Discussion Forum

 The c.d. f  F(x) associated  with p.d.f.

$f(x) =3(1-2x^{2})$. If  0 < x <1. is  $k\left( x-\frac{2x^{3}}{k}\right)$  , then value of k is 

Answer:

Explanation:

 we have , 

$f(x)=\begin{cases}3(1-2x^{2}), & 0<x<1\\0, & otherwise\end{cases}$

$F(x=n)=P(X\leq x)=\int_{0}^{x}f(x) dx $

 =$\int_{0}^{x}3(1-2x^{2}) dx$

 =  $\left[3\left(x-\frac{2x^{3}}{3}\right)\right]_{0}^{x}=3\left(x-\frac{2x^{3}}{3}\right)$

$\therefore$    k=3