My Math Forum Probability Distribution

 Probability and Statistics Basic Probability and Statistics Math Forum

October 9th, 2017, 12:38 PM   #1
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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.
Attached Images
 Screen Shot 2017-10-09 at 4.19.57 PM.jpg (11.1 KB, 6 views)

 October 9th, 2017, 11:47 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,105 Thanks: 1093 $X = \{0,1,2,3,4\}$ 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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