Question

Suppose, X has the following PDF:
f(x)={Cxe^x/2 // 0 x>0 x<=0
a. Find the value of C.
b. Find the mean of ​X​.
c. Find the 2nd Quartile of ​X​.
d. Also, find the standard deviation of ​X

Answer 1

$f(x)=\begin{cases} Cxe^{-x/2}, x>0 \\ 0, x\leq 0 \end{cases}$

(here we have $e^{-x/2}$ because $Cx e^{x/2}$ is unbounded, but $\int \limits_0^{\infin}f(x)=1$ to be PDF)

a.

$\int \limits _0^{\infin}f(x)dx= \int \limits _0^{\infin}Cxe^{-x/2}dx=\\=-C(2x+4)e^{-x/2} \big|_0^{\infin}=4C \ \Rightarrow C=1/4$

Answer: $C=1/4$

b.

$\mu =\int \limits_0^{\infin}xf(x)dx =\int \limits_0^{\infin}1/4 x^2e^{-x/2}dx=\\=-1/2(x^2+4x+8)e^{-x/2}\big|_0^{\infin}=4$

Answer: $\mu=4$

c.

$0.5=F(x)=\int \limits _0^xf(t)dt=\\=\int\limits _0^x 1/4te^{-t/2}dt=-1/2(x+2)e^{-x/2}+1,$

$0.5=-1/2(x+2)e^{-x/2}+1, \ \ (x+2)=e^{x/2}, \ \ \ x\approx 3.356$

Answer: the second quartile is $3.356$.

d.

$\ \sigma ^2=\int\limits _0^{\infin} (x-\mu)^2f(x)dx= \int\limits _0^{\infin} 1/4(x-4)^2x e^{-x/2}dx=\\= -1/2(x^3-2x^2+8x+16)e^{-x/2}\big|_0^{\infin}=8,$

$\sigma=\sqrt{\sigma^2}=\sqrt{8}=2\sqrt{2}.$

Answer: $\sigma =2\sqrt{2}$