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Let $f(x) = \frac{c}{(1 + x^2)}$.(a) For what value of $c$ is $f$ a probability density function?(b) For that value of $c$, find $P (-1 < X < 1)$.

A. $2 c\left(\frac{\pi}{2}\right)=c \pi$B. $\frac{2}{\pi}\left(\frac{\pi}{4}-0\right)=\frac{1}{2}$

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Anna Marie V.

Campbell University

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Boston College

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So we're giving the function. If affection, they will do. See your one plus six square domain Trump. Infinite too minus infinity. We need to find forward value off. See? Is this a problem? DNC function that is forward value of C is the integral minus Infinity, Infinity fixed. E X is equal to one. So this calculate this integral. So we got to see Daniels effects minus infinity to infinity. Legal one. This goose vibrato minus minus by very to equals one, the C p. I equals one. This gives c is equal to do one or by. So the problem into function is on over by one plus X squared. Now we need to calculate the probability ex being between one and minus one. This to be the integral minus 1 to 1. Perfect dx Z equals one over. Pi minus 1 to 1. This is one work by Dan was fixed. My answer Under one manure by five. Therefore minus minus pi before. So this 1/2. Okay,

Anna Marie V.

Campbell University

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Boston College

Lectures

Join Bootcamp