题目
Problem
The function f ( x ) f(x) f ( x )
f ( x ) = { 1 9 ( x + 5 ) ( 3 − x ) , 1 ≤ x ≤ 4 0 , otherwise f(x)=
\begin{cases}
\dfrac{1}{9}(x+5)(3-x), & 1 \leq x \leq 4 \\
0, & \text{otherwise}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 9 1 ( x + 5 ) ( 3 − x ) , 0 , 1 ≤ x ≤ 4 otherwise Albert believes that f ( x ) f(x) f ( x )
(a) Sketch f ( x ) f(x) f ( x )
(3)
The continuous random variable Y Y Y
g ( y ) = { k y ( 12 − y 2 ) , 1 ≤ y ≤ 3 0 , otherwise g(y)=
\begin{cases}
ky(12-y^2), & 1 \leq y \leq 3 \\
0, & \text{otherwise}
\end{cases} g ( y ) = { k y ( 12 − y 2 ) , 0 , 1 ≤ y ≤ 3 otherwise where k k k
(b) Use calculus to find the mode of Y Y Y
(3)
(c) Use algebraic integration to find the value of k k k
(3)
(d) Find the median of Y Y Y
(3)
(e) Describe the skewness of the distribution of Y Y Y
(2)
题目中文翻译
函数 f ( x ) f(x) f ( x )
f ( x ) = { 1 9 ( x + 5 ) ( 3 − x ) , 1 ≤ x ≤ 4 0 , otherwise f(x)=
\begin{cases}
\dfrac{1}{9}(x+5)(3-x), & 1 \leq x \leq 4 \\
0, & \text{otherwise}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 9 1 ( x + 5 ) ( 3 − x ) , 0 , 1 ≤ x ≤ 4 otherwise Albert 认为 f ( x ) f(x) f ( x )
(a) 画出 f ( x ) f(x) f ( x )
连续随机变量 Y Y Y
g ( y ) = { k y ( 12 − y 2 ) , 1 ≤ y ≤ 3 0 , otherwise g(y)=
\begin{cases}
ky(12-y^2), & 1 \leq y \leq 3 \\
0, & \text{otherwise}
\end{cases} g ( y ) = { k y ( 12 − y 2 ) , 0 , 1 ≤ y ≤ 3 otherwise 其中 k k k
(b) 用微积分求 Y Y Y
(c) 用代数积分求 k k k
(d) 求 Y Y Y
(e) 描述 Y Y Y
解答
(a)
解法一
思路
展开
在区间 [ 1 , 4 ] [1,4] [ 1 , 4 ] f ( x ) f(x) f ( x )
答题过程
展开
For 1 ⩽ x ⩽ 4 1\leqslant x\leqslant4 1 ⩽ x ⩽ 4
f ( x ) = 1 9 ( − x 2 − 2 x + 15 ) . f(x)=\frac19(-x^2-2x+15). f ( x ) = 9 1 ( − x 2 − 2 x + 15 ) . The curve is a decreasing quadratic on this interval, passing through
( 1 , 4 3 ) , ( 3 , 0 ) , ( 4 , − 1 ) . \big(1,\tfrac43\big),
\qquad (3,0),
\qquad (4,-1). ( 1 , 3 4 ) , ( 3 , 0 ) , ( 4 , − 1 ) .
For 3 < x ⩽ 4 3<x\leqslant4 3 < x ⩽ 4 f ( x ) < 0 f(x)<0 f ( x ) < 0 f ( x ) f(x) f ( x )
(b)
解法一
思路
展开
众数对应密度函数的最大值。由于 k > 0 k>0 k > 0 g ( y ) g(y) g ( y )
答题过程
展开
For 1 ⩽ y ⩽ 3 1\leqslant y\leqslant3 1 ⩽ y ⩽ 3
g ( y ) = k ( 12 y − y 3 ) . g(y)=k(12y-y^3). g ( y ) = k ( 12 y − y 3 ) . Therefore,
g ′ ( y ) = k ( 12 − 3 y 2 ) . g'(y)=k(12-3y^2). g ′ ( y ) = k ( 12 − 3 y 2 ) . At a stationary point,
k ( 12 − 3 y 2 ) = 0 y 2 = 4. \begin{align*}
k(12-3y^2)=&\,0 \\
y^2=&\,4.
\end{align*} k ( 12 − 3 y 2 ) = y 2 = 0 4. Since 1 ⩽ y ⩽ 3 1\leqslant y\leqslant3 1 ⩽ y ⩽ 3 y = 2 y=2 y = 2
g ′ ′ ( 2 ) = − 6 k ( 2 ) = − 12 k < 0 , g''(2)=-6k(2)=-12k<0, g ′′ ( 2 ) = − 6 k ( 2 ) = − 12 k < 0 , so this is a maximum. Hence the mode of Y Y Y
2 . \boxed{2}. 2 .
