Answer:
![\frac{61}{99}](https://tex.z-dn.net/?f=%5Cfrac%7B61%7D%7B99%7D)
Step-by-step explanation:
let x = 0.616161 (etc.)
make a second equation because you multiply by 100:
100x = 61.616161 (etc.)
subtract from each other
100x = 61.616161
x = 0.616161
you get
99x = 61
solve for x
![\frac{61}{99}](https://tex.z-dn.net/?f=%5Cfrac%7B61%7D%7B99%7D)
Answer:
![E[X^2]= \frac{2!}{2^1 1!}= 1](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%20%5Cfrac%7B2%21%7D%7B2%5E1%201%21%7D%3D%201)
![Var(X^2)= 3-(1)^2 =2](https://tex.z-dn.net/?f=Var%28X%5E2%29%3D%203-%281%29%5E2%20%3D2)
Step-by-step explanation:
For this case we can use the moment generating function for the normal model given by:
![\phi(t) = E[e^{tX}]](https://tex.z-dn.net/?f=%20%5Cphi%28t%29%20%3D%20E%5Be%5E%7BtX%7D%5D)
And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:
![\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx](https://tex.z-dn.net/?f=%5Cphi%28t%29%20%3D%20C%20%5Cint_%7BR%7D%20e%5E%7Btx%7D%20e%5E%7B-%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%20dx%20%3D%20C%20%5Cint_R%20e%5E%7B-%5Cfrac%7Bx%5E2%7D%7B2%7D%20%2Btx%7D%20dx%20%3D%20e%5E%7B%5Cfrac%7Bt%5E2%7D%7B2%7D%7D%20C%20%5Cint_R%20e%5E%7B-%5Cfrac%7B%28x-t%29%5E2%7D%7B2%7D%7Ddx)
And we have that the moment generating function can be write like this:
![\phi(t) = e^{\frac{t^2}{2}](https://tex.z-dn.net/?f=%5Cphi%28t%29%20%3D%20e%5E%7B%5Cfrac%7Bt%5E2%7D%7B2%7D)
And we can write this as an infinite series like this:
![\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...](https://tex.z-dn.net/?f=%5Cphi%28t%29%3D%201%20%2B%28%5Cfrac%7Bt%5E2%7D%7B2%7D%29%2B%5Cfrac%7B1%7D%7B2%7D%20%28%5Cfrac%7Bt%5E2%7D%7B2%7D%29%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bk%21%7D%28%5Cfrac%7Bt%5E2%7D%7B2%7D%29%5Ek%2B%20...)
And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:
![E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%20E%5B1%2B%20tX%20%2B%5Cfrac%7B1%7D%7B2%7D%20%28tX%29%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn%21%7D%28tX%29%5En%20%2B....%5D)
![E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...](https://tex.z-dn.net/?f=E%5Be%5E%7BtX%7D%5D%3D%201%2B%20E%5BX%5Dt%20%2B%5Cfrac%7B1%7D%7B2%7DE%5BX%5E2%5Dt%5E2%20%2B....%2B%5Cfrac%7B1%7D%7Bn1%7DE%5BX%5En%5D%20t%5En%2B...)
and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:
![\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%282k%29%21%7D%20E%5BX%5E%7B2k%7D%5D%20t%5E%7B2k%7D%3D%5Cfrac%7B1%7D%7Bk%21%7D%20%28%5Cfrac%7Bt%5E2%7D%7B2%7D%29%5Ek%20%3D%5Cfrac%7B1%7D%7B2%5Ek%20k%21%7D%20t%5E%7B2k%7D)
And then we have this:
![E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...](https://tex.z-dn.net/?f=E%5BX%5E%7B2k%7D%5D%3D%5Cfrac%7B%282k%29%21%7D%7B2%5Ek%20k%21%7D%2C%20k%3D0%2C1%2C2%2C...)
And then we can find the ![E[X^2]](https://tex.z-dn.net/?f=E%5BX%5E2%5D)
![E[X^2]= \frac{2!}{2^1 1!}= 1](https://tex.z-dn.net/?f=E%5BX%5E2%5D%3D%20%5Cfrac%7B2%21%7D%7B2%5E1%201%21%7D%3D%201)
And we can find the variance like this :
![Var(X^2) = E[X^4]-[E(X^2)]^2](https://tex.z-dn.net/?f=Var%28X%5E2%29%20%3D%20E%5BX%5E4%5D-%5BE%28X%5E2%29%5D%5E2)
And first we find:
![E[X^4]= \frac{4!}{2^2 2!}= 3](https://tex.z-dn.net/?f=E%5BX%5E4%5D%3D%20%5Cfrac%7B4%21%7D%7B2%5E2%202%21%7D%3D%203)
And then the variance is given by:
![Var(X^2)= 3-(1)^2 =2](https://tex.z-dn.net/?f=Var%28X%5E2%29%3D%203-%281%29%5E2%20%3D2)
A + C
(-5, -7) is 7 units to the left of point H
(-7, 7) is more than 7 units up and left of point H
the x axis is at (infinity, 0), being only 7 points up from point H