My Math Forum A die is loaded so that the probability of a face coming up is proportional to the nu

 June 6th, 2017, 05:11 PM #1 Member   Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 A die is loaded so that the probability of a face coming up is proportional to the nu A die is loaded so that the probability of a face coming up is proportional to the number on that face. The die is rolled with outcome X. Find V (X) and D(X). How would you find the Expected value E(x) and the variance V(x) for this problem? For this problem, I knew that the probability for each outcome would be x/21. And I knew that X=1,2,3,4,5,6 so I took the expected value of 1× (1/21)+2× (2/21)+3×(3/21)+4×(4/21)+5×(5/21)+6×(6/21) and got 3, then I tried to find the variance by using V(x)=E(x^2)-(E(x))^2=12, but the answer was 3/4 for V(x).
 June 6th, 2017, 06:26 PM #2 Global Moderator   Joined: May 2007 Posts: 6,378 Thanks: 542 Redo your arithmetic. Mean=91/21. Thanks from poopeyey2
 June 6th, 2017, 06:49 PM #3 Member   Joined: Apr 2017 From: PA Posts: 45 Thanks: 0 So is the procedure right?
June 6th, 2017, 10:19 PM   #4
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Quote:
 Originally Posted by poopeyey2 So is the procedure right?
$E[X] = \sum X p[X] = \sum \limits_{k=1}^6 \dfrac{k}{21}k$

$V[X] = E[X^2] - (E[X])^2 = \left(\sum \limits_{k=1}^6 \dfrac{k}{21}k^2\right) - (E[X])^2$

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