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March 6th, 2013, 10:09 PM  #1 
Newbie Joined: Feb 2013 Posts: 16 Thanks: 0  Probability  PDF and CDF Understanding
Okay so I'm having a hard time understanding PDF and CDF functions, so please correct me if I'm wrong here. PDF takes the total area of a function and it equals one. All the probabilities are over a range, because taking the area of one point is 0. CDF integrates the PDF and makes it cumulative. Thus, the final point on interval [a,b] where F(x) is CDF would be F(b)=1. Thus the probabilities slowly accumulate and you can make claims about values of x before the actual value of x in terms of probability based on the probability at that point. Basically adding up the probabilities to a total of one at F(b)=1. So based on the problem attatched, the CDF's would simply increase at a steady rate to 1. Thus increasing at each k value by 0.2 . However I do not understand the PDF. I understand that the total area of each rectangle added should equal one. But how do I relate these facts? 
March 6th, 2013, 10:17 PM  #2 
Newbie Joined: Feb 2013 Posts: 16 Thanks: 0  Re: Probability  PDF and CDF Understanding
I believe I figured out simply writing out my question, haha. But since P(x g or eq to 2) =0.4 , that means k=1 +k=2 + k=0 must be 0.4 . From what's already calculated based on PDF, 0.5 probability is on graph. thus k =1 =0.1, so k=2 = 0.4. Then k=4 must be 0.2. On the other graph, the values just accumulate by .2 each time right? 
March 7th, 2013, 08:44 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,633 Thanks: 2080 
No. The missing columns in the first diagram should have heights 0.1, 0.2 and 0.2. In the second diagram, the missing columns should have heights 0.1, 0.7, 0.9 and 1.

March 7th, 2013, 12:45 PM  #4  
Newbie Joined: Feb 2013 Posts: 16 Thanks: 0  Re: Quote:
 

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