&\,\frac{\sqrt2}3.
\end{align*} E ( X ) = = = = ∫ 0 2 x ( 2 − x ) d x [ 2 2 x 2 − 3 x 3 ] 0 2 2 − 3 2 2 3 2 . Thus k = 1 3 \boxed{k=\frac13} k = 3 1
(b)
解法一
思路
展开
中位数左侧概率为 1 / 2 1/2 1/2
答题过程
展开
The median m m m P ( X ≤ m ) = 1 / 2 \mathrm P(X\le m)=1/2 P ( X ≤ m ) = 1/2
∫ 0 m ( 2 − x ) d x = 1 2 , 2 m − m 2 2 = 1 2 , m 2 − 2 2 m + 1 = 0. \begin{align*}
\int_0^m(\sqrt2-x)\,\mathrm{d}x=&\,\frac12,\\[4mm]
\sqrt2m-\frac{m^2}2=&\,\frac12,\\[4mm]
m^2-2\sqrt2m+1=&\,0.
\end{align*} ∫ 0 m ( 2 − x ) d x = 2 m − 2 m 2 = m 2 − 2 2 m + 1 = 2 1 , 2 1 , 0. The quadratic formula gives
m = 2 2 ± 8 − 4 2 = 2 ± 1. \begin{align*}
m=&\,\frac{2\sqrt2\pm\sqrt{8-4}}2\\[4mm]
=&\,\sqrt2\pm1.
\end{align*} m = = 2 2 2 ± 8 − 4 2 ± 1. The value 2 + 1 \sqrt2+1 2 + 1 [ 0 , 2 ] [0,\sqrt2] [ 0 , 2 ]
m = 2 − 1 . \boxed{m=\sqrt2-1}. m = 2 − 1 .
解法二
思路
展开
也可用中位数右侧的三角形面积：该三角形的高与底都等于 2 − m \sqrt2-m 2 − m 1 / 2 1/2 1/2
答题过程
展开
For 0 ≤ m ≤ 2 0\le m\le\sqrt2 0 ≤ m ≤ 2 m m m 2 − m \sqrt2-m 2 − m
At the median this area is 1 / 2 1/2 1/2
1 2 ( 2 − m ) 2 = 1 2 . \frac12(\sqrt2-m)^2=\frac12. 2 1 ( 2 − m ) 2 = 2 1 . Since 2 − m ≥ 0 \sqrt2-m\ge0 2 − m ≥ 0 2 − m = 1 \sqrt2-m=1 2 − m = 1
m = 2 − 1 . \boxed{m=\sqrt2-1}. m = 2 − 1 . 