题目
Problem
A continuous random variable X X X has cumulative distribution function F F F given by
F ( x ) = { 0 , x < 1 , 1 5 x + a , 1 ≤ x < 4 , 1 50 x 2 + b , 4 ≤ x ≤ 6 , 1 , x > 6 , F(x)=
\begin{cases}
0, & x < 1,\\
\frac{1}{5}x+a, & 1 \le x < 4,\\
\frac{1}{50}x^2+b, & 4 \le x \le 6,\\
1, & x > 6,
\end{cases} F ( x ) = ⎩ ⎨ ⎧ 0 , 5 1 x + a , 50 1 x 2 + b , 1 , x < 1 , 1 ≤ x < 4 , 4 ≤ x ≤ 6 , x > 6 ,
where a a a and b b b are constants.
(a) Find the value of a a a and the value of b b b .
[2]
(b) Find the probability density function of X X X .
[2]
(c) Given that E ( X ) = 529 150 \mathrm{E}(X)=\frac{529}{150} E ( X ) = 150 529 , find Var ( X ) \operatorname{Var}(X) Var ( X ) .
[3]
(d) Find the 10th and 90th percentiles of X X X .
[3]
题目中文翻译
连续随机变量 X X X 的累积分布函数 F F F 为
F ( x ) = { 0 , x < 1 , 1 5 x + a , 1 ≤ x < 4 , 1 50 x 2 + b , 4 ≤ x ≤ 6 , 1 , x > 6 , F(x)=
\begin{cases}
0, & x < 1,\\
\frac{1}{5}x+a, & 1 \le x < 4,\\
\frac{1}{50}x^2+b, & 4 \le x \le 6,\\
1, & x > 6,
\end{cases} F ( x ) = ⎩ ⎨ ⎧ 0 , 5 1 x + a , 50 1 x 2 + b , 1 , x < 1 , 1 ≤ x < 4 , 4 ≤ x ≤ 6 , x > 6 ,
其中 a a a 和 b b b 为常数。
(a) 求 a a a 和 b b b 的值。
(2)
(b) 求 X X X 的概率密度函数。
(2)
(c) 已知 E ( X ) = 529 150 \mathrm{E}(X)=\frac{529}{150} E ( X ) = 150 529 ,求 Var ( X ) \operatorname{Var}(X) Var ( X ) 。
(3)
(d) 求 X X X 的第 10 百分位数和第 90 百分位数。
(3)
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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连续型随机变量的分布函数在端点连续。用左端累计概率 0 0 0 、右端累计概率 1 1 1 求常数。
答题过程
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At x = 1 x=1 x = 1 , continuity gives 1 / 5 + a = 0 1/5+a=0 1/5 + a = 0 , so a = − 1 / 5 a=-1/5 a = − 1/5 . At x = 6 x=6 x = 6 it gives 36 / 50 + b = 1 36/50+b=1 36/50 + b = 1 , so b = 7 / 25 b=7/25 b = 7/25 . At the interior breakpoint 4 4 4 , both nonconstant pieces then have value 3 / 5 3/5 3/5 , as required.
(b)
解法一
思路
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对分布函数的每个非恒定部分求导。区间外累计概率不变,密度为零。
答题过程
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One version of the density is
f X ( x ) = { 1 / 5 , 1 ≤ x < 4 , x / 25 , 4 ≤ x ≤ 6 , 0 , otherwise . f_X(x)=
\begin{cases}
1/5, & 1\le x<4,\\
x/25, & 4\le x\le6,\\
0, & \text{otherwise}.
\end{cases} f X ( x ) = ⎩ ⎨ ⎧ 1/5 , x /25 , 0 , 1 ≤ x < 4 , 4 ≤ x ≤ 6 , otherwise .
(c)
解法一
思路
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用 x 2 x^2 x 2 乘各段密度再积分,求出二阶矩。最后减去题给均值的平方。
答题过程
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The second moment is
E [ X 2 ] = ∫ 1 4 x 2 5 d x + ∫ 4 6 x 3 25 d x = [ x 3 15 ] 1 4 + [ x 4 100 ] 4 6 = 73 5 . \begin{align*}
&\,\operatorname E[X^2]=\int_1^4\frac{x^2}{5}\,\mathrm{d}x\\
+&\,\int_4^6\frac{x^3}{25}\,\mathrm{d}x\\
=&\,\Big[\frac{x^3}{15}\Big]_1^4\\
+&\,\Big[\frac{x^4}{100}\Big]_4^6\\
=&\,\frac{73}{5}.
\end{align*} + = + = E [ X 2 ] = ∫ 1 4 5 x 2 d x ∫ 4 6 25 x 3 d x [ 15 x 3 ] 1 4 [ 100 x 4 ] 4 6 5 73 .
Thus Var ( X ) = 73 / 5 − ( 529 / 150 ) 2 = 48659 / 22500 ≈ 2.16 \operatorname{Var}(X)=73/5-(529/150)^2=48659/22500\approx2.16 Var ( X ) = 73/5 − ( 529/150 ) 2 = 48659/22500 ≈ 2.16 .
(d)
解法一
思路
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分界点 4 4 4 的累计概率是 0.6 0.6 0.6 ,所以两个百分位数落在不同的分段。
答题过程
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For the tenth percentile, solve x / 5 − 1 / 5 = 0.1 x/5-1/5=0.1 x /5 − 1/5 = 0.1 , giving x = 1.5 x=1.5 x = 1.5 . For the ninetieth, solve x 2 / 50 + 7 / 25 = 0.9 x^2/50+7/25=0.9 x 2 /50 + 7/25 = 0.9 , giving x = 31 ≈ 5.57 x=\sqrt{31}\approx5.57 x = 31 ≈ 5.57 on [ 4 , 6 ] [4,6] [ 4 , 6 ] .