SOLUTION: A random variable X has a CDF such that x/2 0 < x ≤ 1 F(x) = x-1/2 1 < x ≤ 3/2 (a)Graph F(x). (b)Graph the pdf f(x). (c)Find P[X ≤ 1/2]. (d)Find P

A random variable X has a CDF such that
     
 
       
(a)Graph F(x).





(b)Graph the pdf f(x).

 

 

Notice it really is a pdf because the area between it and the x-axis
consists of a 1 by 1/2 rectangle and a 1/2 by 1 rectangle. The sum
of their areas is 1. The green lines below are not part of the graph 
of f(x).  They just show the rectangles that make up the area between 
the graph and the x-axis.



(c)Find P[X ≤ 1/2].

That's the area of the shaded square below, which is a 1/2 by 1/2
rectangle (actually a square) with area (1/2)(1/2) = 1/4.  So,
P[X ≤ 1/2] = 1/4.






(d)Find P[X ≥ 1/2].
That's the area of the shaded area below, which is a 1/2 by 1/2
square plus a 1/2 by 1 rectangle. That's (1/2)(1/2)+((1/2)(1) = 3/4.  So,
P[X > 1/2] = 3/4.






(e)Find P[X ≤ 1.25].
That's the area of the shaded area below, which is a 1 by 1/2 rectangle
plus an 0.25 by 1 rectangle. That's (1)(1/2)+((1/4)(1) = 3/4.  So,
P[X < 1.25] = 3/4.



(f)What is P[X = 1.25]?
LOL!  That is the area of the single black line segment below, which 
is 0. Yes! A line segment has 0 thickness, so its area is 0.  Why does that
make sense?  It's because NO MEASURE CAN BE ABSOLUTELY PERFECT!!.
P[X = 1.25], it means EXACTLY 1.25, not 1.25000000001 and not 1.24999999999.
Nothing can be measured EXACTLY, so to ask for the probability of any
EXACT number, with no tolerance whatever, is in reality impossible, hence
the probability is 0.  
P[X = 1.25] = 0 




Edwin