# Thread: The cumulative distribution function

1. ## The cumulative distribution function

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}$

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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?

2. 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?
You are correct, the book is wrong.

3. hmmmmmmmmm I think so