From this density, the probability that X is between 0.5 and 1.52 is:

Question

Answer 1

This is the continuous uniform distribution with parameters a=0 and b=2, also denoted U(0,2), with expected value E(X)=(a+b)/2 and variance Var(X) (b-a)^2/12. Graphically, the PDF of the uniform distribution is shown as rectangle with Height=1/(b-a) and Base= (b-a). Here the probability that X is between 0.5 and 1.52 is equivalent to the area of the sub-rectangle with one sides starting at x= 0.5 and other side ending at x=1.52. The area of this sub-rectangle is (1.52 - 0.5) * 1/(2-0) = 0.51. Another (advanced) way to show this is by using the following formulae: