Continuous random variables
Concept 5 of 7 in Probability Distributions
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XXX has p.d.f. f(x)=3x2f(x) = 3x^2f(x)=3x2 for 0<x<10 < x < 10<x<1. Then f(0.9)=2.43f(0.9) = 2.43f(0.9)=2.43. How can that be, when no probability is more than 111?
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