The random variable X has probability density function f(x)={ax+bx2 , 0<x<1 . If E(X)=0.6 , find (a)P(X<1/2) and (b)var(x)

Answer 1

The solution of this problem is based on finding numerical values of parameters a and b in the PDF f(x).

Recall that the given PDF is zero over the domain R except that f(x) is non-zero over (0,1). Therefore all integration should be performed from -∞ to +∞ but for simplicity, will be carried out over the interval (0,1).

The area under all PDFs is exactly 1, so
∫f(x)dx (-∞ to +∞)=1
Integration of f(x) from 0 to 1 gives the equation 3a+2b=6.....(1)

Given E(X)=0.6 means
∫x*f(x)dx=0.6.
Performing the integration results in the equation 4a+3b=12*0.6=7.2 .... (2)

Solution of (1) and (2) gives
a=18/5, b=-12/5.

From the values of a and b,
(a) P(X<½)=∫f(x)dx from -∞ to ½
(b) Denoting E(X)=μ
var=∫(x-μ)²f(x)dx over domain R.