Just need some answers checking:
(iii) im not sure what needs doing but i've shown my attempt
A continuous random variable $X$ has cumulative distribution function
$F(x) = \begin{cases} 0 & x<1,\\ x^2(3-2x) & 0\le x \le 1,\\ 1 & x>1. \end{cases} $
$(i)$ Calculate $\Pr(\frac13\ <\ X\ <\ \frac23)$
$(ii)$ Find the probability density function of $X$
$(iii)$ The symmetry of the probability density function on the interval $[0,1]$ implies that $E(X) = \frac12$. Find $\operatorname{var}(X)$
Answers:
(i) $\Pr(a < X < b) = \Pr(X \le b)-\Pr(x \le a) = F_x(b)-F_x(a)$
Given this,
$\Pr(\frac13\ <\ X\ <\ \frac23) = F(\frac23)-F(\frac13) = ((\frac23)^2(3-2(\frac23))) - ((\frac13)^2(3-2(\frac13))) = \cdots = \frac{13}{27}$
So $\Pr(\frac13\ <\ X\ <\ \frac23) = \frac{13}{27}$
(ii) Given $Pr(a\ \leq\ X\ \leq\ b)\ =\ \int_a^b f(x) \, dx$ for finding density function
$\int_a^b x^2(3-2x)\,dx$
$\int_a^b 3x^2-2x^3\,dx$
$= x^3-\frac12x^4$
(iii) Given $E(x) = \frac12,\ (E(x)^2)=\frac14$
To get $E(x^2)$ you do $PDF \cdot x^2:= (x^3 - \frac12x^4) \cdot x^2 = x^6 - \frac12x^8$
$\int_{a}^{b}x^6 - \frac12x^8 = [\frac{x^7}{7} - \frac{x^9}{18}]^1_0$
$Var(x) = [(\frac{x^7}{7} - \frac{x^9}{18})-\frac14]^1_0$
$= \frac{11}{126}-\frac14 = -\frac{41}{252}$