Consider a random variable $X$ having $pdf$
$$f_X(x)=\frac{1}{2\sqrt{x}}I_{(0,1)}(x)$$
(a) Give the $cdf$ of $X$, expressing it as a piecewise function.
(b) Find the median of $X$’s distribution by solving $F_X(η_X)$ = $1\over2$ for $η_X$.
(c) Give the value of $P$($1\over{4}$ $\lt X \lt$ $3\over{4}$).
(d) Give the value of $E(X)$.
Attempted Solution:
(a) $\int\frac{1}{2\sqrt{x}}$ = $\sqrt{x}$ giving a $cdf$ of
$$F_X(x) = \begin{cases} {1} & \text{$x \geq 1$} \\ \sqrt{x} & \text{$0 \lt x \lt 1$} \\{0} & \text{$x \leq 0$}\end{cases}$$
(b) Setting $F_X(η_X)$ $=$ $1\over2$, we get $\sqrt{η_X}$ $=$ $1\over2$ and so $η_X$ = $1\over4$
(c) $P$($1\over{4}$ $\lt X \lt$ $3\over{4}$) = $\sqrt{3\over4} - \sqrt{1\over4} = .366$
(d) $E(X)$ = $\int_0^{1}$ $x$ $f_X(x)$ = $\int_0^{1}$ $x$ $1\over{2\sqrt{x}}$=$\int_0^{1}$ ${\sqrt{x}}\over{2}$ =$1\over3$$x^{3/2}$$|_0^1$ $=$ $1\over3$
Did I do these correctly?