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October 9th, 2017, 01:38 PM  #1 
Newbie Joined: Oct 2017 From: Nowhere Posts: 1 Thanks: 0  Probability Distribution
I have attached the image of the problem. I just do not understand how to proceed. The question says you have "incomplete" distribution function because it only works for 0, 1, 2, and 3. To find the probability for 4, I know you need to apply "complement rule" (subtract all probabilities for 0, 1, 2, and 3 from 1). By definition, F(2) = P(X<=2), you need to be able to calculate P(X<=2) with f(x). But I am very unsure on how to do this correctly. Everything I have tried so far has been wrong. I do know how to get D and E as long as I can figure out how to get C. 
October 10th, 2017, 12:47 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,763 Thanks: 905 
$X = \{0,1,2,3,4\}$ a) $f[X] = \left\{\dfrac{1}{4},~\dfrac{3}{16}~,\dfrac{9}{64}, ~\dfrac{27}{256}\right\},~X=0,1,2,3$ $\displaystyle \sum_{X=0}^4 f[X] = 1 \Rightarrow f[4]=1\dfrac{175}{256}=\dfrac{81}{256}=0.3164$ b) $F(2)=\displaystyle \sum_{X=0}^2~f(X) = \dfrac{1}{4}+\dfrac{3}{16}+\dfrac{9}{64} = \dfrac{37}{64}=0.5781$ c) $E[X] = \displaystyle \sum_{X=0}^4~X f(X)= \dfrac{525}{256}=2.051$ d) $Var[X] = \displaystyle \sum_{X=0}^4~X^2 f(X)  \left(E[X]\right)^2=\dfrac{167511}{65536}=2.556$ e) $SD[X] = \sqrt{Var[X]} = \dfrac{\sqrt{167511}}{256}=1.5988$ 

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