For the question:
Consider a random variable $X$ with probability density function
$$f(x)=\begin{cases} \ell &\text{if }x\in[2,5]\\ 2\ell &\text{if }x\in(5,8] \\0 &\text{otherwise }\end{cases}$$
with $\ell$ as $\frac{1}{9}$
Determine the cumulative distribution function $F_X(t)$.
My answers for the following ranges are as follows:
$i)\text{ } For \text{ }t\in (-\infty,2), \text{ }F_X(t) = 0$
$ii)\text{ } For \text{ }t\in [2,5), \text{ }F_X(t) = \frac{1}{3}$
$iii)\text{ } For \text{ }t\in (5,8], \text{ }F_X(t) = \frac{2}{3}$
$iv)\text{ } For \text{ }t\in (8,-\infty), \text{ }F_X(t) = 1$
All these answers were determined by $F_X(t)=\int_{-\infty}^x f(t)dt$ however the marking system indicates that 2 out of the 4 answers are incorrect.
I suspect that the answers for ii) and iii) are incorrect due to calculating the integrals with the wrong ranges however after consulting the page Expected Values and CDF and rectifying my answers to:
$ii\text{ } rectified)\text{ } For \text{ }t\in [2,5), \text{ }F_X(t) = \frac{x^2-2}{9}$
$iii\text{ } rectified)\text{ } For \text{ }t\in (5,8], \text{ }F_X(t) = \frac{16-2x^2}{9}$
it still marks the answers wrong.
Can somebody please clarify on why these answers are incorrect and let me know how the question should be correctly approached?
Thanks in advance!
You are not integrating correctly $F_X(t)$ for $t\in(2,8)$, $\ell\cdot\int_2^t dx=l(t-2)$, $t\in(2,5]$. You should figure out why you wrongly get a $t^2$.