连续随机变量 $X$ 的概率密度函数 $f$ 如下：

其中 $k$ 是常数。

(a) 证明 $k = \frac{1}{25}$。

(b) 求 $X$ 的中位数。

随机变量 $Y$ 定义为 $Y = \sqrt[3]{X}$。

(c) 求 $Y$ 的概率密度函数。

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。

分别求密度图下的两个面积。两段都是直线与横轴围成的三角形，也可直接积分。

The total integral is

Equating it to $1$ gives $k=1/25$.

两段直线下面各有一个三角形，可直接把面积相加。

The first triangle has base $1$ and height $k$, while the second has base $7$ and height $7k$. Since the total probability is $1$,

Hence $k=1/25$.

$1$ 左侧的概率小于 $1/2$，所以中位数位于第二段。令右侧的三角形面积等于 $1/2$。

We have $F_X(1)=1/50<1/2$, so the median $m$ is in $(1,8)$. Its upper-tail probability is

Setting this equal to $1/2$ gives $8-m=5$, so $m=3$.

也可把中位数左侧的两段积分相加。二次方程的根必须回到密度的支撑区间筛选。

Since $F_X(1)=1/50<1/2$, the median lies in $(1,8)$. Thus

Multiplying by $50$ and rearranging gives

The root $13$ is outside $(1,8)$, so the median is $m=3$.

立方根在支撑区间上递增。将 $X$ 的累积概率换成 $F_X(y^3)$，再用链式法则求密度。

For $0<y<2$, $F_Y(y)=F_X(y^3)$, so $f_Y(y)=3y^2 f_X(y^3)$ away from the breakpoint. The breakpoint $x=1$ corresponds to $y=1$. A density is therefore

也可以先把变换后的累积概率完整写出。前段用左尾面积，后段用右尾的补事件。

For $0\le y<1$, the probability below $y^3$ is $y^6/50$. For $1\le y\le2$, the probability above $y^3$ is $(8-y^3)^2/50$, so the CDF is $1-(8-y^3)^2/50$. It is $0$ below $0$ and $1$ above $2$. The derivatives of the two interior pieces are $3y^5/25$ and $3y^2(8-y^3)/25$, respectively, giving the density above.