Originally Posted by

**geton** Find the value of *a* and the value of *b* so that the following is a valid c.d.f.

$\displaystyle

F(x) = \begin{cases} 0, & \text { $x < 1$}, \\ \frac {1}{b}(2x^3 + ax - 2a), & \text {$1 \le x \le 3$}, \\ 1, & \text {$x > 3$}. \end{cases} $

-------------------

So far I did,

F(3) = 1/b (2x^3 + ax – 2a) = 1

=> 1/b(54+a)=1 ------------------(1)

F(1) = 1/b (2x^3 + ax – 2a) = 0

=> 1/b (2-a) = 0 -----------------(2)

From (1) & (2):

a = 2 & b = 56.

My book’s answers says: a = 2 & b =1/ 56.

Is this method right? Where I did worng?