Answer to Question #131914 in Statistics and Probability for Hadia

Question

A continuous random variable x has density function
f(x)= 3/4x(2-x) , 0<x<2
Find p(x>1/4)

Answer 1

$f(x)=\frac{3x(2-x)}{4}$ , $0<x<2$

Check if it is a pdf

$\int_0^2\frac{3x(2-x)}{4}dx$

$=[\frac{3x^2}{4}-\frac{x^3}{4}]_0^2$

$=(3-2)-0=1$ (it is a pdf)

$P(x>\frac{1}{4})=\int_{\frac{1}{4}}^2\frac{3x(2-x)}{4}dx$

$=[\frac{3x^2}{4}-\frac{x^3}{4}]_{\frac{1}{4}}^2$

$=[3-2]-[\frac{1}{64}(3-\frac{1}{4})]$

$=1-\frac{11}{256}=\frac{245}{256}=0.957$