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

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ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2\[P(x\ge1) = P(\infty\le x\le 1)+P(1\le x \le\infty)\] \[P(x\ge1) = \int_{\infty}^{1}f(x)dx+\int_{1}^{\infty}f(x) dx\] 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 \[f(x)= \frac{3}{32} (4x^2)\] can you do it now?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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. \[\int\limits_{1}^{2} and \int\limits_{2}^{1}?\]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2You need not remove x, that we have already included. Substitute the value of f(x) and evaluate the two integrals

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, 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?

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2see 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you think you could help me with one more example?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I mean, I have another question

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.2I'll try, post it as a new question
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