Answer to Question #202216 in Statistics and Probability for Sonu kumar

Question

If the moment generating function (m.g.f.) of a random variable X is 

M_x(t)=exp(3t+32t²).Find mean and standard derivation of X and also compute 

P(X<3).

Answer 1

Solution:

Given, $M_X(t)=e^{(3t+32t^2)}$ ...(i)

Now, mean$=\mu=M'_X(0)$

On differentiating (i) w.r.t t,

$M'_X(t)=e^{(3t+32t^2)}(3+64t)$ ...(ii)

Put $t=0$

$M'_X(0)=e^{0}(3)=3$

Thus, mean = 3

Again, differentiating (ii) w.r.t. t,

$M''_X(t)=[e^{(3t+32t^2)}]'(3+64t)+e^{(3t+32t^2)}(3+64t)' \\=e^{(3t+32t^2)}(3+64t)^2+e^{(3t+32t^2)}(64) \\=e^{(3t+32t^2)}[64+(3+64t)^2]$

Put $t=0$

$M''_X(0)=e^{0}[64+(3)^2]=1(73)=73$

Now, $\sigma^2=M''_X(0)-[M'_X(0)]^2=73-3^2=64$

Then, $\sigma=8$

Assume $X\sim N(\mu,\sigma)$

$P(X<3)=P(z<\dfrac{3-3}{8})=P(z<0)=0.5$