(c)
解法一
思路
展开
概率密度函数在整个支持区间上的积分等于 1。把 g ( y ) g(y) g ( y ) [ 1 , 3 ] [1,3] [ 1 , 3 ] k k k
答题过程
展开
Since the total probability is 1,
∫ 1 3 k ( 12 y − y 3 ) d y = 1 k [ 6 y 2 − 1 4 y 4 ] 1 3 = 1 k ( 135 4 − 23 4 ) = 1 28 k = 1. \begin{align*}
\int_1^3k(12y-y^3)\,\mathrm{d}y=&\,1 \\
k\big[6y^2-\tfrac14y^4\big]_1^3=&\,1 \\
k\bigg(\frac{135}{4}-\frac{23}{4}\bigg)=&\,1 \\
28k=&\,1.
\end{align*} ∫ 1 3 k ( 12 y − y 3 ) d y = k [ 6 y 2 − 4 1 y 4 ] 1 3 = k ( 4 135 − 4 23 ) = 28 k = 1 1 1 1. Therefore,
k = 1 28 . \boxed{k=\frac1{28}}. k = 28 1 .
(d)
解法一
思路
展开
设中位数为 m m m 0.5 0.5 0.5 k = 1 28 k=\frac1{28} k = 28 1 m m m m 2 m^2 m 2 1 ⩽ m ⩽ 3 1\leqslant m\leqslant3 1 ⩽ m ⩽ 3
答题过程
展开
Let m m m Y Y Y
∫ 1 m 1 28 ( 12 y − y 3 ) d y = 0.5. \int_1^m\frac1{28}(12y-y^3)\,\mathrm{d}y=0.5. ∫ 1 m 28 1 ( 12 y − y 3 ) d y = 0.5. Hence,
1 28 ( 6 m 2 − m 4 4 − 23 4 ) = 1 2 m 4 − 24 m 2 + 79 = 0. \begin{align*}
\frac1{28}
\bigg(6m^2-\frac{m^4}{4}-\frac{23}{4}\bigg)
=&\,\frac12 \\
m^4-24m^2+79=&\,0.
\end{align*} 28 1 ( 6 m 2 − 4 m 4 − 4 23 ) = m 4 − 24 m 2 + 79 = 2 1 0. Treating this as a quadratic in m 2 m^2 m 2
m 2 = 12 ± 65 . m^2=12\pm\sqrt{65}. m 2 = 12 ± 65 . Since 1 ⩽ m ⩽ 3 1\leqslant m\leqslant3 1 ⩽ m ⩽ 3
m = 12 − 65 = 1.98437 … ≈ 1.98 (3 s.f.) . \begin{align*}
m=&\,\sqrt{12-\sqrt{65}} \\
=&\,1.98437\ldots \\
\approx&\,\boxed{1.98}\quad\text{(3 s.f.)}.
\end{align*} m = = ≈ 12 − 65 1.98437 … 1.98 (3 s.f.) .
(e)
解法一
思路
展开
比较 (b) 的众数与 (d) 的中位数。两者都约等于 2，几乎重合，因此分布没有明显偏态。
答题过程
展开
The mode is 2 2 2 1.98437 … 1.98437\ldots 1.98437 …
median ≈ mode , \text{median}\approx\text{mode}, median ≈ mode , the distribution has no appreciable skew; it is approximately symmetric.
解法二
思路
展开
官方评分资料也允许用均值参与比较。由期望积分求出均值约为 1.99 1.99 1.99
答题过程
展开
The mean is
E ( Y ) = ∫ 1 3 y [ 1 28 y ( 12 − y 2 ) ] d y = 1 28 [ 4 y 3 − 1 5 y 5 ] 1 3 = 139 70 = 1.9857 … . \begin{align*}
E(Y)
=&\,\int_1^3y\bigg[\frac1{28}y(12-y^2)\bigg]
\,\mathrm{d}y \\
=&\,\frac1{28}
\big[4y^3-\tfrac15y^5\big]_1^3 \\
=&\,\frac{139}{70} \\
=&\,1.9857\ldots.
\end{align*} E ( Y ) = = = = ∫ 1 3 y [ 28 1 y ( 12 − y 2 ) ] d y 28 1 [ 4 y 3 − 5 1 y 5 ] 1 3 70 139 1.9857 … . Thus the mean, median and mode are all approximately 2 2 2
=&\,\int_1^3y\bigg[\frac1{28}y(12-y^2)\bigg] 