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$Old$ January 6th, 2009, 04:40 PM

Pringles $Pringles is offline$

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$Unhappy$ Continuous Random Variable with Distribution!

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

Evaluate k

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#2

$Old$ January 6th, 2009, 04:43 PM

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

Evaluate k

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

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

Can you take it from here?

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$Old$ January 6th, 2009, 04:58 PM

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$Default$ I'll try

Quote:

Originally Posted by Chris L T521 $View Post$

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

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

Can you take it from here?

I'll try.Kindly check on me again.thanks

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$Old$ January 6th, 2009, 05:08 PM

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$Smile$ Asswer?

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

[1-1/4]-[0]

=3/4

I hope it is right.WHAT ABOUT THE PRESENTATION

thanks

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$Old$ January 6th, 2009, 05:36 PM

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

[1-1/4]-[0]

=3/4

I hope it is right.WHAT ABOUT THE PRESENTATION

thanks

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

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

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$Old$ January 6th, 2009, 05:47 PM

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$Default$ K=4/3 ?

so K=4/3

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$Old$ January 6th, 2009, 05:48 PM

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Originally Posted by Pringles $View Post$

so K=4/3

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