Let X be a random variable with density function fx(s) such that .pdf

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Let X be a random variable with density function fx(s) such that: fx(s) = ke-s for s > 2 and and fx(s) = 0 for s 2. Here k > 0 designates a constant. a) Determine constant k. b) Calculate P(0 X 5). c) Calculate E[X]. Solution integration (ke^-s ds) = -ke^-s applying limits we get ke^-2 the integral should be 1 therefore k = e^2 P(0 = X = 5) integral integration (ke^-s ds) from 2 to 5 -ke^-s linits 2 to 5 we get k(e^-2 - e^-5) = e-e^-3 E(x) = integration (ske^-s ds) from 2 to infinity is 3 E(x) = 3.

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