Question

here the pmf:

f(x)={ cxe^x/2 ;x>0
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

$a) \int_{-\infty}^{\infty}f(x)dx=1\\ \int_{0}^{\infty}Cxe^{-\frac{x}{2}}dx=1\\ \int_{0}^{\infty}xe^{-\frac{x}{2}}dx=( u=x; dv=e^{-\frac{x}{2}}; du=dx; v=-2e^{-\frac{x}{2}})=\\ -2xe^{-\frac{x}{2}}\bigg|^{\infty}_{0}+2\int_{0}^{\infty}e^{-\frac{x}{2}}dx=2(-2)e^{-\frac{x}{2}}\bigg|^{\infty}_{0}=4.\\ C=\frac{1}{4}\\ b) MX= \int_{-\infty}^{\infty}xf(x)dx= \int_{0}^{\infty}\frac{1}{4}x^2e^{-\frac{x}{2}}dx.\\ \int_{0}^{\infty}x^2e^{-\frac{x}{2}}dx=(u=x^2; dv=e^{-\frac{x}{2}}; du=2xdx; v=-2e^{-\frac{x}{2}})=-2x^2e^{-\frac{x}{2}}\bigg|^{\infty}_{0}+4\int_{0}^{\infty}xe^{-\frac{x}{2}}dx=0+4\cdot 4=16.\\ MX=\frac{1}{4}16=4.\\ d) DX=MX^2-(MX)^2\\ MX^2=\int_{0}^{\infty}\frac{1}{4}x^3e^{-\frac{x}{2}}dx=\frac{1}{4}\cdot 6\cdot 16=24.\\ DX=24-4^2=8\\ \sigma X=2\sqrt{2}.\\ с) F(x)=\int_{-\infty}^{x}p(t)dt\\ F(x)=\int_{0}^{x}\frac{1}{4}te^{-\frac{t}{2}}dt\\ P\{X\leq Q_2\}=0.5=F(Q_2)\\ \int_{0}^{Q_2}te^{-\frac{t}{2}}dt=-2Q_2e^{-\frac{Q_2}{2}}-4e^{-\frac{Q_2}{2}}+4 \text{ (by parts)}.\\ F(Q_2)=-\frac{1}{2}Q_2e^{-\frac{Q_2}{2}}-e^{-\frac{Q_2}{2}}+1.\\ -\frac{1}{2}Q_2e^{-\frac{Q_2}{2}}-e^{-\frac{Q_2}{2}}+1=\frac{1}{2}.\\ -\frac{1}{2}Q_2e^{-\frac{Q_2}{2}}-e^{-\frac{Q_2}{2}}=-\frac{1}{2}.\\ e^{-\frac{Q_2}{2}}(\frac{1}{2}Q_2+1)=\frac{1}{2}.\\ 2e^{-\frac{Q_2}{2}}(\frac{1}{2}Q_2+1)=1.\\ \ln 2e^{-\frac{Q_2}{2}}+\ln (\frac{1}{2}Q_2+1)=0.\\ \ln 2-\frac{Q_2}{2}+\ln (\frac{1}{2}Q_2+1)=0.\\ \ln (Q_2+2)=\frac{Q_2}{2}.\\ \ln (Q_2+2)^2=Q_2.\\ \ln (Q_2+2)^2=\ln e^{Q_2}.\\ \ln\frac{(Q_2+2)^2}{e^{Q_2}}=0.\\ \frac{(Q_2+2)^2}{e^{Q_2}}=1.\\ (Q_2+2)^2=e^{Q_2}.$We have 3 solutions of this equation (if we look at the graphs of (x+2)^2 and e^x). Two of them are negative and one is positive. But in our case $Q_2>0.$

Using numerical methods we find that $Q_2\approx 3.3567.$