Answer to Question #187978 in Statistics and Probability for Gaurav Goyal

Question

The probability density function of a random variable X is f(x) = C|x|. Find C, and the value of x° such that Fx(x°) = 3/4, where F is the CDF.

Answer 1

The pdf is-

$f(x)=Cx, 0\le x\le 2$

$0 , \text{ othervise }$

As we know that for pdfs-

$\int_{-\infty}^{\infty}f(x)dx=1$

$\int_0^2Cxdx=1$

$\dfrac{Cx^2}{2}|_0^2=1$

$\dfrac{4C}{2}=1\Rightarrow C=\dfrac{1}{2}$

Also CDF is calculating by integrating the PDf-

$F_X(x^o)=\int_{-\infty}^{x^o}\dfrac{x}{2}dx$

$F_X(x^o)=\int_{0}^{x^o}\dfrac{x}{2}dx$

$\dfrac{3}{4}=\dfrac{x^2}{4}|_0^{x^o}\Rightarrow \dfrac{(x^o)^2}{4}=\dfrac{3}{4}\Rightarrow x^o=\sqrt{3}$