Thread: Continuous Random Variable with Distribution!

1. Continuous Random Variable with Distribution!

P(x)={k(1-x^3) 0< or = x < or = 1
0 otherwise

Evaluate k

2. Originally Posted by Pringles
P(x)={k(1-x^3) 0< or = x < or = 1
0 otherwise

Evaluate k
For this to be a pdf, $\displaystyle \int_{-\infty}^{\infty}P\left(x\right)\,dx=1$

Thus, solve the equation $\displaystyle \int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1$ for k.

Can you take it from here?

3. I'll try

Originally Posted by Chris L T521
For this to be a pdf, $\displaystyle \int_{-\infty}^{\infty}P\left(x\right)\,dx=1$

Thus, solve the equation $\displaystyle \int_{-\infty}^{\infty} k\left(1-x^3\right)\,dx=1\implies \int_0^1 k\left(1-x^3\right)\,dx=1$ for k.

Can you take it from here?
I'll try.Kindly check on me again.thanks

4. Asswer?

[x-1/4X^4]0 1

[1-1/4]-[0]

=3/4

I hope it is right.WHAT ABOUT THE PRESENTATION

thanks

5. Originally Posted by Pringles
[x-1/4X^4]0 1

[1-1/4]-[0]

=3/4

I hope it is right.WHAT ABOUT THE PRESENTATION

thanks
Recall that $\displaystyle k{\color{red}\int_0^1 \left(1-x^3\right)\,dx}=1$

You got the correct value for the part in red. Now what's k?

6. K=4/3 ?

so K=4/3

7. Originally Posted by Pringles
so K=4/3