You are expecting a parcel to arrive between 7:00 and 8:00 tomorrow. Let x be the time passed after 7:00 until the package arrives in hours. The density function of x is $$f(x)=\begin{cases}\frac{3-3x^{2}}{2} & 0<x<1\\0 & else\end{cases}$$
1.show that f(x) is a density function.
2.Find the expectation and variance of the arrival time of the parcel.
3.Suppose that you realize that your doorbell is not working and the postman will put the parcel in front of your door. You decide to check whether the parcel arrived twice exactly at 7:30am and 8:00. Let T be the total time the parcel waits in front of the door (in hours) until you pick it. Find the density function of T.
4.Calculate the expectation of T.
For question 1 is it just simply integral it and if it equals to 1 then it is a density function and the question 2 is just integral with xf(x) find the expected value and variance is just integral $$x^{2}f(x)$$and using the formula to calculate it . for the last two question is it just equal to f(0.5x)?
I think you mean $0<x<1$. Also, you should not name the random variable exactly like the variable you use in the density since that may cause confusion.
No, that's not right; in fact, this function is no density since its integral over the intervall $[0,2]$ equals 2.
Observe that T cannot take values below 0 or above 0.5. So henceforth, assume $t\in[0,0.5]$ (and the density will be 0 everywhere else). Also, observe that $T\leq t$ is equivalent to "the parcel arrived either between 7:30-$t$ and 7:30 or between 8:00-$t$ and 8:00". With this and the given density function, you can easily calculate the distribution of T: \begin{align*} \mathbf{P}[T\leq t]&=\int_{0.5-t}^{0.5}f(x)\,dx+\int_{1-t}^1f(x)\,dx \end{align*} Evaluating this expression and taking the derivative with respect to $t$, you get the desired density.