# Thread: are my solutions is true???

1. ## are my solutions is true???

let X has distribution function given by :

F(x)=1- e^(-4x) , x>0

find :

(1) p(x<=1)

(2) p(x=4)

(3) the mean of X

(4) p (x>=25)

(5) the second moment about the mean

my solutions :

(1) 1-(e^-4)

(2) zero since the distribution is continuous

(3)mu=(1\16)

(4)p (x>=25)= 1 - p (x<=25)

= 1- { 1 - e^(-4*25) }

(5) moment generating function is :

( 1 - (t\4) )^-1 we differentiate it twice and substitute t=0

ops: ops: :P

2. Originally Posted by flower3
let X has distribution function given by :

F(x)=1- e^(-4x) , x>0

find :

(1) p(x<=1)

(2) p(x=4)

(3) the mean of X

(4) p (x>=25)

(5) the second moment about the mean

my solutions :

(1) 1-(e^-4) Mr F says: Correct.

(2) zero since the distribution is continuous Mr F says: Correct.

(3)mu=(1\16) Mr F says: I get 1/4. Note that the pdf is $\displaystyle {\color{red}f(x) = 4 e^{-4x}}$, x > 0, and zero otherwise. Note also the result you get when you differentiate the moment generating function (below) and substitute t = 0 ......

(4)p (x>=25)= 1 - p (x<=25)

= 1- { 1 - e^(-4*25) } Mr F says: Correct. And I'm sure you can simplify it.

(5) moment generating function is :

( 1 - (t\4) )^-1 Mr F says: Correct.

we differentiate it twice and substitute t=0 Mr F says: This gives you the second moment about the origin. However, the mean is not equal to zero and so this will not equal the second moment about the mean.
[snip]
Since the mean is equal to 1/4, the second moment about the mean is $\displaystyle E\left[ \left(X - \frac{1}{4}\right)^2\right]$.

In fact, the second moment about the mean is equal to the variance of X. And you know that $\displaystyle Var(X) = E(X^2) - \mu^2 = E(X^2) - \frac{1}{16}$ where $\displaystyle E(X^2)$ is the second moment about the origin ......

3. thanks for you