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geno678
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HELP!Sums of Random Variables problem: Statistics
3. Assume that Y = 3 X1+5 X2+4 X3+6 X4 and X1, X2, X3 and X4 are random variables that represent the dice rolls of a 6 sided, 8 sided, 10 sided and 12 sided dice, respectively.
a. If all four dice rolls yield a 3, what is the sum of these random variables?
b. What is the expected value of the sum of these random variables?
=11+22+⋯+
1,2,… represent coefficients
1,2,…, represent variables
If =21+82
1 = First dice roll ; 2 = Second dice roll
Lets say we roll 3, then 5
Y = 2(3) + 8(5) = 46
for part (a)
for part(b)
for expected value
==11+22+⋯+
this part i don't understand.
Ok so I attempted part (a). I have no clue if I'm correct though.
So I was given Y = 3(X1) + 5 (X2) + 4(X3) + 6(X4)
Here's where I'm confused. The problems says that all four dice yield a 3.
I made X1 = 1/6, X2 = 1/8, X3 = 1/10, X4 = 1/12
I plugged these values into my equation.
Y = 3(1/6) + 5 (1/8) + 4 (1/10) + 6(1/12) = 2.025
However, in the example that my professor gave me I'm thinking that
X1, X2, X3, and X4 = 3.
Is the answer really Y = 3(3) + 5 (3) + 4(3) + 6(3) = 54
Homework Statement
3. Assume that Y = 3 X1+5 X2+4 X3+6 X4 and X1, X2, X3 and X4 are random variables that represent the dice rolls of a 6 sided, 8 sided, 10 sided and 12 sided dice, respectively.
a. If all four dice rolls yield a 3, what is the sum of these random variables?
b. What is the expected value of the sum of these random variables?
Homework Equations
=11+22+⋯+
1,2,… represent coefficients
1,2,…, represent variables
If =21+82
1 = First dice roll ; 2 = Second dice roll
Lets say we roll 3, then 5
Y = 2(3) + 8(5) = 46
for part (a)
for part(b)
for expected value
==11+22+⋯+
this part i don't understand.
The Attempt at a Solution
Ok so I attempted part (a). I have no clue if I'm correct though.
So I was given Y = 3(X1) + 5 (X2) + 4(X3) + 6(X4)
Here's where I'm confused. The problems says that all four dice yield a 3.
I made X1 = 1/6, X2 = 1/8, X3 = 1/10, X4 = 1/12
I plugged these values into my equation.
Y = 3(1/6) + 5 (1/8) + 4 (1/10) + 6(1/12) = 2.025
However, in the example that my professor gave me I'm thinking that
X1, X2, X3, and X4 = 3.
Is the answer really Y = 3(3) + 5 (3) + 4(3) + 6(3) = 54