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Unformatted text preview: 1. f(x) = 3e3x for 0x< and 0 elsewhere verify 2. f(y) = (1/12)[4+2yy 2 ] for 0y3 and 0 elsewhere verify In this situation, we have probabilities for intervals of values, but assume that the probability of any individual point is 0. 1. in number 1 above P[.1X.4] P[X>.5] P[X=3] = 0 2. in number 2 above P[0<Y<1] P[Y>.5] The expected value of a continuous random variable is the average value that you would get if you observed the random variable many (infinitely many) times. E[X] = xf(x)dx, where the integral is over the range of possible values of the variable X. Examples from above continous probability...
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This note was uploaded on 02/01/2012 for the course STAT 490 taught by Professor Boyer during the Spring '11 term at Kansas State University.
 Spring '11
 Boyer
 Probability

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