Question: 1. Let X be a random variable with the density function f(x): = if x > 0 0 elsewhere Find Var(X12). = 3 and
1. Let X be a random variable with the density function f(x): = if x > 0 0 elsewhere Find Var(X12). = 3 and 0 = 9. Thus, Solution: X is a Weibull random variable with parameters Y Therefore, X3 is an exponential random variable Y with parameter 0 = 9. Var(X12) = VarY4 = EY8 (EY4) = 08.8! (04.4!) = 39744 98. - 9.
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