Consider the following game. Everybody has to press a button on a machine which results
in one of three possible outcomes:
The result 1 has probability p1 = 1/4.
The result 2 has probability p2 = 1/3.
The result 3 has probability p3 = 5/12.
Now everybody presses the button three times and adds up the results. This is the result
of the game.
1. Calculate the expected value for the game described above.
is 6.5 correct?
Thx a lot in advance......
We will assume that is the case, although it is not clear.
The outcomes are
The outcomes can happen in one way each.
Using this generating polynomial you can read of the coefficients.
The term tells us happen seven ways.
I already did it like this and my solution would be 6.5...
can someone affirm this?
And then the following was asked:
Calculate the the variance and the third moment around the expected value for the
game described above.
So for the variance I used this formula....
Discrete random variable
If the random variable X is discrete with probability mass function x1 7!
p1; : : : ; xn 7! pn, then
Var(X) = Sum of pi times (xi - ev)^2
where pi=probability for i
xi = value for i
ev= expected value
So started with 1/64*(3-6,5)^2 + 1/16*(4-6,5)^2 and so on...
my result is 23/12....
I hope this is correct....
But how do I calculate the skewness now?
for μ I would again take the expected value, but what do I insert fpr X and E???
THX in advance...