Answer to Question #147289 in Statistics and Probability for Arshad Ali

Question

A continuous random variable has a density function f (x) = 2 (5-x)/5, where 2<x<3. Calculate the probability

Answer 1

At first, we will check that $f(x)$ it is a density. We calculate $\int_2^32\frac{5-x}{5}dx=(2x-\frac{x^2}5)|_2^3=6-\frac{9}{5}-4+\frac{4}{5}=1$. Thus, it is indeed a probability density function. We denote by $X$ a random variable that has a density $f(x)$ . We calculate:

$P(X\leq a)=\left\{\begin{matrix} 1,a\geq3, \\ (2x-\frac{x^2}5)|_2^a=2a-\frac{a^2}{5}-\frac{16}5,\,\,\,2<a<3, \\ 0,\quad a<2. \end{matrix}\right.$