Question

If X1, X2, X3 are the number of spots on the up-faces of the three dice, then X = X1 + X2 + X3. Use this fact to find the mean and the standard deviation without finding the distribution of X (Start with the distribution of each of the X1).

Answer 1

Can we assume independence?
I am assuming they would roll the dice independently right, and not in some strange way?

Answer 2

yeah

Answer 3

You can then say that the distriution of each of X1, X2 and X3 is a discrete uniform distribution having each probability 1/6 . The expected value of a discrete uniform with minimum value a=1 and maximum value b=6 is (a+b)/2

The variance of that distribution is then ((b-a+1)^2 - 1)/12
(If you do not know these formulas, you might have to prove them..)

So then you can say:
. you can then use the property of expectation and variance to find these vales for X1+X2+X3....
E(X1+X2+X3) = E(X1)+E(X2)+E(X3)
Var(X1+X2+X3) = Var(X1)+Var(X2)+Var(X3) by independence

Answer 4

and since each die is identical, the variance of X1, X2, X3 and expectations of X1, X2 and X3 are the same

Answer 5

so apparently the answer for the mean is 3 (3.5) which is 10.5 and the variance of the sum is 3 (2.917) which is 8.751. Could you please explain where these numbers came from?

Answer 6

So the distribution of a discrete uniform distribution can be written generally as:

\[P(X=x) = \frac{1}{b-a+1}\] for x=a, a+1, ..., b
Here a = 1, b = 6 , so the pmf reduces to 1/6 for values x = 1, 2, ..., 6

Then you use the formula for expectation for any random variable X:
\[E(X) = \sum_{\text{all }x}x*P(X=x)\]

Thus, you find \[E(X) = sum_{x=a}^b x*\frac{1}{b-a+1}\]

And similary you use the definition of variance
\(Var(X) = E(X^2) + [E(X))]^2 \)

So you first find \(E(X^2)\) by doing:
\[E(X^2) = sum_{x=a}^b x^2*\frac{1}{b-a+1}\

Then you find the variance using the result of E(X) that you found above to compute
the variance Var(X)

Doing this will give you those formulas I mentioned above

Answer 7

Correction:
**Thus, you find \[E(X) = \sum_{x=a}^b x*\frac{1}{b-a+1}\]
**So you first find \(E(X^2)\) by doing:
\[E(X^2) = \sum_{x=a}^b x^2*\frac{1}{b-a+1}\
sorry about that

Answer 8

If the LaTeX formulas don't display, you can try copy/pasting them into this site:
http://www.codecogs.com/latex/eqneditor.php