Let X be the number of heads obtained in 5 tosses of a fair coin. Let -(negative infinity) < t < positive infinity and define g(X)=e^tx.
a. Compute M(t)=E(e^tx)=Summation g(x)f(x)= Summation x=0 to x=5 (5 choose x) (1/2)^5 (e^tx)=
b. M(0)
c. M'(t)
d.M'(0)=E(X)=
e. M''(t)
f. M''(0)=E(X^2)