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Let X be a discrete random variables with PMF given by : P_X(x)= x/15 ,{x=1,2,3,4,5} 0 , otherwise b)let Y=(X-3)^2, find range Y, S_Y , PMF of Y

Mathematics
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Other answers:

P_X(x)= x/15 ,{x=1,2,3,4,5} 0 , otherwise so to find range using Y y= (X-3)^2 ((x/15)-3)^2 y= (1/15 -3)^2 (-44/45)^2 to 0
am I right?
no
x takes the values 1,2,3,4,5 y=(x-3)^2 plug those x values into the above equation
for the range(support)
oh, it is X, not x
yes...so X takes the values 1,2,3,4,5 Y=(X-3)^2 plug those X values into the above equation :)
1 Attachment
ok...do what I wrote
(1-3)^2,(2-3)^2,(3-3)^2,(4-3)^2,(5-3)^3 4,1,0,1,4
so 0,1,4
that is the range (or support)
now you can find the pmf
right?
so y=(x-3)^2 do we solve for x?
no
\[P_{Y}(0)=P_{X}(3)\] \[P_{Y}(1)=P_{X}(2)+P_{X}(4)\] \[P_{Y}(4)=P_{X}(1)+P_{X}(5)\]
can you explain me how you got that?
\[P_{Y}(1)=P(Y=1)=P(X=2\text{ or }X=4)\] \[=P(X=2)+P(X=4)=P_{X}(2)+P_{X}(4)\]
ok?
still figuring out the last post
ok
I understand you sum because it is 'or' but why is P(y=1) = P(x=2 or x=4)
\[Y=(X-3)^2\] \[1=(2-3)^2\] \[1=(4-3)^2\] if I tell you Y=1 then either X=2 or X=4
got it
good
so that's just P(Y=1) do we find Y=1,2,3,4,5?
Y only takes the values 0,1,4
got it , thanks; I wish you were my probability professor

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