How Should I Calculate P(X<1.23) Using a Moment Generating Function?

psycho007
Messages
1
Reaction score
0
given m(t) = (1-p+p*e^t)^5
what is probability P(x<1.23)

i know that m(t) = e^tx * f(x)
m'(0) = E(X)
and m''(0) , can find the var(x)
should i calculate it using a normal table?
 
Physics news on Phys.org
psycho007 said:
given m(t) = (1-p+p*e^t)^5
what is probability P(x<1.23)

i know that m(t) = e^tx * f(x)
m'(0) = E(X)
and m''(0) , can find the var(x)
should i calculate it using a normal table?

No, m(t) is not e^tx * f(x), but it is ∫{e^(tx) f(x) dx: x=0..∞}.

Why would you use a normal table when the random variable is very far from normal? You can actually work out explicitly what is the distribution by expanding out the power and collecting terms in e^(kt) for k = 0,1,2,3,4,5.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top