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anonymous

  • 4 years ago

The probability density function\[f(x)=\left(\begin{matrix}c(4-x^2) \\ 0\end{matrix}\right)\] C = constant = 3/32\[-2 \le x \le 2\]otherwise. Find \[P(|x| \ge 1).\]

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  1. ash2326
    • 4 years ago
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    \[P(|x|\ge1) = P(-\infty\le x\le -1)+P(1\le x \le\infty)\] \[P(|x|\ge1) = \int_{\infty}^{-1}f(x)dx+\int_{1}^{\infty}f(x) dx\] since f(x) in non zero for \(-2\le x \le 2\) we'll have \[P(|x|\ge1) = \int_{-2}^{-1}f(x)dx+\int_{1}^{2}f(x) dx\] here \[f(x)= \frac{3}{32} (4-x^2)\] can you do it now?

  2. anonymous
    • 4 years ago
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    I understand it, but I don't know how to remove |x| to make it x. Is it just adding the - integrals to the positive integrals? i.e. \[\int\limits_{-1}^{2} and \int\limits_{2}^{1}?\]

  3. anonymous
    • 4 years ago
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    *-2

  4. ash2326
    • 4 years ago
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    You need not remove |x|, that we have already included. Substitute the value of f(x) and evaluate the two integrals

  5. anonymous
    • 4 years ago
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    Ok, yes, I understand how to do the integration from then on. But I'm not sure how you got the -2 < x < -1 and 1 < x < 2 in the first place?

  6. ash2326
    • 4 years ago
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    see we want probability in the range \(|x|\ge 1\) this is equivalent to \(-\infty \le x\le 1\) and \( 1 \le x\le \infty\) therefore I split the integral in two ranges. f(x) is non zero in the range \(-2 \le x\le 2\), therefore it boils down to two integrals from -2 to -1 and from 1 to 2

  7. anonymous
    • 4 years ago
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    Ah Ok. Thank you!

  8. anonymous
    • 4 years ago
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    Do you think you could help me with one more example?

  9. anonymous
    • 4 years ago
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    I mean, I have another question

  10. ash2326
    • 4 years ago
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    I'll try, post it as a new question

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