题目
Problem
A random variable X X X
f ( x ) = { 3 16 ( 4 x − x 2 ) 2 ≤ x ≤ 4 , 0 otherwise . \mathrm{f}(x)=\begin{cases}
\dfrac3{16}(4x-x^2) & 2\leq x\leq4,\\
0 & \text{otherwise}.
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 16 3 ( 4 x − x 2 ) 0 2 ≤ x ≤ 4 , otherwise . (a) Show that E ( X ) = 11 4 \mathrm{E}(X)=\dfrac{11}{4} E ( X ) = 4 11
(3)
(b) Find V a r ( X ) \mathrm{Var}(X) Var ( X )
(3)
(c) Given that the median of X X X m m m P ( m < X < 3 ) \mathrm{P}(m<X<3) P ( m < X < 3 )
(3)
题目中文翻译
随机变量 X X X
f ( x ) = { 3 16 ( 4 x − x 2 ) 2 ≤ x ≤ 4 , 0 其他情况。 \mathrm{f}(x)=\begin{cases}
\dfrac3{16}(4x-x^2) & 2\leq x\leq4,\\
0 & \text{其他情况。}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 16 3 ( 4 x − x 2 ) 0 2 ≤ x ≤ 4 , 其他情况。 (a) 证明 E ( X ) = 11 4 \mathrm{E}(X)=\dfrac{11}{4} E ( X ) = 4 11
(3)
(b) 求 V a r ( X ) \mathrm{Var}(X) Var ( X )
(3)
(c) 已知 X X X m m m P ( m < X < 3 ) \mathrm{P}(m<X<3) P ( m < X < 3 )
(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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将x乘密度积分，明确代入4与2的精确结果。
答题过程
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E ( X ) = 3 16 ∫ 2 4 ( 4 x 2 − x 3 ) d x = 3 16 [ 4 x 3 3 − x 4 4 ] 2 4 = 3 16 ( 64 3 − 20 3 ) = 11 4 , \begin{align*}
\mathrm{E}(X)=&\,\frac3{16}\int_2^4(4x^2-x^3)\,\mathrm{d}x\\[4mm]
=&\,\frac3{16}\bigg[\frac{4x^3}{3}-\frac{x^4}{4}\bigg]_2^4\\[4mm]
=&\,\frac3{16}\bigg(\frac{64}{3}-\frac{20}{3}\bigg)\\[4mm]
=&\,\boxed{\frac{11}{4}},
\end{align*} E ( X ) = = = = 16 3 ∫ 2 4 ( 4 x 2 − x 3 ) d x 16 3 [ 3 4 x 3 − 4 x 4 ] 2 4 16 3 ( 3 64 − 3 20 ) 4 11 , as required.
(b)
解法一
思路
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积分求二阶矩，再减去上一问均值的平方。
答题过程
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E ( X 2 ) = 3 16 ∫ 2 4 ( 4 x 3 − x 4 ) d x = 3 16 [ x 4 − x 5 5 ] 2 4 = 3 16 ( 256 5 − 48 5 ) = 39 5 . \begin{align*}
\mathrm{E}(X^2)=&\,\frac3{16}\int_2^4(4x^3-x^4)\,\mathrm{d}x\\[4mm]
=&\,\frac3{16}\bigg[x^4-\frac{x^5}{5}\bigg]_2^4\\[4mm]
=&\,\frac3{16}\bigg(\frac{256}{5}-\frac{48}{5}\bigg)\\[4mm]
=&\,\frac{39}{5}.
\end{align*} E ( X 2 ) = = = = 16 3 ∫ 2 4 ( 4 x 3 − x 4 ) d x 16 3 [ x 4 − 5 x 5 ] 2 4 16 3 ( 5 256 − 5 48 ) 5 39 . Hence
V a r ( X ) = 39 5 − ( 11 4 ) 2 = 19 80 . \mathrm{Var}(X)=\frac{39}{5}-\bigg(\frac{11}{4}\bigg)^2
=\boxed{\frac{19}{80}}. Var ( X ) = 5 39 − ( 4 11 ) 2 = 80 19 .
(c)
解法一
思路
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中位数左侧概率为一半。先求3左侧概率11/16，再相减即可，不需求出m。
答题过程
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P ( X < 3 ) = 3 16 ∫ 2 3 ( 4 x − x 2 ) d x = 3 16 [ 2 x 2 − x 3 3 ] 2 3 = 3 16 [ ( 18 − 9 ) − ( 8 − 8 3 ) ] = 11 16 > 1 2 . \begin{align*}
\mathrm{P}(X<3)
=&\,\frac3{16}\int_2^3(4x-x^2)\,\mathrm{d}x\\[4mm]
=&\,\frac3{16}\bigg[2x^2-\frac{x^3}{3}\bigg]_2^3\\[4mm]
=&\,\frac3{16}\bigg[(18-9)-\bigg(8-\frac83\bigg)\bigg]\\[4mm]
=&\,\frac{11}{16}>\frac12.
\end{align*} P ( X < 3 ) = = = = 16 3 ∫ 2 3 ( 4 x − x 2 ) d x 16 3 [ 2 x 2 − 3 x 3 ] 2 3 16 3 [ ( 18 − 9 ) − ( 8 − 3 8 ) ] 16 11 > 2 1 . Thus the median m m m P ( X < m ) = 1 2 \mathrm{P}(X<m)=\frac12 P ( X < m ) = 2 1
P ( m < X < 3 ) = 11 16 − 1 2 = 3 16 . \mathrm{P}(m<X<3)=\frac{11}{16}-\frac12=\boxed{\frac3{16}}. P ( m < X < 3 ) = 16 11 − 2 1 = 16 3 . 