Thread: Expectation of a Variable

dice has faces numbered 1 to 6 and is weighted so that the probability of
throwing n (n = 1; 2; :::; 6) in a single throw is proportional to n, i.e., equals
Cn for some constant C.
(a) Find C .
(b) Let X be equal to the number shown when the dice is thrown.
i) Compute E(X^3).

I've calculated C as 1/21.

However, i thought the expectation of E(x^3) would thus be:


E(X^3) = n^3/21
so: 1/21 + 2^3/21 + 3^3/21 etc....

EDIT: Is Latex now working? I keep getting errors whilst using it.

Thankyou

Use the TEX tags instead of the MATH tags, for now. Personally, i don't know hte difference, except that one works.

Let's try that Expectation again.

p(1) = 1/21
p(2) = 2/21
p(3) = 3/21
etc.

E(X^3) = 1*p(1) + 8*p(2) + 27*p(3) + ... + 216*p(6) -- There aren't any 'n's left in there.