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The moment generating function of $X$ is $$m(t)= \exp( -6t+32t^2)$$ Find

A. $\;\;P(-4<X<16)$

B. $\;\;P(-10<X<0)$

I have known that the first derivative for moment generating function is the expected value of $X.$

amWhy
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Smith Pay
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  • Maybe you should look at the form of $m(t)$. Do you see something familiar? – Karl Dec 10 '14 at 15:01
  • @Karl is right. It is more about knowing usual random variables and their respective moment generating function than a calculus problem. Unless you have seen the inverse Laplace transform. – Pi89 Dec 10 '14 at 15:18
  • I might have the idea that p(1)= -6 and p(2)=32 but I'm not even sure about that , if so how can I use it for my qaustion – Smith Pay Dec 10 '14 at 15:24
  • examples of moment generating functions here: http://en.wikipedia.org/wiki/Moment-generating_function#Examples – ir7 Dec 10 '14 at 15:26
  • If that a normal continuous distribution , so the mean is -6 and the variance is 64 ? If so , I integrate the f(x) of the normal between the probability boundary? – Smith Pay Dec 10 '14 at 15:34
  • and the important property suggested in the other comments to use: http://en.wikipedia.org/wiki/Moment-generating_function#Important_properties – ir7 Dec 10 '14 at 15:35
  • @AhmadTalafha yes, once you know your probability distribution function or cumulative distribution function, the calculations of your probabilities should be standard. – ir7 Dec 10 '14 at 15:38

1 Answers1

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HINT: Observe that the mgf of $X\sim\mathcal{N}(\mu,\sigma^2)$ is $M_X(t)=\textrm{e}^{\mu t+\frac{1}{2}\sigma^2t^2}$. So your random variable is $X\sim\mathcal{N}(-6,8^2)$.

alexjo
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