If an impulse is modeled by a random variable x with this density function, how to calculate the prob that... If an impulse is modeled by a random variable x with this density function,
a) how to calculate the probability that exactly $3$ of $90$ independent sample are $< 0.3$?
b) What is the medium intensity of an impulse?
$$ f(x) = \begin{cases}(2/9)*x &\mbox{if } x \in (0,3)\\ 0&\mbox{else}\end{cases}$$ 
For the point B, i think that I could use the the integral from $0$ to $3$ of $x\cdot f(x)$. Should I?
For the point A, what is the best strategy to use? Should I answer this question using the Bernoulli experiment?
 A: a) Here, the binomial distribution is the right approach: we have $90$ Bernoulli trials.
The probability that an impulse is $\lt 0.3$ is equal to
$$\int_0^{0.3}\frac{2}{9}x\,dx.$$
Calculate. The integral turns  out to be $0.01$.
The probability that exactly $3$ in a sample of $90$ are less that $0.3$ is therefore equal to
$$\binom{90}{3}(0.01)^3 (0.99)^{87}.$$
As a slight complication, we can use the Poisson distribution to approximate the Binomial. If you are familiar with the Poisson, you could use as an approximation the probability that a Poisson with parameter $\lambda=(90)(0.01)$ takes on value $3$. This is $e^{-\lambda}\frac{\lambda^3}{3!}$.  
However, the Binomial gives the exact answer.
b) What the second problem asks for is not clear from the wording. There are two standard measures of "average" location, the mean and the median. You used the word "medium," which is not a standard term.
The mean is given by the expression
$$\int_0^3 xf(x)\,dx,$$
where $f(x)$ is your density function. 
If $X$ is your random variable, then the median of $X$ is the number $m$ such that $\Pr(X\le m)=\frac{1}{2}$. So we want
$$\int_0^m \frac{2}{9}x\,dx=\frac{1}{2}.$$
Integrate. We get $\frac{m^2}{9}=\frac{1}{2}$, which yields $m=\frac{3}{\sqrt{2}}$.
