You are playing a board game in which severity of a plenty is determined by rolling three dice and adding spots on the up-face. The dice are all balanced so that each face is equally likely, and the three dice fall independently.
If X1, X2, and X3 are the number of spots on the up-face of the three dice, then X= X1+X2+X3.
Use this fact to find the mean μx and the standard deviation σx without finding the distribution of of X.
(Start with the distribution of each of the Xi.)
3 answers
assistance needed
What you have here is a "Distribution of Linear Combinations" problem.
Lets start with the expected mean of Xi. There is a 1/6 prob of rolling either a 1,2,..6. So mean of X1 is (1/6)*(1+2+3+4+5+6) = 3.5. Since all the dice are the same, mean X1= mean X2 = mean X3 = 3.5 Since each die is thrown the same number of times, the expected mean for X = (mean X1)+(mean X2) + (mean X3) = 3.5+3.5+3.5 = 10.5
Now for the variance of X1. The expected variace for X1 is ((1-3.5)^2+(2-3.5)^2 + ...(6-3.5)^2 ) divided by 6. I get Var X1 = 2.9167 and SD X1=1.708.
Since X1,X2,X3 are independent and since each die contributes 1/3 towards the total, SD X = sqrt(Var X1 + Var X2 + Var X3) = sqrt(3*2.9169) = 2.958
Lets start with the expected mean of Xi. There is a 1/6 prob of rolling either a 1,2,..6. So mean of X1 is (1/6)*(1+2+3+4+5+6) = 3.5. Since all the dice are the same, mean X1= mean X2 = mean X3 = 3.5 Since each die is thrown the same number of times, the expected mean for X = (mean X1)+(mean X2) + (mean X3) = 3.5+3.5+3.5 = 10.5
Now for the variance of X1. The expected variace for X1 is ((1-3.5)^2+(2-3.5)^2 + ...(6-3.5)^2 ) divided by 6. I get Var X1 = 2.9167 and SD X1=1.708.
Since X1,X2,X3 are independent and since each die contributes 1/3 towards the total, SD X = sqrt(Var X1 + Var X2 + Var X3) = sqrt(3*2.9169) = 2.958
(I wonder how I got that strange looking title to my first response???)