题目
Problem
The diagram shows the continuous random variable X X X with probability density function f f f given by
f ( x ) = { 1 128 ( 4 a x − b x 3 ) , 0 ≤ x ≤ 4 , c , 4 ≤ x ≤ 6 , 0 , otherwise , f(x)=
\begin{cases}
\frac{1}{128}(4ax-bx^3), & 0 \le x \le 4,\\
c, & 4 \le x \le 6,\\
0, & \text{otherwise},
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 128 1 ( 4 a x − b x 3 ) , c , 0 , 0 ≤ x ≤ 4 , 4 ≤ x ≤ 6 , otherwise ,
where a a a , b b b and c c c are constants.
The upper quartile of X X X is equal to 4.
(a) Show that c = 1 8 c=\frac{1}{8} c = 8 1 and find the values of a a a and b b b .
[4]
(b) Find the exact value of the median of X X X .
[3]
(c) Find E ( X ) \operatorname{E}(\sqrt{X}) E ( X ) , giving your answer correct to 2 decimal places.
[3]
题目中文翻译
图中显示连续随机变量 X X X ,其概率密度函数 f f f 为
f ( x ) = { 1 128 ( 4 a x − b x 3 ) , 0 ≤ x ≤ 4 , c , 4 ≤ x ≤ 6 , 0 , 其他情况 , f(x)=
\begin{cases}
\frac{1}{128}(4ax-bx^3), & 0 \le x \le 4,\\
c, & 4 \le x \le 6,\\
0, & \text{其他情况},
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 128 1 ( 4 a x − b x 3 ) , c , 0 , 0 ≤ x ≤ 4 , 4 ≤ x ≤ 6 , 其他情况 ,
其中 a a a 、b b b 和 c c c 为常数。
X X X 的上四分位数等于 4。
(a) 证明 c = 1 8 c=\frac{1}{8} c = 8 1 ,并求 a a a 和 b b b 的值。
(4)
(b) 求 X X X 的中位数的精确值。
(3)
(c) 求 E ( X ) \operatorname{E}(\sqrt{X}) E ( X ) ,答案精确到 2 位小数。
(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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上四分位数把右段的概率定为 1 / 4 1/4 1/4 。再用左段的面积和两段在 x = 4 x=4 x = 4 的接合条件求系数。
答题过程
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Since 4 4 4 is the upper quartile, the area from 4 4 4 to 6 6 6 is 1 / 4 1/4 1/4 . It equals 2 c 2c 2 c , so c = 1 / 8 c=1/8 c = 1/8 . The area from 0 0 0 to 4 4 4 gives
1 128 [ 2 a x 2 − b 4 x 4 ] 0 4 = 32 a − 64 b 128 = 3 4 , \begin{align*}
&\,\frac1{128}\Big[2ax^2-\frac b4x^4\Big]_0^4\\
=&\,\frac{32a-64b}{128}=\frac34,
\end{align*} = 128 1 [ 2 a x 2 − 4 b x 4 ] 0 4 128 32 a − 64 b = 4 3 ,
so a − 2 b = 3 a-2b=3 a − 2 b = 3 . The two density expressions agree at 4 4 4 , giving ( 16 a − 64 b ) / 128 = 1 / 8 (16a-64b)/128=1/8 ( 16 a − 64 b ) /128 = 1/8 , or a − 4 b = 1 a-4b=1 a − 4 b = 1 . Subtraction yields b = 1 b=1 b = 1 and a = 5 a=5 a = 5 .
(b)
解法一
思路
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中位数小于上四分位数 4 4 4 ,所以用左段积分。令 u = m 2 u=m^2 u = m 2 把方程降成二次方程。
答题过程
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The median m m m is between 0 0 0 and 4 4 4 . It satisfies
1 128 ∫ 0 m ( 20 x − x 3 ) d x = 5 m 2 64 − m 4 512 = 1 2 . \begin{align*}
&\,\frac1{128}\int_0^m(20x-x^3)\,\mathrm{d}x\\
=&\,\frac{5m^2}{64}-\frac{m^4}{512}=\frac12.
\end{align*} = 128 1 ∫ 0 m ( 20 x − x 3 ) d x 64 5 m 2 − 512 m 4 = 2 1 .
Thus m 4 − 40 m 2 + 256 = 0 m^4-40m^2+256=0 m 4 − 40 m 2 + 256 = 0 , or ( m 2 − 8 ) ( m 2 − 32 ) = 0 (m^2-8)(m^2-32)=0 ( m 2 − 8 ) ( m 2 − 32 ) = 0 . Only m 2 = 8 m^2=8 m 2 = 8 gives 0 < m < 4 0<m<4 0 < m < 4 . Hence the exact median is m = 2 2 m=2\sqrt2 m = 2 2 .
(c)
解法一
思路
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在两段上分别积分 x \sqrt x x 乘密度,最后将结果取到两位小数。
答题过程
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The expectation is
E [ X ] = 1 128 ∫ 0 4 ( 20 x 3 / 2 − x 7 / 2 ) d x + 1 8 ∫ 4 6 x 1 / 2 d x = 1 128 [ 8 x 5 / 2 − 2 9 x 9 / 2 ] 0 4 + 1 12 [ x 3 / 2 ] 4 6 = 10 9 + 6 6 − 8 12 = 4 9 + 6 2 . \begin{align*}
&\,\operatorname E[\sqrt X]\\
=&\,\frac1{128}\int_0^4(20x^{3/2}-x^{7/2})\,\mathrm{d}x\\
+&\,\frac18\int_4^6x^{1/2}\,\mathrm{d}x\\
=&\,\frac1{128}\Big[8x^{5/2}-\frac29x^{9/2}\Big]_0^4\\
+&\,\frac1{12}\Big[x^{3/2}\Big]_4^6\\
=&\,\frac{10}{9}+\frac{6\sqrt6-8}{12}\\
=&\,\frac49+\frac{\sqrt6}{2}.
\end{align*} = + = + = = E [ X ] 128 1 ∫ 0 4 ( 20 x 3/2 − x 7/2 ) d x 8 1 ∫ 4 6 x 1/2 d x 128 1 [ 8 x 5/2 − 9 2 x 9/2 ] 0 4 12 1 [ x 3/2 ] 4 6 9 10 + 12 6 6 − 8 9 4 + 2 6 .
This is 1.67 1.67 1.67 to two decimal places.