Question

X has the following PDF:
f(x)={Cxe^(-x/2) , x>0 and 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

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

a. Find the value of C. 

$$1=\displaystyle\int_{-\infin}^\infin f(x)dx=\displaystyle\int_{0}^\infin Cxe^{-x/2}dx=$$ $$=\lim\limits_{A\rarr\infin}\bigg(\displaystyle\int_{0}^\infin Cxe^{-x/2}dx\bigg)$$

$$\int{xe^{-x/2}}dx=-2xe^{-x/2}+2\int{e^{-x/2}}dx=$$ $$=-2xe^{-x/2}-4e^{-x/2}+c_1$$

$$1=C\lim\limits_{A\rarr\infin}\big[-2xe^{-x/2}-4e^{-x/2}\big]\begin{matrix} A\\ 0 \end{matrix}=$$

$$=C(-0-0+0+4)=4C=>C={1\over 4}$$

$$f(x)= \begin{cases} {1\over 4}xe^{-x/2} & x>0 \\ 0 & x\leq 0 \end{cases}$$

b. Find the mean of X. 

$$mean=E(X)=\displaystyle\int_{-\infin}^\infin xf(x)dx=$$ $$=\displaystyle\int_{0}^\infin {1\over 4}x^2 e^{-x/2}dx$$

$$\int{x^2e^{-x/2}}dx=-2x^2 e^{-x/2}+4\int{xe^{-x/2}}dx=$$ $$=-2x^2 e^{-x/2}-8xe^{-x/2}-16e^{-x/2}+c_2$$

$$mean=E(X)=$$ $$={1\over 4}\lim\limits_{A\rarr\infin}\big[-2x^2 e^{-x/2}-8xe^{-x/2}-16e^{-x/2}\big]\begin{matrix} A\\ 0 \end{matrix}=$$ $$={1\over 4}(-0-0-0+0+0+16)=4$$

c. Find the 2nd Quartile of X. 

$$F(x)=\displaystyle\int_{-\infin}^x f(t)dt$$

$$\displaystyle\int_{0}^x {1\over 4}te^{-t/2}dt={1\over 4}\big[-2te^{-t/2}-4e^{-t/2}\big]\begin{matrix} x\\ 0 \end{matrix}=$$

$$={1\over 4}(-2xe^{-x/2}-4e^{-x/2}+0+4)$$

$F(0)=0$

Then

$$F(x)= \begin{cases} -{1\over 2}xe^{-x/2}-e^{-x/2}+1 & x>0 \\ 0 & x\leq 0 \end{cases}$$

$$F(x_m)={1\over 2}=> -{1\over 2}x_me^{-x_m/2}-e^{-x_m/2}+1={1\over 2}=>$$

$$=> (x_m+2)e^{-x_m/2}=1$$

$median=2^{nd}Quartile=x_m\approx3.357$

d. Also, find the standard deviation of X.

$$Var(X)=\sigma^2=E(X^2)-[E(X)]^2$$

$$E(X^2)=\displaystyle\int_{-\infin}^\infin x^2f(x)dx=$$

$$=\displaystyle\int_{0}^\infin {1\over 4}x^3 e^{-x/2}dx$$

$$\int{x^3e^{-x/2}}dx=-2x^3 e^{-x/2}+6\int{x^2e^{-x/2}}dx=$$

$$=-2x^3 e^{-x/2}+6(-2x^2 e^{-x/2}-8xe^{-x/2}-16e^{-x/2})=$$

$$=-2x^3 e^{-x/2}-12x^2 e^{-x/2}-48xe^{-x/2}-96e^{-x/2}$$

$$E(X^2)={1\over 4}\lim\limits_{A\rarr\infin}\big[-2x^3 e^{-x/2}-12x^2 e^{-x/2}-48xe^{-x/2}-96e^{-x/2}\big]\begin{matrix} A\\ 0 \end{matrix}=$$

$$={1\over 4}(-0-0-0+0+0+96)=24$$

$$Var(X)=\sigma^2=E(X^2)-[E(X)]^2=24-(4)^2=8$$

$$\sigma=\sqrt{8}=2\sqrt{2}\approx2.8284$$