1. Consider the random quantity X, with PDF fx(x) = {4/3(2x-x^2), 0

Given that,

$f(x)={3\over4}(2x-x^2), 0\lt x\lt 2\\0,elsewhere$

$a)$ verify that it is a pdf.

$\displaystyle\int^2_0f(x)dx=\displaystyle\int^2_0{3\over4}(2x-x^2)dx ={3\over4}(x^2-{x^3\over3})|^2_0={3\over4}(4-{8\over3})={3\over4}\times{4\over3}=1.0$

Therefore, it is a valid pdf.

$b)$

$i)$ Expectation of $X$

It is given as,

$E(x)=\displaystyle\int^2_0 xf(x)dx=\displaystyle\int^2_0{3\over4}(2x^2-x^3)dx={3\over4}({2\over3}x^3-{x^4\over4})|^2_0=1$

$ii)$ Variance of $X$

The variance of $X$ is given as,

$Var(x)=E(x^2)-(E(x))^2$

We determine,

$E(X^2)=\displaystyle\int^2_0 x^2f(x)dx=\displaystyle\int^2_0{3\over4}(2x^3-x^4)dx={3\over4}({x^4\over2}-{x^5\over5})|^2_0={6\over5}$

Therefore,

$var(x)={6\over5}-1^2={1\over5}$

$c)$

The cumulative distribution for $X$ is,

$F(x)=\displaystyle\int^x_0f(u)du=\displaystyle\int^x_0{3\over4}(2u-u^2)du={3\over4}(u^2-{u^3\over3})|^x_0={3\over4}(x^2-{x^3\over3})$

It can be written as,

$F(x)={3\over4}(x^2-{x^3\over3}), 0\lt x\lt 2\\0,elsewhere$

$d)$

To find the median, we find a value $y$ such that,

$\displaystyle\int^y_0f(x)dx=0.5$

So,

$\displaystyle\int^y_0{3\over4}(2x-x^2)dx={3\over4}(x^2-{x^3\over3})|^y_0={3\over4}(y^2-{y^3\over3})=0.5$

We need to solve for the value of $y$

${3\over4}(y^2-{y^3\over3})=0.5\implies (y^2-{y^3\over3})={2\over3}\implies y^2(1-{y\over3})={2\over3}\implies y=0.82 \space or\space 1$

The value of the median must be around the middle of the range (0,2). So the most appropriate value of the median is y=1